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174 CHAPTER 7. CHANGE OF VARIABLES The obstacle is what to do about the dA = dx dy part of the integral. For this, we need to see how small pieces of area in the xy-plane are related to those in the rθ-plane. We return to this example later. 7.1 Change of variables for double integrals Let RR D f(x, y) dx dy be a given double integral. We want to describe how to convert the integral to an equivalent one with a different domain of integration and a correspondingly different integrand. The setup is to assume that there is a smooth function T : D∗→D, where D∗is a subset of R2. We think of T as transforming D∗onto D, so we assume that T(D∗) = D and that T is one-to-one, except possibly along the boundary of D∗. This last part means that whenever a and b are distinct points of D∗, not on the boundary, then the values T(a) and T(b) are distinct, too: a ̸= b ⇒T(a) ̸= T(b). (For the purposes of integration, anomalous behavior confined to the boundary can be allowed typically without messing things up. See the remarks on page 130.) The new integral will be an integral over D∗. It helps keep things straight if the coordinates in the domain and codomain have different names, so we think of T as a transformation from the uv-plane to the xy-plane and write T(u, v) = (x(u, v), y(u, v)). Thus we want to convert the given integral over D with respect to x and y to an integral over D∗with respect to u and v. We outline the main idea. It is similar to what we did in Section 5.5 to define surface integrals with respect to surface area using a parametrization, though of course there is a big difference between motivating a definition and justifying a theorem. Another difference is that, in the interim, we have learned about the derivative of a vector-valued function. Figure 7.2: Change of variables: x and y as functions of u and v via transformation T To begin, subdivide D∗into small subrectangles of dimensions △ui by △vj, as in Figure 7.2. The transformation T sends a typical subrectangle of area △ui △vj in the uv-plane to a small curvy quadrilateral in the xy-plane. Let △Aij denote the area of the curvy quadrilateral. If we choose a sample point pij in each curvy quadrilateral, we can form a Riemann sum-like approximation of the integral over D: ZZ D f(x, y) dx dy ≈ X i,j f(pij) △Aij. (7.1) To turn this into a Riemann sum over D∗, we need to relate the areas of the subrectangles and the curvy quadrilaterals. To do so, we use the first-order approximation of T. Choose a point aij in the (i, j)th subrect- angle of D∗such that T(aij) = pij. Then the first-order approximation says that, for points u in | Multivariable_Calculus_Shimamoto_Page_186_Chunk3401 |
7.1. CHANGE OF VARIABLES FOR DOUBLE INTEGRALS 175 D∗near aij: T(u) ≈T(aij) + DT(aij) · (u −aij) ≈pij + DT(aij) · (u −aij) ≈ | Multivariable_Calculus_Shimamoto_Page_187_Chunk3402 |
176 CHAPTER 7. CHANGE OF VARIABLES Returning to the first-order approximation (7.2) of T, we conclude that multiplication by DT(aij) transforms the subrectangle of area △ui △vj that contains aij to a parallelogram of area | det DT(aij)| △ui △vj. Hence, after additional translation, the first-order approximation (7.2) sends the subrectangle containing aij to a parallelogram that: • approximates the curvy quadrilateral of area △Aij that contains pij and • has area | det DT(aij)| △ui △vj. Thus the Riemann sum approximation (7.1) becomes: ZZ D f(x, y) dx dy ≈ X i,j f(pij) △Aij ≈ X i,j f(T(aij)) | det DT(aij)| △ui △vj. Letting △ui and △vj go to zero, this becomes an integral over D∗, giving the following major result. Theorem 7.3 (Change of variables theorem for double integrals). Let D and D∗be bounded subsets of R2, and let T : D∗→D be a smooth function such that T(D∗) = D and that is one-to-one, except possibly on the boundary of D∗. If f is integrable on D, then: ZZ D f(x, y) dx dy = ZZ D∗f(T(u, v)) | det DT(u, v)| du dv. In other words, x and y are replaced in terms of u and v in the function f using (x, y) = T(u, v), and dx dy is replaced by | det DT(u, v)| du dv = det ∂x ∂u ∂x ∂v ∂y ∂u ∂y ∂v du dv. The determinant of the derivative, det DT(u, v), is sometimes referred to as the Jacobian determinant. By comparison, in first-year calculus, if f(x) is a real-valued function of one variable and x is expressed in terms of another variable u, say as x = T(u), then the method of substitution says that: Z T(b) T(a) f(x) dx = Z b a f(T(u)) T ′(u) du. In this case, it seems simplest to understand the result in terms of antiderivatives using the chain rule and the fundamental theorem of calculus rather than through Riemann sums, though we won’t present the details of the argument. In particular, substituting for x in terms of u includes the substitution dx = T ′(u) du = dx du du. The substitution dx dy = | det DT(u, v)| du dv in the previous paragraph is the analogue in the double integral case. We say more about the principle of substitution in the next section. Returning to double integrals, in the case of polar coordinates, the substitutions for x and y are given by (x, y) = T(r, θ) = (r cos θ, r sin θ). Then DT(r, θ) = cos θ −r sin θ sin θ r cos θ , and: | det DT(r, θ)| = |r cos2 θ + r sin2 θ| = |r| = r. By the change of variables theorem, this means that “dx dy = r dr dθ” and that polar coordinates transform integrals as follows. Corollary 7.4. In polar coordinates: ZZ D f(x, y) dx dy = ZZ D∗f(r cos θ, r sin θ) r dr dθ, where D∗is the region in the rθ-plane that describes D in polar coordinates. | Multivariable_Calculus_Shimamoto_Page_188_Chunk3403 |
7.2. A WORD ABOUT SUBSTITUTION 177 We now complete Example 7.1, which asked to evaluate RR D p x2 + y2 dx dy, where D is the disk x2 + y2 ≤4 in the xy-plane. As noted earlier, D is described in polar coordinates by the rectangle D∗given by 0 ≤r ≤2, 0 ≤θ ≤2π in the rθ-plane. The polar coordinate transformation T is shown in Figure 7.4. Figure 7.4: A change of variables in polar coordinates We verify that T satisfies the hypotheses of the theorem, though we shall grow less meticulous about writing out the details on this point after a couple more examples. All the points on the left side of D∗, where r = 0, are mapped to the origin, that is, T(0, θ) = (0, 0) for 0 ≤θ ≤2π. Also, along the top and bottom of D∗, θ = 0 and θ = 2π correspond to the same thing in the xy-plane: T(r, 0) = T(r, 2π) for 0 ≤r ≤2. Apart from this, distinct points in D∗are mapped to distinct values in D. In other words, T is one-to-one away from the boundary of D∗, so the change of variables theorem applies. Since p x2 + y2 = r, this gives: ZZ D p x2 + y2 dx dy = ZZ D∗r · r dr dθ = Z 2π 0 Z 2 0 r2 dr dθ = Z 2π 0 1 3r3 r=2 r=0 dθ = Z 2π 0 8 3 dθ = 8 3θ 2π 0 = 16 3 π. 7.2 A word about substitution The change of variables theorem concerns how to convert an integral with respect to x and y into an integral with respect to two other variables, say u and v. Another name for this is substitution: we require expressions to substitute for x, y, and dA = dx dy in terms of u and v. Substituting for x and y means writing them as functions of u and v: x = x(u, v) and y = y(u, v). This is equivalent to having a function T(u, v) = (x(u, v), y(u, v)) from the uv-plane to the xy-plane. The logistics can be confusing and perhaps counterintuitive: the transformation of the integral goes from being with respect to x and y to being with respect to u and v, but, in order to accomplish this, the | Multivariable_Calculus_Shimamoto_Page_189_Chunk3404 |
178 CHAPTER 7. CHANGE OF VARIABLES geometric transformation goes the other way, from (u, v) to (x, y). This is an inherent aspect of substitution. Sometimes, the original integrand is denoted by an expression like η = f(x, y) dx dy, and then the transformed integrand f(T(u, v)) | det DT(u, v)| du dv is denoted by T ∗(η). Another appropriate notation for it might be η∗. With this notation, the change of variables theorem becomes: ZZ T(D∗) η = ZZ D∗T ∗(η). (7.3) Actually, what we just said is not entirely accurate and the situation is a little more complicated,7 but it serves to illustrate the general principle: in a substitution, the geometry is “pushed forward,” and the integral is “pulled back.” 7.3 Examples: linear changes of variables, symmetry We illustrate the change of variables theorem with three further examples. Example 7.5. Let D be the region in R2 given by x2 9 + y2 4 ≤1, y ≥0. Evaluate RR D(x2 +y2) dx dy. Here, the integrand f(x, y) = x2 + y2 is fairly simple, but the region of integration poses some problems. D is the region inside the upper half of an ellipse. The endpoints of the iterated integral with respect to x and y are somewhat hard to work with, and polar coordinates don’t help either— the relationship between r and θ needed to describe D is a little complicated. On the other hand, we can think of D as a transformed semicircular disk, which is easy to describe in polar coordinates. Thus we attempt to map the half-disk D∗of radius 1 given in the uv-plane by u2 + v2 ≤1, v ≥0, over to D and hope that this does not complicate the function to be integrated too much. Figure 7.5: Changing an integral over a half-ellipse to an integral over a half-disk The x and y-intercepts of the ellipse are ±3 and ±2, respectively, so to achieve the transforma- tion, as illustrated in Figure 7.5, we stretch horizontally by a factor of 3 and vertically by a factor of 2. In other words, define: T(u, v) = (3u, 2v). In effect, we are making the substitutions x = 3u and y = 2v. Then T transforms D∗to D, that is, if (u, v) ∈D∗, then T(u, v) ∈D. This is because if (u, v) satisfies u2 + v2 ≤1 and v ≥0, then (x, y) = T(u, v) satisfies x2 9 + y2 4 = (3u)2 9 + (2v)2 4 = u2 + v2 ≤1 and y = 2v ≥0. Moreover, T is one-to-one on the whole uv-plane. This seems reasonable geometrically, or, in terms of equations, if (u1, v1) ̸= (u2, v2), then (3u1, 2v1) ̸= (3u2, 2v2). Perhaps it is more readable 7The correct expressions are η = f(x, y) dx ∧dy and T ∗(η) = f(T(u, v)) det DT(u, v) du ∧dv. We shall learn more about integrands like this and, more importantly, how to integrate them in the last two chapters. In particular, for the correct version of equation (7.3), see Exercise 4.20 in Chapter 11. | Multivariable_Calculus_Shimamoto_Page_190_Chunk3405 |
7.3. EXAMPLES: LINEAR CHANGES OF VARIABLES, SYMMETRY 179 to phrase this in the contrapositive: if (3u1, 2v1) = (3u2, 2v2), then (u1, v1) = (u2, v2). This is clear. For instance, 3u1 = 3u2 ⇒u1 = u2. Finally, DT(u, v) = [ 3 0 0 2 ], so | det DT(u, v)| = 6, and f(T(u, v)) = (3u)2 + (2v)2 = 9u2 + 4v2. Thus, by the change of variables theorem: ZZ D (x2 + y2) dx dy = ZZ D∗(9u2 + 4v2) · 6 du dv. We convert this to polar coordinates in the uv-plane using u = r cos θ, v = r sin θ, and du dv = r dr dθ. The half-disk D∗is described in polar coordinates by 0 ≤r ≤1, 0 ≤θ ≤π, in other words, by the rectangle [0, 1] × [0, π] in the rθ-plane. Thus: ZZ D (x2 + y2) dx dy = ZZ D∗in polar (9r2 cos2 θ + 4r2 sin2 θ) · 6 · r dr dθ = 6 Z π 0 Z 1 0 r3(9 cos2 θ + 4 sin2 θ) dr dθ. (7.4) At this point, we pause to put on the record a couple of facts from the exercises that we shall use freely from now on (see Exercise 2.3 in Chapter 5 and Exercise 1.1 in this chapter). • For a function F whose variables separate into two independent continuous factors, F(x, y) = f(x)g(y), the integral over a rectangle [a, b] × [c, d] is given by: Z d c Z b a f(x)g(y) dx dy = Z b a f(x) dx Z d c g(y) dy . Note that integrals over rectangles are characterized by the property that all the limits of integration are constant. • R π 0 cos2 θ dθ = R π 0 sin2 θ dθ = π 2 . (An easy way to remember this is to note that R π 0 cos2 θ dθ = R π 0 sin2 θ dθ from the graphs of the sine and cosine functions and that R π 0 (cos2 θ + sin2 θ) dθ = R π 0 1 dθ = θ π 0 = π.) Applying these facts to the present integral (7.4) gives: ZZ D (x2 + y2) dx dy = 6 Z 1 0 r3 dr Z π 0 (9 cos2 θ + 4 sin2 θ) dθ = 6 · | Multivariable_Calculus_Shimamoto_Page_191_Chunk3406 |
180 CHAPTER 7. CHANGE OF VARIABLES Figure 7.6: A linear change of variables Then T maps the unit square D∗= [0, 1] × [0, 1] in the uv-plane onto D. See Figure 7.6. We leave for the exercises the verification that T is one-to-one on R2 (Exercise 3.3), so we can use the change of variables theorem to pull back the original integral to an integral over D∗. The substitutions given by T are x = 3u −v and y = 2u + v. Also, DT(u, v) = 3 −1 2 1 , and | det DT(u, v)| = |3 + 2| = 5. Hence: ZZ D xy dx dy = ZZ D∗(3u −v)(2u + v) · 5 du dv = 5 Z 1 0 Z 1 0 (6u2 + uv −v2) dv du = 5 Z 1 0 (6u2v + 1 2uv2 −1 3v3) v=1 v=0 du = 5 Z 1 0 (6u2 + 1 2u −1 3) du = 5 | Multivariable_Calculus_Shimamoto_Page_192_Chunk3407 |
7.3. EXAMPLES: LINEAR CHANGES OF VARIABLES, SYMMETRY 181 Figure 7.7: Symmetry in the y-axis We apply the change of variables theorem with f(x, y) = x so that f(T(x, y)) = f(−x, y) = −x. Then: ZZ D x dx dy = ZZ D∗=D f(T(x, y)) | det DT(x, y)| dx dy = ZZ D −x · 1 dx dy = − ZZ D x dx dy. As a result, 2 RR D x dx dy = 0, so RR D x dx dy = 0. (b) Here, the intuition is that the contributions to the Riemann sums P y △x △y for RR D y dx dy are the same for (x, y) and (−x, y), hence the sums converge to the same value on L and on R. More formally, again let T(x, y) = (−x, y). Then T(L) = R, i.e., R∗= L, and | det DT(x, y)| = 1 as before. Now, f(x, y) = y, so f(T(x, y)) = f(−x, y) = y. Consequently: ZZ R y dx dy = ZZ R∗=L f(T(x, y)) | det DT(x, y)| dx dy = ZZ L y · 1 dx dy = ZZ L y dx dy. This is exactly what we wanted to show. This example can be generalized. Namely, let D be symmetric in the y-axis, and let f : D →R be an integrable real-valued function. • If f(−x, y) = −f(x, y) for all (x, y) in D, then: ZZ D f(x, y) dx dy = 0. • If f(−x, y) = f(x, y) for all (x, y) in D, then: ZZ L f(x, y) dx dy = ZZ R f(x, y) dx dy. The justifications use essentially the same reasoning as in the example. | Multivariable_Calculus_Shimamoto_Page_193_Chunk3408 |
182 CHAPTER 7. CHANGE OF VARIABLES 7.4 Change of variables for n-fold integrals The change of variables theorem for double integrals has a natural analogue for functions of n variables. Let T be a smooth transformation that sends a subset W ∗of Rn onto a subset W of Rn and that is one-to-one, except possibly on the boundary of W ∗. See Figure 7.8. We think of T as a function from (u1, u2, . . . , un)-space to (x1, x2, . . . , xn)-space, so: T(u1, u2, . . . , un) = | Multivariable_Calculus_Shimamoto_Page_194_Chunk3409 |
7.4. CHANGE OF VARIABLES FOR n-FOLD INTEGRALS 183 • Spherical coordinates. This time, T(ρ, ϕ, θ) = (x, y, z) with conversions: x = ρ sin ϕ cos θ y = ρ sin ϕ sin θ z = ρ cos ϕ. See Section 5.4.3. Then DT(ρ, ϕ, θ) = sin ϕ cos θ ρ cos ϕ cos θ −ρ sin ϕ sin θ sin ϕ sin θ ρ cos ϕ sin θ ρ sin ϕ cos θ cos ϕ −ρ sin ϕ 0 . We leave as an exercise the calculation that | det DT(ρ, ϕ, θ)| = ρ2 sin ϕ (Exercise 2.8 in Chapter 6), i.e.: dx dy dz = ρ2 sin ϕ dρ dϕ dθ. Example 7.8. Find the volume of a closed ball of radius a in R3, that is, the volume of: W = {(x, y, z) ∈R3 : x2 + y2 + z2 ≤a2}. See Figure 7.9. Figure 7.9: A ball of radius a in R3 We begin with Vol (W) = RRR W 1 dx dy dz. (See page 144.) As we are about to discover, W is easy to describe in spherical coordinates, so we convert to an integral in spherical coordinates. The endpoints depend on the order of integration, and here we choose dρ dϕ dθ. Thus the first integration is with respect to ρ. For fixed ϕ and θ, ρ varies along a radial line segment from ρ = 0 to ρ = a. Then, for fixed θ, ϕ rotates all the way down from the positive z-axis, ϕ = 0, to the negative z-axis, ϕ = π. Finally, θ varies from θ = 0 to θ = 2π. In other words, W is described in ρϕθ-space by the three-dimensional box W ∗= [0, a] × [0, π] × [0, 2π]. Hence: Vol (W) = ZZZ W 1 dx dy dz = ZZZ W ∗1 · ρ2 sin ϕ dρ dϕ dθ = Z 2π 0 Z π 0 Z a 0 ρ2 sin ϕ dρ dϕ dθ. The variables of the integrand separate into independent factors and all endpoints are constant, so, | Multivariable_Calculus_Shimamoto_Page_195_Chunk3410 |
184 CHAPTER 7. CHANGE OF VARIABLES as in the two-dimensional case, we can separate the integrals as well: Vol (W) = Z a 0 ρ2 dρ Z π 0 sin ϕ dϕ Z 2π 0 1 dθ = 1 3ρ3 a 0 −cos ϕ π 0 θ 2π 0 = 1 3a3 · | Multivariable_Calculus_Shimamoto_Page_196_Chunk3411 |
7.4. CHANGE OF VARIABLES FOR n-FOLD INTEGRALS 185 Figure 7.10: The region bounded by a circular cylinder of radius 2, a sphere of radius 3, and the xy-plane the base of W, which is a disk of radius 2. Thus 0 ≤r ≤2 and 0 ≤θ ≤2π. Therefore: ZZZ W (x2 + y2)z dx dy dz = Z 2π 0 Z 2 0 Z √ 9−r2 0 r3z dz dr dθ = Z 2π 0 Z 2 0 1 2r3z2 z= √ 9−r2 z=0 dr dθ = Z 2π 0 Z 2 0 1 2r3(9 −r2) dr dθ = 1 2 Z 2 0 r3(9 −r2) dr Z 2π 0 1 dθ = 1 2 9 4r4 −1 6r6 2 0 θ 2π 0 = 1 2 · | Multivariable_Calculus_Shimamoto_Page_197_Chunk3412 |
186 CHAPTER 7. CHANGE OF VARIABLES Figure 7.11: An ice cream cone-shaped solid, capped by a sphere of radius 2 √ 2 order of integration dρ dϕ dθ, for fixed ϕ and θ, ρ goes along a radial segment from ρ = 0 to ρ = 2 √ 2, the radius of the spherical cap. Then, for fixed θ, the angle ϕ varies from ϕ = 0 to ϕ = π 4 . Lastly, θ goes all the way around from θ = 0 to θ = 2π. Hence: Vol (W) = ZZZ W 1 dx dy dz = ZZZ W in spherical 1 · ρ2 sin ϕ dρ dϕ dθ = Z 2π 0 Z π 4 0 Z 2 √ 2 0 ρ2 sin ϕ dρ dϕ dθ = Z 2 √ 2 0 ρ2 dρ Z π 4 0 sin ϕ dϕ Z 2π 0 1 dθ = 1 3ρ3 2 √ 2 0 −cos ϕ π 4 0 θ 2π 0 = 16 √ 2 3 · − √ 2 2 −(−1) · 2π = 32 3 ( √ 2 −1)π. Similarly, substituting the conversion z = ρ cos ϕ: ZZZ W z dx dy dz = ZZZ W in spherical ρ cos ϕ · ρ2 sin ϕ dρ dϕ dθ = Z 2π 0 Z π 4 0 Z 2 √ 2 0 ρ3 cos ϕ sin ϕ dρ dϕ dθ = Z 2 √ 2 0 ρ3 dρ Z π 4 0 cos ϕ sin ϕ dϕ Z 2π 0 1 dθ = 1 4ρ4 2 √ 2 0 1 2 sin2 ϕ π 4 0 θ 2π 0 = 16 · 1 4 · 2π = 8π. | Multivariable_Calculus_Shimamoto_Page_198_Chunk3413 |
7.4. CHANGE OF VARIABLES FOR n-FOLD INTEGRALS 187 As a result, z = 1 32 3 ( √ 2−1)π · 8π = 3 4( √ 2−1), and the centroid of W is the point: (x, y, z) = | Multivariable_Calculus_Shimamoto_Page_199_Chunk3414 |
188 CHAPTER 7. CHANGE OF VARIABLES (x1, x2, x3, x4) is in W. This is satisfied by all points in the unit disk x2 3 + x2 4 ≤1. In fact, given such a (x3, x4), the corresponding points (x1, x2) are those such that x2 1 + x2 2 ≤1 −x2 3 −x2 4. We write this way of describing W as: Vol (W) = ZZ x2 3+x2 4≤1 ZZ x2 1+x2 2≤1−x2 3−x2 4 1 dx1 dx2 dx3 dx4. See Figure 7.13 (more nonsense). Figure 7.13: The volume of a four-dimensional ball as a double integral of a double integral, RR ( RR . . . dx1 dx2) dx3 dx4 The inner integral represents the area of a two-dimensional disk of radius a = p 1 −x2 3 −x2 4 in the x1x2-plane. Hence RR x2 1+x2 2≤1−x2 3−x2 4 1 dx1 dx2 = πa2 = π(1 −x2 3 −x2 4), whence: Vol (W) = ZZ x2 3+x2 4≤1 π(1 −x2 3 −x2 4) dx3 dx4. This is an integral over the unit disk in the x3x4-plane, so it can be evaluated using polar coordinates with dx3 dx4 = r dr dθ: Vol (W) = Z 2π 0 Z 1 0 π(1 −r2) · r dr dθ = π Z 1 0 r(1 −r2) dr Z 2π 0 1 dθ = π 1 2r2 −1 4r4 1 0 θ 2π 0 = π · 1 4 · 2π = π2 2 . We leave for the exercises the problems of finding the volume of a five-dimensional ball and, more broadly, a strategy for approaching the volume of an n-dimensional ball in general. See Exercises 4.9 and 4.10. | Multivariable_Calculus_Shimamoto_Page_200_Chunk3415 |
7.5. EXERCISES FOR CHAPTER 7 189 7.5 Exercises for Chapter 7 Section 1 Change of variables for double integrals 1.1. This exercise involves two standard one-variable integrals that appear regularly enough that it seems like a good idea to get out into the open how they can be evaluated quickly. Namely, we compute the integrals R nπ 0 cos2 x dx and R nπ 0 sin2 x dx, where n is a positive integer. They can be found using trigonometric identities, but there is another way that is easier to reproduce on the spot. (a) First, sketch the graphs y = cos2 x and y = sin2 x for x in the interval [0, nπ], and use them to illustrate that R nπ 0 cos2 x dx = R nπ 0 sin2 x dx. (b) This is an optional exercise for those who are uneasy about drawing the conclusion in part (a) based only on a picture. (i) Show that R nπ 0 sin2 x dx = R nπ+ π 2 π 2 cos2 x dx. (Hint: Use the substitution u = x + π 2 and the identity sin(θ −π 2 ) = −cos θ.) (ii) Show that R π 2 0 cos2 x dx = R nπ+ π 2 nπ cos2 x dx. (Hint: cos(θ −π) = −cos θ.) Deduce that R nπ 0 cos2 x dx = R nπ 0 sin2 x dx. (c) Integrate the identity cos2 x + sin2 x = 1, and use part (a) (or (b)) to show that: Z nπ 0 cos2 x dx = nπ 2 and Z nπ 0 sin2 x dx = nπ 2 . 1.2. Find RR D y2 dx dy, where D is the upper half-disk described by x2 + y2 ≤1, y ≥0. 1.3. Find RR D xy dx dy, where D is the region in the first quadrant lying inside the circle x2+y2 = 4 and below the line y = x. 1.4. Find RR D cos(x2 + y2) dx dy, where D is the region described by 4 ≤x2 + y2 ≤16. 1.5. Let D be the region in the xy-plane satisfying x2 + y2 ≤2, y ≥x, and x ≥0. See Figure 7.14. Figure 7.14: The region x2 + y2 ≤2, y ≥x, and x ≥0 Consider the integral ZZ D p 2 −x2 dx dy. (a) Write an expression for the integral using the order of integration dy dx. (b) Write an expression for the integral using the order of integration dx dy. | Multivariable_Calculus_Shimamoto_Page_201_Chunk3416 |
190 CHAPTER 7. CHANGE OF VARIABLES (c) Write an expression for the integral in polar coordinates in whatever order of integration you prefer. (d) Evaluate the integral using whichever approach seems best. 1.6. Let W be the region in R3 lying above the xy-plane, inside the cylinder x2 + y2 = 1, and below the plane x + y + z = 2. Find the volume of W. 1.7. Let W be the region in R3 lying above the xy-plane, inside the cylinder x2 + y2 = 1, and below the plane x + y + z = 1. Find the volume of W. 1.8. In this exercise, we evaluate the improper one-variable integral R ∞ −∞e−x2 dx by following the unlikely strategy of relating it to an improper double integral that turns out to be more tractable. Let a be a positive real number. (a) Let Ra be the rectangle Ra = [−a, a] × [−a, a]. Show that: ZZ Ra e−x2−y2 dx dy = Z a −a e−x2 dx 2 . (b) Let Da be the disk x2 + y2 ≤a2. Use polar coordinates to evaluate RR Da e−x2−y2 dx dy. (c) Note that, as a goes to ∞, both Ra and Da fill out all of R2. It is true that both lim a→∞ RR Ra e−x2−y2 dx dy and lim a→∞ RR Da e−x2−y2 dx dy exist and that they are equal. Their common value is the improper integral RR R2 e−x2−y2 dx dy. Use this information along with your answers to parts (a) and (b) to show that: Z ∞ −∞ e−x2 dx = √π. Section 3 Examples: linear changes of variables, symmetry 3.1. Let D be the set of points (x, y) in R2 such that (x −3)2 + (y −2)2 ≤4. (a) Sketch and describe D. (b) Let T : R2 →R2 be the translation T(u, v) = (u, v) + (3, 2) = (u + 3, v + 2). Describe the region D∗in the uv-plane such that T(D∗) = D. (c) Use the change of variables in part (b) to convert the integral ZZ D (x + y) dx dy over D to an integral over D∗. Then, evaluate the integral using whatever techniques seem best. 3.2. Let a and b be positive real numbers. Find the area of the region x2 a2 + y2 b2 ≤1 inside an ellipse by using the change of variables T(u, v) = (au, bv). (Recall from Chapter 5 that Area (D) = RR D 1 dA.) 3.3. Let T : R2 →R2 be the linear change of variables T(u, v) = (3u −v, 2u + v) of Example 7.6. (a) Show that the only point (u, v) that satisfies T(u, v) = 0 is (u, v) = 0. (b) Show that T is one-to-one on R2. (Hint: Start by assuming that T(a) = T(b), and use the linearity of T and part (a).) 3.4. Find RR D(2x + 4y) dx dy if: | Multivariable_Calculus_Shimamoto_Page_202_Chunk3417 |
7.5. EXERCISES FOR CHAPTER 7 191 (a) D is the parallelogram with vertices (0, 0), (3, 1), (5, 5), and (2, 4), (b) D is the triangular region with vertices (3, 1), (5, 5), and (2, 4). (Hint: Take advantage of the work you’ve already done in part (a). You should be able to use quite a bit of it.) 3.5. Find ZZ D (x + y) ex2−y2 dx dy, where D is the rectangle with vertices (0, 0), (1, −1), (3, 1), and (2, 2). 3.6. Find ZZ D p 12 + x2 + 3y2 dx dy, where D is the region in R2 described by x2 + 3y2 ≤12 and 0 ≤y ≤ 1 √ 3 x. 3.7. This problem concerns the integral: ZZ D y x exy dx dy, (7.6) where D is the region in the first quadrant of the xy-plane that is bounded by the curves y = x, y = 4x, xy = 1, and xy = 2. See Figure 7.15, left. The main idea is to simplify the integrand by making the substitutions u = xy and v = y x. (a) With these substitutions, solve for x and y as functions of u and v: (x, y) = T(u, v). (b) Describe the region D∗in the uv-plane such that T(D∗) = D. (c) Evaluate the integral (7.6). D Figure 7.15: The region bounded by y = x, y = 4x, xy = 1, and xy = 2 (left) and the region x2 −4xy + 8y2 ≤4 (right) 3.8. Let D be the region in R2 described by x2 −4xy + 8y2 ≤4. See Figure 7.15, right. Find ZZ D y2 dx dy. (Hint: x2 −4xy + 8y2 = (x −2y)2 + 4y2.) 3.9. Find ZZ D (x −y)2 (x + y)2 dx dy, where D is the square with vertices (2, 0), (4, 2), (2, 4), and (0, 2). 3.10. Let T : R2 →R2 be a one-to-one linear transformation from the uv-plane to the xy-plane represented by a matrix A = a b c d , i.e., (x, y) = T(u, v) = A · [ uv ]. Let D∗and D be bounded regions in R2 of positive area such that T maps D∗onto D, as shown in Figure 7.16. (a) Let k = | det A|. Use the change of variables theorem to show that: Area (D) = k · | Multivariable_Calculus_Shimamoto_Page_203_Chunk3418 |
192 CHAPTER 7. CHANGE OF VARIABLES Figure 7.16: A linear change of variables (b) Let (u, v) and (x, y) denote the centroids of D∗and D, respectively. Show that: (x, y) = T(u, v). In other words, linear transformations map centroids to centroids. 3.11. A region D in the xy-plane is called symmetric in the line y = x if, whenever a point (x, y) is in D, so is its reflection (y, x). If D is such a region, prove that its centroid lies on the line y = x. 3.12. Let D be a subset of R2 that is symmetric about the origin, that is, whenever (x, y) is in D, so is (−x, −y). (a) Let f : D →R be an integrable function on D. Find a condition on f that ensures that RR D f(x, y) dx dy = 0. Justify your answer. (b) Find an example of such a function other than the constant function f = 0. Section 4 Change of variables for n-fold integrals 4.1. Find ZZZ W (x2 + y2 + z2) dx dy dz if W is the region in R3 that lies above the cone z = p x2 + y2 and below the plane z = 2. 4.2. Find ZZZ W x2z dx dy dz if W is the region in R3 described by x2 + y2 + z2 ≤4, y ≥0, and z ≥0. 4.3. Find ZZZ W z (3 + x2 + y2)2 dx dy dz if W is the region in R3 given by x2 + y2 + z2 ≤1, z ≥0. (Hint: The proper coordinate system and order of integration can make a difference.) 4.4. A class of enthusiastic multivariable calculus students celebrates the change of variables theo- rem by drilling a cylindrical hole of radius 1 straight through the center of the earth. Assuming that the earth is a three-dimensional closed ball of radius 2, find the volume of the portion of the earth that remains. 4.5. Find the centroid of the half-ball of radius a in R3 that lies inside the sphere x2 +y2 +z2 = a2 and above the xy-plane. 4.6. Let W be the wedge-shaped solid in R3 that lies under the plane z = y, inside the cylinder x2 + y2 = 1, and above the xy-plane. Find the centroid of W. | Multivariable_Calculus_Shimamoto_Page_204_Chunk3419 |
7.5. EXERCISES FOR CHAPTER 7 193 4.7. Let W be the region in R3 that lies inside the three cylinders x2 + z2 = 1, y2 + z2 = 1, x2+y2 = 1, and above the xy-plane. Find the volume of W. (Hints: The integral R 1 cos2 θ dθ = R sec2θ dθ = tan θ + C might be helpful. Also, sin3 θ = sin2 θ · sin θ = (1 −cos2 θ) sin θ.) 4.8. Within the solid ball x2 + y2 + z2 ≤a2 of radius a in R3, find the average distance to the origin. 4.9. Find the volume of the five-dimensional closed unit ball: W = {(x1, x2, x3, x4, x5) ∈R5 : x2 1 + x2 2 + x2 3 + x2 4 + x2 5 ≤1}. (Hint: Organize the calculation as a double integral of a triple integral: RR ( RRR . . . ).) 4.10. If a > 0, let Wn(a) denote the closed ball x2 1 + x2 2 + · · · + x2 n ≤a2 in Rn of radius a centered at the origin, and let Vol (Wn(a)) denote its n-dimensional volume. For instance, we showed in Example 7.11 that Vol (W4(1)) = π2 2 . (a) Use the change of variables theorem to show that Vol (Wn(a)) = an Vol (Wn(1)). (b) By thinking of Vol (Wn(1)) as a double integral of an (n −2)-fold integral, find a rela- tionship between Vol (Wn(1)) and Vol (Wn−2(1)). (c) Verify that your answer to part (b) predicts a correct relationship between Vol (W4(1)) and Vol (W2(1)). (d) Find Vol (W6(1)), the volume of the six-dimensional closed unit ball. | Multivariable_Calculus_Shimamoto_Page_205_Chunk3420 |
194 CHAPTER 7. CHANGE OF VARIABLES | Multivariable_Calculus_Shimamoto_Page_206_Chunk3421 |
Part V Integrals of vector fields 195 | Multivariable_Calculus_Shimamoto_Page_207_Chunk3422 |
Chapter 8 Vector fields Now that we know what it means to differentiate vector-valued functions of more than one variable, we turn to integrating them. The functions that we integrate, however, are of a special type. The integrals we consider are rather specialized, too. The functions are called vector fields, and this short chapter is devoted to introducing them. The integration begins in the next chapter. 8.1 Examples of vector fields The domain of a vector field F is a subset U of Rn. The values of F are in Rn as well, but we don’t think so much of F as transforming U into its image. Rather, for a point x of U, we visualize the value F(x) as an arrow translated to start at x. For example, F might represent a force exerted on an object, and the force might depend on where the object is. Then F(x) could represent the force exerted if the object is at the point x. Or some medium might be flowing through U, and F(x) could represent the velocity of the medium at x. By taking a sample of points throughout U and drawing the corresponding arrows translated to start at those points, one can see how some vector quantity acts on U. For instance, with the aid of the arrows, one might be able to visualize the path an object would follow under the influence of a force or a flowing medium. With this prelude, the actual definition is quite simple. Definition. Let U be a subset of Rn. A function F: U →Rn is called a vector field on U. A distinctive feature of vector fields is that both the elements x of the domain and their values F(x) are in Rn, as opposed to a function in general from Rn to Rm. Example 8.1. Let F: R2 →R2 be given by F(x, y) = (1, 0) for all (x, y) in R2. This vector field assigns the vector (1, 0) = i to every point of the plane. To visualize the vector field, we draw the vector (1, 0), or rather a translated copy of it, at a representative collection of points. That is, at each such point, we draw an arrow of length 1 pointing to the right. This is a constant vector field. It is sketched on the left of Figure 8.1. As the figure shows, the sample vectors often end up running into each other, making the picture a little muddled. As a result, we usually depict a vector field by drawing a constant positive scalar multiple of the vectors F(x), where the scalar factor is chosen to improve the readability of the picture. For the vector field in the last example, this is done on the right of the figure. The lengths of the vectors are scaled down, but the arrows still point in the correct direction. 197 | Multivariable_Calculus_Shimamoto_Page_209_Chunk3423 |
198 CHAPTER 8. VECTOR FIELDS Figure 8.1: The constant vector field F(x, y) = (1, 0) = i literally (left) and scaled down (right) Example 8.2. Let F: R2 →R2 be given by F(x, y) = (−y, x). For example, F(1, 0) = (−0, 1) = (0, 1) and F(0, 1) = (−1, 0), as in Figure 8.2. Figure 8.2: The vector field F(x, y) = (−y, x) at the points (1, 0) and (0, 1) We can get some qualitative information of how F acts by noting, for example, that in the first quadrant, where both x and y are positive, F(x, y) = (−y, x) has a negative first component and positive second component, i.e., F(+, +) = (−, +), so the arrow points to the left and up. Similarly, in the second quadrant F(−, +) = (−, −) (to the left and down), in the third quadrant F(−, −) = (+, −) (to the right and down), and in the fourth F(+, −) = (+, +) (to the right and up). Figure 8.3: The vector field F(x, y) = (−y, x) In fact, we can identify both the length and direction of F(x, y) more precisely: | Multivariable_Calculus_Shimamoto_Page_210_Chunk3424 |
8.1. EXAMPLES OF VECTOR FIELDS 199 • ∥F(x, y)∥= p (−y)2 + x2 = ∥(x, y)∥, • F(x, y) · (x, y) = (−y, x) · (x, y) = −yx + xy = 0. In other words, the length of F(x, y) equals the distance of (x, y) to the origin, and its direction is orthogonal to (x, y) in the counterclockwise direction. After drawing in a sample of arrows, one can imagine F as describing a circular counterclockwise flow, or vortex, around the origin, growing in magnitude as one moves further out, as in Figure 8.3. Example 8.3. Let W: R2 −{(0, 0)} →R2 be given by W(x, y) = | Multivariable_Calculus_Shimamoto_Page_211_Chunk3425 |
200 CHAPTER 8. VECTOR FIELDS Figure 8.5: An inverse square vector field To find a formula for G, we write G(x, y, z) = c ∥(x,y,z)∥2 u, where u is the unit vector in the direction from (x, y, z) to (0, 0, 0), that is, u = −(x,y,z) ∥(x,y,z)∥. Hence G(x, y, z) = − c ∥(x,y,z)∥2 · (x,y,z) ∥(x,y,z)∥= − c ∥(x,y,z)∥3 (x, y, z). This is called an inverse square field. To summarize, it is the vector field G: R3−{(0, 0, 0)} → R3 given variously by: G(x, y, z) = − c ∥(x, y, z)∥3 (x, y, z) = − c (x2 + y2 + z2)3/2 (x, y, z) = − cx (x2 + y2 + z2)3/2 , − cy (x2 + y2 + z2)3/2 , − cz (x2 + y2 + z2)3/2 , for some positive constant c. In the future, we take the constant of proportionality to be c = 1 and refer to the resulting vector field G as “the” inverse square field. These examples shall often serve as test cases for the concepts we are about to develop. 8.2 Exercises for Chapter 8 In Exercises 1.1–1.6, sketch the given vector field F. To get started, it may be helpful to get an idea of the general direction of the arrows F(x, y) at points on the x and y-axes and in each of the four quadrants. 1.1. F(x, y) = (x, y) 1.2. F(x, y) = (2x, y) 1.3. F(x, y) = (y, −x) 1.4. F(x, y) = (−2y, x) 1.5. F(x, y) = (y, x) 1.6. F(x, y) = (−x, y) | Multivariable_Calculus_Shimamoto_Page_212_Chunk3426 |
8.2. EXERCISES FOR CHAPTER 8 201 1.7. (a) Find a smooth vector field F on R2 such that, at each point (x, y), F(x, y) is a unit vector normal to the parabola of the form y = x2 + c that passes through that point. (b) Find a smooth vector field F on R2 such that, at each point (x, y), F(x, y) is a unit vector tangent to the parabola of the form y = x2 + c that passes through that point. 1.8. (a) Find a smooth vector field F on R2 such that ∥F(x, y)∥= ∥(x, y)∥and, at each point (x, y) other than the origin, F(x, y) is a vector normal to the curve of the form xy = c that passes through that point. (b) Find a smooth vector field F on R2 such that ∥F(x, y)∥= ∥(x, y)∥and, at each point (x, y) other than the origin, F(x, y) is a vector tangent to the curve of the form xy = c that passes through that point. Let F be a vector field on an open set U in Rn. If we think of F as the velocity of a medium flowing through U, then an object dropped into U will follow a path along which the velocity at each point is the value of the vector field at the point. More precisely, a smooth path α : I →U defined on an interval I is called an integral path of F if α′(t) = F(α(t)) for all t in I. The curve traced out by α is called an integral curve. For example, let F(x, y) = (−y, x) be the circulating vector field from Example 8.2. If a > 0, let α: R →R2 be the parametrization of a circle of radius a given by: α(t) = (a cos t, a sin t). (8.1) Then α′(t) = (−a sin t, a cos t), and F(α(t)) = F(a cos t, a sin t) = (−a sin t, a cos t). Since the two are equal, α is an integral path of F. The integral curves are circles centered at the origin. This reinforces our sense that F describes counterclockwise circular flow around the origin. 1.9. Let F(x, y) = (−2y, x). (This vector field also appears in Exercise 1.4.) (a) If a is a real number, show that α(t) = | Multivariable_Calculus_Shimamoto_Page_213_Chunk3427 |
202 CHAPTER 8. VECTOR FIELDS as | Multivariable_Calculus_Shimamoto_Page_214_Chunk3428 |
Chapter 9 Line integrals We begin integrating vector fields in the case that the domain of integration is a curve. The integrals are called line integrals. By comparison, we learned in Section 2.3 about the integral over a curve C of a real-valued function f. By definition, R C f ds = R b a f(α(t)) ∥α′(t)∥dt, where α : [a, b] →Rn is a smooth parametrization of C. This is called the integral with respect to arclength. The integral of a vector field is a new concept—for example, we don’t simply integrate each of the real-valued component functions of the vector field independently using the integral with respect to arclength. In the grand scheme of things, line integrals are a richer subject than integrals with respect to arclength. In part, this is because of the ways in which the line integral can be interpreted and applied, many of which originate in physics (for instance, see Exercise 3.19 at the end of the chapter). But in addition, on a more abstract level, line integrals provide us with a first peek into a line of unifying results that relate various types of integrals to one another. By the end of the book, we shall be in a position to see that these connections run quite deep. We open the chapter by defining the line integral and discussing what it represents. There can be more than one option for how to compute a given line integral, and we identify the circumstances under which the different approaches apply. These computational techniques, however, follow from theoretical principles that are interesting in their own right. For instance, for vector fields in the plane, some line integrals can be expressed as double integrals. This result is known as Green’s theorem. Throughout the chapter, the discussion reveals an active interplay between geometric features of the curve, the vector field, and, on occasion, the geometry of the domain on which the vector field is defined as well. 9.1 Definitions and examples Let F: U →Rn be a vector field on an open set U in Rn, and let C be a curve in U. Roughly speaking, the integral of F over C is meant to represent the “effectiveness” of F in contributing to motion along C. By this, we mean the degree of alignment of F with the direction of motion. For instance, suppose that F is a force field and that α(t) is the position of a moving object at time t. Then the line integral of F over C is called the work done by F. If F points in the same direction as the direction of motion, then F helps move the object along. If it points in the opposite direction, then it impedes the motion. If it is orthogonal, it neither helps nor hinders. In general, we measure the contribution of F to the motion at each point of C by finding the scalar component Ftan of F in the direction tangent to the curve. See Figure 9.1. Then, we integrate with respect to arclength, obtaining R C Ftan ds. Note that Ftan = ∥F∥cos θ, where θ is the angle between F and the tangent direction. Hence, 203 | Multivariable_Calculus_Shimamoto_Page_215_Chunk3429 |
204 CHAPTER 9. LINE INTEGRALS Figure 9.1: The component Ftan of a vector field in the tangent direction if T = α′ ∥α′∥is the unit tangent vector, then: Ftan = ∥F∥cos θ = ∥F∥∥T∥cos θ = F · T. By the definition of the integral with respect to arclength with f = Ftan, we obtain: Z C Ftan ds = Z C F · T ds = Z b a | Multivariable_Calculus_Shimamoto_Page_216_Chunk3430 |
9.1. DEFINITIONS AND EXAMPLES 205 One final word about notation: we have chosen to write the line integral as an integral over α rather than over the underlying curve C to acknowledge that the definition makes use of the specific parametrization α. This is an important point, and we elaborate on it momentarily. Just as in first-year calculus a function can be integrated over many intervals, so can a vector field be integrated over many paths. The relation between the vector field and the path is reflected in the value of the integral. Example 9.1. Let F be the vector field on R2 given by F(x, y) = (−y, x). This is the vector field from Chapter 8 that circulates around the origin, shown again in Figure 9.2. Figure 9.2: The circulating vector field F(x, y) = (−y, x) Integrate F over the following paths. (a) α1(t) = (cos t, sin t), 0 ≤t ≤2π (b) α2(t) = (t, t), 0 ≤t ≤3 (c) α3(t) = (cos t, sin t), 0 ≤t ≤π/2 (d) α4(t) = (1 −t, t), 0 ≤t ≤1 (e) α5(t) = (cos2 t, sin2 t), 0 ≤t ≤π/2 In differential form notation, the integrals R α F · ds we seek are written R α −y dx + x dy. Figure 9.3: The vector field F(x, y) = (−y, x) along the unit circle (left), the line y = x (middle), and the line x + y = 1 (right) (a) The path α1 traces out the unit circle counterclockwise. The direction of the vector field and the direction of motion are aligned at every point. See Figure 9.3 at left. As a result, F · T is always positive, and we expect the integral to be positive as well. In fact, for this parametrization, | Multivariable_Calculus_Shimamoto_Page_217_Chunk3431 |
206 CHAPTER 9. LINE INTEGRALS x = cos t and y = sin t, so substitution gives: Z α1 −y dx + x dy = Z 2π 0 | Multivariable_Calculus_Shimamoto_Page_218_Chunk3432 |
9.1. DEFINITIONS AND EXAMPLES 207 Note that each of the paths α3, α4, and α5 above goes from (1, 0) to (0, 1). The values of the line integral are not all equal, however. Thus changing the path between the endpoints can change the integral. On the other hand, α4 and α5 not only go between the same two points but also trace out the same curve, just parametrized in different ways. The integrals over these two paths are equal. Is this a coincidence? In general, suppose that α: [a, b] →U and β : [c, d] →U are smooth parametrizations of the same curve C. To help distinguish between them, we use t to denote the parameter of α and u for the parameter of β. It simplifies the argument somewhat to assume that α is one-to-one, except possibly at the endpoints, where α(a) = α(b) is allowed, and likewise for β. This is not strictly necessary, though it has been true of all the examples of parametrizations we have encountered thus far. We want to compare the integrals R α F · ds = R b a F(α(t)) · α′(t) dt and R β F · ds = R d c F(β(u)) · β′(u) du obtained using the respective parametrizations. Consider first the case that α and β trace out C in the same direction, that is, both start at a point p = α(a) = β(c) and end at a point q = α(b) = β(d), as in Figure 9.4. Let x be a point of C other than the endpoints. Then x = α(t) for some value of t and x = β(u) for some value of u. We write this as t = g(u), that is, for a given value of u, g(u) is the value of t such that α(t) = β(u). We also set g(c) = a and g(d) = b. Thus: β(u) = α(g(u)) for all u in [c, d]. (9.3) For instance, in the case of the line segments α4 and α5 above, α5(u) = (cos2 u, sin2 u) = | Multivariable_Calculus_Shimamoto_Page_219_Chunk3433 |
208 CHAPTER 9. LINE INTEGRALS In other words, both parametrizations give the same value of the line integral. If the parametrizations traverse C in opposite directions so that, say p = α(a) = β(d) and q = α(b) = β(c), then the only difference is that the endpoints are reversed. That is, g(d) = a and g(c) = b, so: Z β F · ds = Z g(d) g(c) F(α(t)) · α′(t) dt (just as before, but . . . ) = Z a b F(α(t)) · α′(t) dt = − Z b a F(α(t)) · α′(t) dt (9.5) = − Z α F · ds. The integrals differ by a sign. We summarize this as follows. Proposition 9.2. If α and β are smooth parametrizations of the same curve C, then: Z β F · ds = ± Z α F · ds, where the choice of sign depends on whether they traverse C in the same or opposite direction. This allows us to define a line integral over a curve, as opposed to over a parametrization. Definition. An oriented curve is a curve C with a specified direction of traversal. Given such a C, define: Z C F · ds = Z α F · ds, where α is any smooth parametrization that traverses C in the given direction. By the preceding proposition, any two such parametrizations give the same answer. The point is that the line integral depends only on the underlying curve and the direction in which it is traversed. This is critical in that it allows us to think of the line integral as an integral over an oriented geometric set, independent of the parametrization that is used to trace it out. This is an issue that we overlooked previously when we discussed integrals over curves with respect to arclength and integrals over surfaces with respect to surface area. In both of those cases, the integrals were defined using parametrizations, but we did not check that the answers were independent of the particular parametrization used to trace out the curve or surface. In the case of the integral with respect to arclength, the story is that R α f ds = R β f ds for any smooth parametrizations α and β of C. The new twist is that the integral is not only independent of the parametrization but also of the direction in which it traverses C. The proof proceeds as before, but a difference arises because the definition R β f ds = R d c f(β(u)) ∥β′(u)∥du involves the norm ∥β′(u))∥, not β′(u) itself. This introduces a factor of |g′(u)| after applying the chain rule as in equation (9.4) and taking norms. If α and β traverse C in the same direction, then g′(u) > 0, so |g′(u)| = g′(u) and the argument proceeds essentially unchanged. If they traverse C in opposite directions, then g′(u) < 0, so |g′(u)| = −g′(u). This minus sign is counteracted by the minus sign that appears in this case when reversing the limits of integration in (9.5), so the two minus signs cancel. We leave the details for the exercises (Exercise 1.12). The role of parametrizations in surface integrals is discussed in Chapter 10. Having cleared the air on that lingering matter, we now return to studying line integrals. | Multivariable_Calculus_Shimamoto_Page_220_Chunk3434 |
9.1. DEFINITIONS AND EXAMPLES 209 Example 9.3. Evaluate R C x dx+2y dy+3z dz if C is the oriented curve consisting of three quarter- circular arcs on the unit sphere in R3 from (1, 0, 0) to (0, 1, 0) to (0, 0, 1) and then back to (1, 0, 0). See Figure 9.5. Figure 9.5: A curve consisting of three arcs C1, C2, and C3 The vector field being integrated here is given by F(x, y, z) = (x, 2y, 3z). Since C consists of three arcs C1, C2, C3, we can imagine traversing each of them in succession and breaking up the integral as R C F · ds = R C1 F · ds + R C2 F · ds + R C3 F · ds. Moreover, we may parametrize each arc however we like as long as it is traversed in the proper direction. In this way, we can avoid having to parametrize all of C over a single parameter interval. For C1, we use the usual parametrization of the unit circle in the xy-plane from (1, 0, 0) to (0, 1, 0): α1(t) = (cos t, sin t, 0), 0 ≤t ≤π 2 . Z C1 x dx + 2y dy + 3z dz = Z π 2 0 | Multivariable_Calculus_Shimamoto_Page_221_Chunk3435 |
210 CHAPTER 9. LINE INTEGRALS smooth oriented curves, that is, curves of the form C = C1 ∪C2 ∪· · ·∪Ck, where C1, C2, . . . , Ck are traversed in succession and each of them has a smooth parametrization. Such a curve C is called piecewise smooth. Under these conditions, we define: Z C F · ds = Z C1 F · ds + Z C2 F · ds + · · · + Z Ck F · ds. This allows us to deal with curves that have corners, as in the example, without worrying about whether the entire curve can be parametrized in a smooth way. 9.2 Another word about substitution Before moving on, we comment on the approach we took to defining the line integral. A curve is one-dimensional in the sense that it can be traced out using one parameter. Thus we might think of an integral over a curve as essentially a one-variable integral. The one-variable integrals that we are used to are those from first-year calculus, integrals over intervals in the real line. The integral of a vector field F in Rn over a curve C is a new type of integral, however. F itself is a function of n variables x1, x2, . . . , xn, but, along C, each of these variables can be expressed in terms of the parameter t. This is exactly the information provided by the parametrization α(t) = (x1(t), x2(t), . . . , xn(t)), a ≤t ≤b. So the idea is that substituting for x1, x2, . . . , xn in terms of t converts an integral over C to an integral over the interval I = [a, b], where things are familiar. This fits into a typical pattern. Whereas the parametrization α transforms the interval I into the curve C, substitution has the effect of pulling back an integral over C to an integral over I. For instance, after substitution, the component function Fi(x1, x2, . . . , xn) of F(x1, x2, . . . , xn) is converted to the function Fi(x1(t), x2(t), . . . , xn(t)) = Fi(α(t)). We say that α pulls back Fi to Fi ◦α and write: α∗(Fi) = Fi ◦α. Similarly, the expression dxi is pulled back to dxi dt dt, and we write: α∗(dxi) = dxi dt dt. Actually, the xi’s on each side of this equation mean slightly different things (coordinate label vs. function of t) and it would be better to write α∗(dxi) = dαi dt dt, but hopefully our cavalier attitude will not lead us astray. For a line integral, by pulling back the individual terms, the entire integrand F1 dx1 + F2 dx2 + · · · + Fn dxn in terms of x1, x2, . . . , xn is pulled back to the expression: α∗(F1 dx1 + F2 dx2 + · · · + Fn dxn) = (F1 ◦α) dx1 dt + (F2 ◦α) dx2 dt + · · · + (Fn ◦α) dxn dt dt in terms of t. If we denote F1 dx1 + F2 dx2 + · · · + Fn dxn by ω, then the definition of the line integral reads: Z α(I) ω = Z I α∗(ω). (9.6) This looks a lot like the change of variables theorem. For instance, compare with equation (7.3) of Chapter 7. We verified that the value of the integral is independent of the parametrization α so | Multivariable_Calculus_Shimamoto_Page_222_Chunk3436 |
9.3. CONSERVATIVE FIELDS 211 that it is well-defined as an integral over the oriented curve C = α(I). The chain rule played an essential, though easily missed, role in the verification. See equation (9.4). This approach recurs more generally. In order to define a new type of integral, we use substitu- tion to pull the integral back to a familiar situation. This is done in a way that allows the change of variables theorem and the chain rule to conspire to make the process legitimate. 9.3 Conservative fields If C is a smooth oriented curve in an open set U in Rn from a point p to a point q as in Figure 9.6, the value of the line integral R C F · ds does not depend on how C is parametrized as long as it is traversed in the correct direction. We have seen, however, that the value does depend on the actual curve C: choosing a different curve from p to q can change the integral. Are there situations where even changing the curve doesn’t matter? Figure 9.6: An oriented curve C from p to q By definition, R C F · ds = R b a F(α(t)) · α′(t) dt, where α: [a, b] →U is a smooth parametrization that traces out C with the proper orientation. Now, suppose—and this is a big assumption—that F is the gradient of a smooth real-valued function f, that is, F = ∇f, or, in terms of components, (F1, F2, . . . , Fn) = ( ∂f ∂x1 , ∂f ∂x2 , . . . , ∂f ∂xn ). Then F(α(t)) · α′(t) = ∇f(α(t)) · α′(t) = (f ◦α)′(t), where the last equality is the Little Chain Rule. In other words, the integrand in the definition of the line integral is a derivative. So: Z C F · ds = Z b a (f ◦α)′(t) dt = (f ◦α)(t) b a = f(α(b)) −f(α(a)) = f(q) −f(p). This depends only on the endpoints p and q and not at all on the curve that connects them. In order to apply the Little Chain Rule, this argument requires that C have a smooth parametri- zation. If C is piecewise smooth instead, say C = C1 ∪C2 ∪· · · ∪Ck, where each Ci is smooth and C1 goes from p to p1, C2 from p1 to p2, and so on until Ck goes from pk−1 to q, then applying the previous result to the pieces gives a telescoping sum in which the intermediate endpoints appear twice with opposite signs: Z C F · ds = Z C1 F · ds + Z C2 F · ds + · · · + Z Ck F · ds = | Multivariable_Calculus_Shimamoto_Page_223_Chunk3437 |
212 CHAPTER 9. LINE INTEGRALS We have proven the following. Theorem 9.4 (Conservative vector field theorem). If F is a conservative vector field on U with potential function f, then for any piecewise smooth oriented curve C in U: Z C F · ds = f(q) −f(p), where p and q are the starting and ending points, respectively, of C. A couple of other ways to write this are: Z C ∇f · ds = f(q) −f(p) or Z C ∂f ∂x1 dx1 + ∂f ∂x2 dx2 + · · · + ∂f ∂xn dxn = f(q) −f(p). We shall give some examples to illustrate the theorem shortly, but first we present an immediate general consequence that also helps explain in part why the word conservative is used. (See Exercise 3.19 for another reason.) Definition. A curve C is called closed if it starts and ends at the same point. See Figure 9.7 for some examples. Figure 9.7: Examples of closed curves: an ellipse (left) and a figure eight (middle) in R2 and a knot in R3 (right) Corollary 9.5. If F is conservative on U, then for any piecewise smooth oriented closed curve C in U: Z C F · ds = 0. Example 9.6. Let F be the vector field on R3 given by F(x, y, z) = (x, 2y, 3z). Is F a conservative field? We look to see if there is a real-valued function f such that ∇f = F, that is, ∂f ∂x = x, ∂f ∂y = 2y, and ∂f ∂z = 3z. By inspection, it’s easy to see that f(x, y, z) = 1 2x2 + y2 + 3 2z2 works. Thus F is conservative with f as potential function. It follows that R C x dx + 2y dy + 3z dz = 0 for any piecewise smooth oriented closed curve C in R3. This explains the answer of 0 that we obtained in Example 9.3 when we integrated F around the closed curve consisting of three circular arcs (Figure 9.5). Similarly, if C is any curve from (1, 0, 0) to (0, 1, 0), such as the arc C1 in the xy-plane of Example 9.3, then by the conservative vector field theorem: Z C x dx + 2y dy + 3z dz = f(0, 1, 0) −f(1, 0, 0) = (0 + 1 + 0) −(1 2 + 0 + 0) = 1 2. Again, this agrees with what we got before. More importantly, it is a much simpler way to evaluate the integral. | Multivariable_Calculus_Shimamoto_Page_224_Chunk3438 |
9.3. CONSERVATIVE FIELDS 213 Example 9.7. Let G be the inverse square field: G(x, y, z) = − x (x2 + y2 + z2)3/2 , − y (x2 + y2 + z2)3/2 , − z (x2 + y2 + z2)3/2 , where (x, y, z) ̸= (0, 0, 0). Find R C G · ds if C is the portion of the helix parametrized by α(t) = (cos t, sin t, t), 0 ≤t ≤3π. We could calculate the integral explicitly using the definition and the parametrization, but, inspired by the previous example, we check first to see if G is conservative. After a little trial- and-error, we find that g(x, y, z) = (x2 + y2 + z2)−1/2 = 1 ∥(x,y,z)∥works as a potential funcion. For instance, ∂g ∂x = −1 2(x2 + y2 + z2)−3/2 · 2x = − x (x2+y2+z2)3/2 , and so on. Since C goes from p = α(0) = (1, 0, 0) to q = α(3π) = (−1, 0, 3π), the conservative vector field theorem implies: Z C G · ds = g(−1, 0, 3π) −g(1, 0, 0) = 1 √ 1 + 9π2 −1. The truth is that evaluating this integral directly using the parametrization is actually fairly straightforward. Try it—it turns out that it’s not as bad as it might look. This is often the case with conservative fields. Instead, perhaps the most meaningful takeaway from the approach we took is that we can say that R C G · ds = 1 √ 1+9π2 −1 for any piecewise smooth curve in the domain of G, R3 −{(0, 0, 0)}, from p = (1, 0, 0) to q = (−1, 0, 3π). That the given curve is a helix is irrelevant. Indeed, more generally, if C is any piecewise smooth curve in R3 −{(0, 0, 0)} from a point p to a point q, then using the potential function g(x) = 1 ∥x∥above gives: Z C G · ds = 1 ∥q∥− 1 ∥p∥. In particular, the integral of the inverse square field around any piecewise smooth oriented closed curve that avoids the origin is 0. Example 9.8. Let F be the counterclockwise circulation field F(x, y) = (−y, x) on R2 (Figure 9.2). Is F a conservative field? This time, the answer is no, and we give a couple of ways of looking at it. First, if F were conservative, then its integral around any closed curve would be zero. But just by looking at the way the field circulates around the origin, one can see that the integral around a circle centered at the origin and oriented counterclockwise is positive. In fact, we calculated earlier in Example 9.1 that, if C is the unit circle traversed counterclockwise, then R C −y dx + x dy = 2π. Since C is closed and the integral is not zero, F is not conservative. A more direct approach is to try to guess a potential function f. Here, we want ∇f = F = (−y, x), that is: ( ∂f ∂x = −y ∂f ∂y = x. (9.7) To satisfy the first condition, we might try f(x, y) = −xy, but then ∂f ∂y = −x, not x, and there doesn’t seem to be any way to fix it. This shows that our first guess didn’t work, but it’s possible that we are simply bad guessers. To show that (9.7) is truly inconsistent, we bring in the second-order partials. For if there were a function f that satisfied (9.7), then: ∂2f ∂y ∂x = ∂ ∂y | Multivariable_Calculus_Shimamoto_Page_225_Chunk3439 |
214 CHAPTER 9. LINE INTEGRALS This last argument is easily generalized. Theorem 9.9 (Mixed partials theorem). Let F be a smooth vector field on an open set U in Rn, written in terms of components as F = (F1, F2, . . . , Fn). If F is conservative, then: ∂Fi ∂xj = ∂Fj ∂xi for all i, j = 1, 2, . . . , n. Equivalently, if ∂Fi ∂xj ̸= ∂Fj ∂xi for some i, j, then F is not conservative. Proof. Assume that F is conservative, and let f be a potential function. Then F = ∇f, that is, (F1, F2, . . . , Fn) = ( ∂f ∂x1 , ∂f ∂x2 , . . . , ∂f ∂xn ). Therefore Fi = ∂f ∂xi and Fj = ∂f ∂xj , and so ∂Fi ∂xj = ∂2f ∂xj ∂xi and ∂Fj ∂xi = ∂2f ∂xi ∂xj . By the equality of mixed partials, these last two partial derivatives are equal. We look at what the mixed partials theorem says for R2 and R3. For a vector field F = (F1, F2) on a subset of R2, there is just one mixed partials pair: ∂F1 ∂y and ∂F2 ∂x . Thus: In R2, if F = (F1, F2) is conservative, then ∂F1 ∂y = ∂F2 ∂x . Next, for F = (F1, F2, F3) on a subset of R3, there are three mixed partials pairs: (i) ∂F1 ∂y and ∂F2 ∂x , (ii) ∂F1 ∂z and ∂F3 ∂x , and (iii) ∂F2 ∂z and ∂F3 ∂y . We describe a slick way to compare the partial derivatives in each pair. Let ∇= ( ∂ ∂x, ∂ ∂y, ∂ ∂z). This is not a legitimate vector, but we think of it as an “operator.” For instance, it acts on a real-valued function f(x, y, z) by scalar multiplication: ∇f = ( ∂ ∂x, ∂ ∂y, ∂ ∂z) f = (∂f ∂x, ∂f ∂y , ∂f ∂z ). This agrees with the usual notation for the gradient. For a vector field F = (F1, F2, F3), ∇acts by the cross product: ∇× F = det i j k ∂ ∂x ∂ ∂y ∂ ∂z F1 F2 F3 = ∂F3 ∂y −∂F2 ∂z , − | Multivariable_Calculus_Shimamoto_Page_226_Chunk3440 |
9.4. GREEN’S THEOREM 215 Having raised the suspicion, we compute the curl: ∇× F = det i j k ∂ ∂x ∂ ∂y ∂ ∂z 2xy3z4 + x 3x2y2z4 + 3 4x2y3z3 + y2 = | Multivariable_Calculus_Shimamoto_Page_227_Chunk3441 |
216 CHAPTER 9. LINE INTEGRALS Similarly, for C3: Z C3 F1 dx + F2 dy = Z b a F1(x, d) dx. For the vertical segments C2 and C4, we use y as parameter, c ≤y ≤d, keeping x fixed. Hence dx dy = 0 and dy dy = 1, which gives: Z C2 F1 dx + F2 dy = Z d c F2(b, y) dy and Z C4 F1 dx + F2 dy = Z d c F2(a, y) dy. Substituting these four calculations into equation (9.8) results in: Z C F1 dx + F2 dy = Z b a F1(x, c) dx + Z d c F2(b, y) dy − Z b a F1(x, d) dx − Z d c F2(a, y) dy = Z d c | Multivariable_Calculus_Shimamoto_Page_228_Chunk3442 |
9.4. GREEN’S THEOREM 217 Note that, for the boundaries of R1, R2, R3, R4, the portions in the interior of D appear as part of two different Cj traversed in opposite directions. Thus these portions of the line integrals cancel out, leaving only the parts left exposed around the boundary of the original region D. Hence, if C denotes the boundary of D, we have: ZZ D | Multivariable_Calculus_Shimamoto_Page_229_Chunk3443 |
218 CHAPTER 9. LINE INTEGRALS We have proven Green’s theorem only in very special cases. The argument given for a rectangle with a hole, however, can be extended easily to any domain D that is a union of finitely many rectangles that intersect at most along common portions of their boundaries. For a general region D, one strategy for giving a proof is to show that D can be approximated by such a union of rectangles so that Green’s theorem is approximately true for D. By taking a limit of such approximations, the theorem can be proven. This requires considerable technical expertise which we leave for another course. Example 9.12. Let F(x, y) = (x −y2, xy). Find R C F · ds if C is the oriented simple closed curve consisting of the semicircular arc in the upper half-plane from (2, 0) to (−2, 0) followed by the line segment from (−2, 0) to (2, 0). See Figure 9.11. Figure 9.11: An integral around the boundary of a half-disk Let D be the filled-in half-disk bounded by C so that C = ∂D. By Green’s theorem: Z C=∂D (x −y2) dx + xy dy = ZZ D | Multivariable_Calculus_Shimamoto_Page_230_Chunk3444 |
9.5. THE VECTOR FIELD W 219 Figure 9.12: Integrating the circulating vector field F(x, y) = (−y, x) around a simple closed curve C may also be places where C moves in the clockwise direction around the origin. For instance, if C does not encircle the origin, this happens at points nearest the origin. The tangental component is negative along such places. This means that there may be two types of contributions to the integral that counteract one another. The magnitude of F increases as you move away from the origin, however, so we might expect the positive contribution to be greater. In fact, using Green’s theorem, we can measure precisely how much greater it is. Let D be the filled-in region bounded by C. Then, by Green’s theorem: Z C −y dx + x dy = ZZ D | Multivariable_Calculus_Shimamoto_Page_231_Chunk3445 |
220 CHAPTER 9. LINE INTEGRALS Figure 9.13: The vector field W Figure 9.14: The circle and disk of radius a: Ca = ∂Da By the quotient rule: ∂W2 ∂x = ∂ ∂x x x2 + y2 = (x2 + y2) · 1 −x · 2x (x2 + y2)2 = y2 −x2 (x2 + y2)2 and ∂W1 ∂y = ∂ ∂y − y x2 + y2 = −(x2 + y2) · 1 −y · 2y (x2 + y2)2 = y2 −x2 (x2 + y2)2 . Hence ∂W2 ∂x = ∂W1 ∂y , so Green’s theorem gives: Z Ca − y x2 + y2 dx + x x2 + y2 dy = ZZ Da ∂W2 ∂x −∂W1 ∂y dx dy = ZZ Da 0 dx dy = 0. Approach B. We calculate the line integral directly using a parametrization of Ca, say α(t) = (a cos t, a sin t), 0 ≤t ≤2π. This gives: Z Ca − y x2 + y2 dx + x x2 + y2 dy = Z 2π 0 − a sin t a2 cos2 t + a2 sin2 t · (−a sin t) + a cos t a2 cos2 t + a2 sin2 t · (a cos t) dt = Z 2π 0 a2 sin2 t + a2 cos2 t a2 dt = Z 2π 0 a2 a2 dt = Z 2π 0 1 dt = 2π. | Multivariable_Calculus_Shimamoto_Page_232_Chunk3446 |
9.5. THE VECTOR FIELD W 221 In other words, 0 = 2π. Evidently, there’s been a mistake. The truth is that Approach A is flawed. The disk Da contains the origin, and the origin is not in the domain of W. Thus Green’s theorem does not apply to W and Da. Instead, the correct answer is B: Z Ca − y x2 + y2 dx + x x2 + y2 dy = 2π. (9.10) This calculation has some noteworthy consequences. • The value of the integral is 2π regardless of the radius of the circle. • W is not conservative, as integrating it around a closed curve does not always give 0. • Nevertheless, all possible mixed partials pairs are equal: ∂W2 ∂x = ∂W1 ∂y . Hence the converse of the mixed partials theorem can fail. We broaden the analysis by considering any piecewise smooth simple closed curve C in the plane, oriented counterclockwise, that does not pass through the origin, meaning that (0, 0) /∈C. This doesn’t seem like very much to go on, but it turns out that we can say a lot about the integral of W over C. There are two cases, corresponding to the fact that C separates the plane into two regions, which we refer to as the interior and the exterior. Case 1. (0, 0) lies in the exterior of C. See Figure 9.15. Figure 9.15: The origin lies in the exterior of C: ∂D = C. Then the region D bounded by C does not contain the origin, so Approach A above is valid and Green’s theorem applies with C = ∂D. Since the mixed partials of W are equal, it follows that: Z C − y x2 + y2 dx + x x2 + y2 dy = ZZ D ∂W2 ∂x −∂W1 ∂y dx dy = 0. Case 2. (0, 0) lies in the interior of C. See Figure 9.16. As we saw with the flaw in Approach A, we can no longer fill in C and use Green’s theorem. Instead, let Cϵ be a circle of radius ϵ centered at the origin and oriented counterclockwise, where ϵ is chosen small enough that Cϵ is contained entirely in the interior of C. Let D be the region inside of C but outside of Cϵ. The origin does not belong to D, so Green’s theorem applies. Since the mixed partials of W are equal, this gives: Z ∂D − y x2 + y2 dx + x x2 + y2 dy = ZZ D | Multivariable_Calculus_Shimamoto_Page_233_Chunk3447 |
222 CHAPTER 9. LINE INTEGRALS Figure 9.16: The origin lies in the interior of C: ∂D = C ∪(−Cϵ). On the other hand, taking orientation into account, ∂D = C ∪(−Cϵ), so: Z ∂D − y x2 + y2 dx + x x2 + y2 dy = Z C − y x2 + y2 dx + x x2 + y2 dy − Z Cϵ − y x2 + y2 dx + x x2 + y2 dy. Combining these results gives: Z C − y x2 + y2 dx + x x2 + y2 dy = Z Cϵ − y x2 + y2 dx + x x2 + y2 dy. But we showed in equation (9.10) that the integral of W around any counterclockwise circle about the origin, such as Cϵ, is 2π. Hence the value around C must be the same. To summarize, we have proven: Z C − y x2 + y2 dx + x x2 + y2 dy = ( 0 if (0, 0) lies in the exterior of C, 2π if (0, 0) lies in the interior of C. (9.11) If C is oriented clockwise instead, then the change in orientation reverses the sign of the integral, so the corresponding values in the two cases are 0 and −2π, respectively. That there are exactly three possible values of the integral of W over all piecewise smooth oriented simple closed curves in R2 −{(0, 0)} may be somewhat surprising. 9.6 The converse of the mixed partials theorem We now discuss briefly and informally conditions regarding when a vector field F = (F1, F2, . . . , Fn) is conservative. On the one hand, we have seen that, if F is conservative, then: C1. the integral of F over any piecewise smooth oriented curve C depends only on the endpoints of C, C2. the integral of F over any piecewise smooth oriented closed curve is 0, and C3. all mixed partial pairs are equal: ∂Fi ∂xj = ∂Fj ∂xi for all i, j. So these conditions are necessary for F to be conservative. Are any of them sufficient? It turns out that the converses corresponding to C1 and C2 are true, but we know from the example of W in the previous section that the converse of C3 is not. Computing the mixed partials and seeing if they are equal is the simplest of the three conditions to check, however, so we could ask if there are special circumstances under which the converse of C3 does hold. We investigate this next and toss in a few remarks about the other two conditions along the way. | Multivariable_Calculus_Shimamoto_Page_234_Chunk3448 |
9.6. THE CONVERSE OF THE MIXED PARTIALS THEOREM 223 Definition. A subset U of Rn is called simply connected if: • every pair of points in U can be joined by a piecewise smooth curve in U and • every closed curve in U can be continuously contracted within U to a point. For example, Rn itself is simply connected. So are rectangles and balls. A sphere in R3 is simply connected as well. A closed curve in any of them can be collapsed to a point within the set rather like a rubber band shrinking in on itself. See Figure 9.17. Figure 9.17: Examples of simply connected sets: a disk, a rectangle, and a sphere The punctured plane R2 −{(0, 0)} is not simply connected, however. A closed curve that goes around the origin cannot be shrunk to a point without passing through the origin, i.e., without leaving the set. Similarly, an annulus in R2, that is, the ring between two concentric circles, is not simply connected. An example of a non-simply connected surface is the surface of a donut, which is called a torus. A loop that goes through the hole cannot be contracted to a point while remaining on the surface. These examples are shown in Figure 9.18. Figure 9.18: Examples of non-simply connected sets and closed curves that cannot be contracted within the set: the punctured plane, an annulus, and a torus Theorem 9.15 (Converse of the mixed partials theorem). Let F = (F1, F2, . . . , Fn) be a smooth vector field on an open set U in Rn. If • ∂Fi ∂xj = ∂Fj ∂xi for all i, j = 1, 2, . . . , n and • U is simply connected, then F is a conservative vector field on U. A rigorous proof of this theorem is quite hard, so we postpone even any discussion of it until after we look at an example. Example 9.16. Consider again the vector field W(x, y) = | Multivariable_Calculus_Shimamoto_Page_235_Chunk3449 |
224 CHAPTER 9. LINE INTEGRALS For example, let U be the right half-plane, U = {(x, y) ∈R2 : x > 0}. This is a simply connected set, so a potential function for W must exist on U. Indeed, one can check that w(x, y) = arctan( y x) works. For instance, ∂w ∂x = 1 1+ | Multivariable_Calculus_Shimamoto_Page_236_Chunk3450 |
9.6. THE CONVERSE OF THE MIXED PARTIALS THEOREM 225 Figure 9.20: Oriented closed curves with winding number 1 (left), 2 (middle), and −1 (right) One reason that this discussion is interesting is that, in first-year calculus, we tend to focus on the integral of a function as telling us mainly about the function. The winding number, on the other hand, is an instance where the integral is telling us about the geometry of the domain, i.e., the curve, over which the integral is taken. We close by going over some of the ideas behind the proof of Theorem 9.15, the converse of the mixed partials theorem. We do this only in the case of a vector field F = (F1, F2) on an open set U in R2. Thus we assume that ∂F2 ∂x = ∂F1 ∂y and that U is simply connected, and we want to prove that F is conservative on U, that is, that it has a potential function f. We think of f as an “anti-gradient” of F. Then the method of finding f is a lot like what is done in first-year calculus for the fundamental theorem, where an antiderivative is constructed by integrating. Choose a point a of U. If x is any point of U, then, as part of the definition of simply connected, there is a piecewise smooth curve C in U from a to x. The idea is to show that the line integral R C F · ds = R C F1 dx + F2 dy is the same regardless of which curve C is chosen. For suppose that C1 is another piecewise smooth curve in U from a to x. See Figure 9.21. We want to prove that: Z C F1 dx + F2 dy = Z C1 F1 dx + F2 dy. Figure 9.21: Two curves C and C1 from a to x and the region D that fills in the closed curve C2 = C ∪(−C1) By reversing the orientation on C1 and appending it to C, we obtain a closed curve C2 = C ∪(−C1) in U. Here’s the hard part. Since U is simply connected, a closed curve such as C2 can be shrunk within U to a point. The idea is that, in the process of the contraction, C2 sweeps out a filled-in region D such that C2 = ±∂D. It is difficult to turn this intuition into a rigorous proof. Assuming it has been done, we apply Green’s theorem together with the assumption of equal mixed partials: Z C2 F1 dx + F2 dy = ± ZZ D ∂F2 ∂x −∂F1 ∂y dx dy = ± ZZ D 0 dx dy = 0. (9.12) | Multivariable_Calculus_Shimamoto_Page_237_Chunk3451 |
226 CHAPTER 9. LINE INTEGRALS At the same time, since C2 = C ∪(−C1), we have: Z C2 F1 dx + F2 dy = Z C F1 dx + F2 dy − Z C1 F1 dx + F2 dy. (9.13) By combining equations (9.12) and (9.13), we find that R C F1 dx + F2 dy = R C1 F1 dx + F2 dy, as desired. As a result, if x ∈U, we may define f(x) = R C F1 dx + F2 dy, where C is any piecewise smooth curve in U from a to x. We write this more suggestively as: f(x) = Z x a F1 dx + F2 dy. (9.14) Note that, whether U is simply connected or not, as long as condition C1 above holds—namely, the line integral depends only on the endpoints of a curve—then the formula in equation (9.14) describes a well-defined function f. The reasoning above also contains the argument that, if con- dition C2 holds—that the integral around any closed curve is 0—that too is enough to show that the function f in (9.14) is well-defined. This basically follows from equation (9.13). Thus, to show that any of C1, C2, or, on simply connected domains, C3 suffice to imply that F is conservative, it remains only to confirm that f in (9.14) is a potential function for F: ∂f ∂x = F1 and ∂f ∂y = F2. This part is actually not so bad, and we leave it for the exercises (Exercise 6.4). 9.7 Exercises for Chapter 9 Section 1 Definitions and examples 1.1. Let F be the vector field on R2 given by F(x, y) = (x, y). (a) Sketch the vector field. (b) Based on your drawing, describe some paths over which you expect the integral of F to be positive. Find an example of such a path, and calculate the integral over it. (c) Describe some paths over which you expect the integral to be 0. Find an example of such a path, and calculate the integral over it. 1.2. Find R C F · ds if F(x, y) = (−y, x) and C is the circle x2 + y2 = a2 of radius a, traversed counterclockwise. 1.3. Find R C F·ds if F(x, y) = (cos2 x, sin y cos y) and C is the curve parametrized by α(t) = (t, t2), 0 ≤t ≤π. 1.4. Find R C F · ds if F(x, y, z) = (x + y, y + z, x + z) and C is the curve parametrized by α(t) = (t, t2, t3), 0 ≤t ≤1. 1.5. Find R C F·ds if F(x, y, z) = (−y, x, z) and C is the curve parametrized by α(t) = (cos t, sin t, t), 0 ≤t ≤2π. 1.6. Consider the line integral R C (x + y) dx + (y −x) dy. (a) What is the vector field that is being integrated? (b) Evaluate the integral if C is the line segment parametrized by α(t) = (t, t), 0 ≤t ≤1, from (0, 0) to (1, 1). | Multivariable_Calculus_Shimamoto_Page_238_Chunk3452 |
9.7. EXERCISES FOR CHAPTER 9 227 (c) Evaluate the integral if C consists of the line segment from (0, 0) to (1, 0) followed by the line segment from (1, 0) to (1, 1). (d) Evaluate the integral if C consists of the line segment from (0, 0) to (0, 1) followed by the line segment from (0, 1) to (1, 1). 1.7. Consider the line integral R C xy dx + yz dy + xz dz. (a) What is the vector field that is being integrated? (b) Evaluate the integral if C is the line segment from (1, 2, 3) to (4, 5, 6). (c) Evaluate the integral if C is the curve from (1, 2, 3) to (4, 5, 6) consisting of three consec- utive line segments, each parallel to one of the coordinate axes: from (1, 2, 3) to (4, 2, 3) to (4, 5, 3) to (4, 5, 6). 1.8. Find R C (x −y2) dx + 3xy dy if C is the portion of the parabola y = x2 from (−1, 1) to (2, 4). 1.9. Find R C y dx + z dy + x dz if C is the curve of intersection in R3 of the cylinder x2 + y2 = 1 and the plane z = x + y, oriented counterclockwise as viewed from high above the xy-plane, looking down. For the line integrals in Exercises 1.10–1.11, (a) sketch the curve C over which the integral is taken and (b) evaluate the integral. (Hint: Look for ways to reduce, or eliminate, messy calculation.) 1.10. R C 8z e(x+y)2 dx + (2x2 + y2)2 dy + 6xy2z dz, where C is parametrized by: α(t) = | Multivariable_Calculus_Shimamoto_Page_239_Chunk3453 |
228 CHAPTER 9. LINE INTEGRALS where Tα(t) is the unit tangent vector and vα(t) is the speed. We have embellished our original notation with the subscript α to emphasize our interest in the effect of the parametrization. We assume that vα(t) ̸= 0 for all t so that κα(t) is defined. Let β : J →R3 be another such parametrization of C. As in the text, we assume that α and β are one-to-one, except possibly at the endpoints of their respective domains. As in equation (9.3), there is a function g: J →I, assumed to be smooth, such that β(u) = α(g(u)) for all u in J. In other words, g(u) is the value of t such that β(u) = α(t). (a) Show that the velocities of α and β are related by vβ(u) = vα | Multivariable_Calculus_Shimamoto_Page_240_Chunk3454 |
9.7. EXERCISES FOR CHAPTER 9 229 3.9. Let p be a positive real number, and let F be the vector field on R3 −{(0, 0, 0)} given by: F(x, y, z) = − x (x2 + y2 + z2)p , − y (x2 + y2 + z2)p , − z (x2 + y2 + z2)p . For instance, the inverse square field is the case p = 3/2. (a) The norm of F is given by ∥F(x, y, z)∥= ∥(x, y, z)∥q for some exponent q. Find q in terms of p. (b) For which values of p is F a conservative vector field? The case p = 1 may require special attention. 3.10. Find R C ey dx + xey dy if C is the curve parametrized by α(t) = (et2, t3), 0 ≤t ≤1. 3.11. Let a and b be real numbers. If C is a piecewise smooth oriented closed curve in the punctured plane R2 −{(0, 0)}, show that: Z C ax x2 + y2 dx + by x2 + y2 dy = Z C (b −a)y x2 + y2 dy. 3.12. Let G(x, y, z) = | Multivariable_Calculus_Shimamoto_Page_241_Chunk3455 |
230 CHAPTER 9. LINE INTEGRALS 3.18. For a vector field F = (F1, F2, F3) on an open set in R3, is the cross product ∇× F, i.e., the curl, necessarily orthogonal to F? Either prove that it is, or find an example where it isn’t. 3.19. Newton’s second law of motion, F = ma, relates the force F acting on an object to the object’s mass m and acceleration a. Let F be a smooth force field on an open set of R3, and assume that, under the influence of F, an object of mass m travels so that its motion is described by a smooth path α: [a, b] →R3 with velocity v(t) and acceleration a(t). (a) The quantity K(t) = 1 2m∥v(t)∥2 is called the kinetic energy of the object. Use New- ton’s second law to show that: Z C F · ds = K(b) −K(a), where C is the oriented curve parametrized by α. In other words, the work done by the force equals the change in the object’s kinetic energy. (b) Assume, in addition, that F is a conservative field with potential function f. Show that the function E(t) = 1 2m∥v(t)∥2 −f(α(t)) is constant. (The function −f is called a potential energy for F. The fact that, for a conservative force field F, the sum of the kinetic and potential energies along an object’s path is constant is known as the principle of conservation of energy.) 3.20. Let F be a smooth vector field on an open set U in Rn. A smooth path α: I →U defined on an interval I is called an integral path of F if α′(t) = F(α(t)) for all t. In other words, at time t, when the position is the point α(t), the velocity equals the value of the vector field at that point. (See page 201.) (a) If α: [a, b] →U is a nonconstant integral path of F and C is the oriented curve parametrized by α, show that R C F · ds > 0. (Hint: The assumption that α is not constant means that there is some point t in [a, b] where α′(t) ̸= 0.) (b) Let f : U →R be a smooth real-valued function on U, and let F = ∇f. If there is a nonconstant integral path of F going from a point p to a point q, show that f(q) > f(p). (c) Show that a conservative vector field F cannot have a nonconstant closed integral path. 3.21. Let U be an open set in Rn with the property that every pair of points in U can be joined by a piecewise smooth curve in U. Let f : U →R be a smooth real-valued function defined on U. If ∇f(x) = 0 for all x in U, use line integrals to give a simple proof that f is a constant function. Section 4 Green’s theorem 4.1. Find R C (x2 −y2) dx + 2xy dy if C is the boundary of the square with vertices (0, 0), (1, 0), (1, 1), and (0, 1), oriented counterclockwise. 4.2. Find R C (2xy2 −y3) dx + (x2y + x3) dy if C is the unit circle x2 + y2 = 1, oriented counter- clockwise. 4.3. Find R C −y cos x dx + xy dy if C is the boundary of the parallelogram with vertices (0, 0), (2π, π), (3π, 3π), and (π, 2π), oriented counterclockwise. | Multivariable_Calculus_Shimamoto_Page_242_Chunk3456 |
9.7. EXERCISES FOR CHAPTER 9 231 Figure 9.22: An oval track 4.4. Find R C (ex+2y −3y) dx + (4x + 2ex+2y) dy if C is the track shown in Figure 9.22 consisting of two straightaways joined by semicircles at each end, oriented counterclockwise. 4.5. (a) Use the parametrization α(t) = (a cos t, a sin t), 0 ≤t ≤2π, and Corollary 9.14 to verify that the area of the disk x2 + y2 ≤a2 of radius a is πa2. (b) Let a and b be positive real numbers. By modifying the parametrization in part (a), use Corollary 9.14 to find the area of the region x2 a2 + y2 b2 ≤1 inside an ellipse. In Exercises 4.6–4.11, evaluate the line integral using whatever methods seem best. 4.6. R C (x+y) dx+(x−y) dy, where C is the curve in R2 consisting of the line segment from (0, 1) to (1, 0) followed by the line segment from (1, 0) to (2, 1) 4.7. R C ex x2+y2 dx− x x2+y2 dy, where C is the unit circle x2 +y2 = 1 in R2, oriented counterclockwise 4.8. R C (3x2y −xy2 −3x2y2) dx + (x3 −2x3y + x2y) dy, where C is the closed triangular curve in R2 with vertices (0, 0), (1, 1), and (0, 1), oriented counterclockwise 4.9. R C yz2 dx+x4z dy +x2y2 dz, where C is the curve parametrized by α(t) = (t, t2, t3), 0 ≤t ≤1 4.10. R C (6x2−4y+2xy) dx+(2x−2 sin y+3x2) dy, where C is the diamond-shaped curve |x|+|y| = 1 in R2, oriented counterclockwise 4.11. R C yz dx+xz dy +xy dz, where C is the curve of intersection in R3 of the cylinder x2 +y2 = 1 and the saddle surface z = x2 −y2, oriented counterclockwise as viewed from high above the xy-plane, looking down Let U be an open set in R2. A smooth real-valued function h: U →R is called harmonic on U if: ∂2h ∂x2 + ∂2h ∂y2 = 0 at all points of U. Exercises 4.12–4.15 concern harmonic functions. 4.12. Show that h(x, y) = e−x sin y is a harmonic function on R2. 4.13. Let D be a bounded subset of U whose boundary consists of a finite number of piecewise smooth simple closed curves. If h is harmonic on U, prove that: Z ∂D −h∂h ∂y dx + h∂h ∂x dy = ZZ D ∥∇h(x, y)∥2 dx dy. | Multivariable_Calculus_Shimamoto_Page_243_Chunk3457 |
232 CHAPTER 9. LINE INTEGRALS 4.14. Let D be a region as in the preceding exercise, and suppose that every pair of points in D can be joined by a piecewise smooth curve in D. If h is harmonic on U and if h = 0 at all points of ∂D, prove that h = 0 on all of D. (Hint: In addition to the preceding exercise, see Exercises 2.4 of Chapter 5 and 3.21 of this chapter.) 4.15. Continuing with the previous exercise, prove that a harmonic function is completely deter- mined on D by its values on the boundary in the sense that, if h1 and h2 are harmonic on U and if h1 = h2 at all points of ∂D, then h1 = h2 on all of D. Section 5 The vector field W 5.1. Let F = (F1, F2) be a smooth vector field that: • is defined everywhere in R2 except at three points p, q, r and • satisfies the mixed partials condition ∂F2 ∂x = ∂F1 ∂y everywhere in its domain. Let C1, C2, C3 be small circles centered at p, q, r, respectively, oriented counterclockwise as shown in Figure 9.23. Assume that: Z C1 F1 dx + F2 dy = 12, Z C2 F1 dx + F2 dy = 10, and Z C3 F1 dx + F2 dy = 15. Figure 9.23: Three circles C1, C2, and C3 and a fish-shaped curve C (a) Find R C F1 dx + F2 dy if C is the oriented fish-shaped curve shown in the figure. (b) Draw a piecewise smooth oriented closed curve K such that R K F1 dx + F2 dy = 1. Your curve need not be simple. Include in your drawing the direction in which K is traversed. Section 6 The converse of the mixed partials theorem 6.1. (a) Let C be the curve in R2 −{(0, 0)} parametrized by α(t) = (cos t, −sin t), 0 ≤t ≤2π. Describe C geometrically, and calculate its winding number using the definition. (b) More generally, let n be an integer (positive, negative, or zero allowed), and let Cn be the curve traced out by the path α(t) = (cos nt, sin nt), 0 ≤t ≤2π. Give a geometric description of Cn, and calculate its winding number using the definition. | Multivariable_Calculus_Shimamoto_Page_244_Chunk3458 |
9.7. EXERCISES FOR CHAPTER 9 233 6.2. Let C be a piecewise smooth simple closed curve in R2, oriented counterclockwise, such that the origin lies in the exterior of C. What are the possible values of the winding number of C? 6.3. Let F = (F1, F2) be a smooth vector field on the punctured plane U = R2 −{(0, 0)} such that ∂F2 ∂x = ∂F1 ∂y on U, and let C be a piecewise smooth oriented closed curve in U. (a) If C does not intersect the x-axis, explain why R C F · ds = 0. (b) Suppose instead that C may intersect the positive x-axis but not the negative x-axis. Is it necessarily still true that R C F · ds = 0? Explain. 6.4. (a) Let c = (c, d) be a point of R2, and let B = B(c, r) be an open ball centered at c. For the moment, we think of R2 as the uv-plane to avoid having x and y mean too many different things later on. Let F(u, v) = (F1(u, v), F2(u, v)) be a continuous vector field on B. Given any point x = (x, y) in B, let C1 be the curve from (c, d) to (x, y) consisting of a vertical line segment followed by a horizontal line segment, as indicated on the left of Figure 9.24. Define a function f1 : B →R by: f1(x, y) = Z C1 F1 du + F2 dv. Show that ∂f1 ∂x (x, y) = F1(x, y) for all (x, y) in B. (Hint: Parametrize C1 and use the fundamental theorem of calculus, keeping in mind what a partial derivative is.) Figure 9.24: The curves of integration C1 and C2 for functions f1 and f2, respectively (b) Similarly, let C2 be the curve from (c, d) to (x, y) consisting of a horizontal segment followed by a vertical segment, as shown on the right of Figure 9.24, and define: f2(x, y) = Z C2 F1 du + F2 dv. Show that ∂f2 ∂y (x, y) = F2(x, y) for all (x, y) in B. (c) Now, let U be an open set in R2 such that every pair of points in U can be joined by a piecewise smooth curve in U. Let a be a point of U. Assume that F = (F1, F2) is a smooth vector field on U such that the function f : U →R given in equation (9.14) is well-defined. That is: f(x) = Z x a F1 du + F2 dv, | Multivariable_Calculus_Shimamoto_Page_245_Chunk3459 |
234 CHAPTER 9. LINE INTEGRALS where the notation means R C F1 du + F2 dv for any piecewise smooth curve C in U from a to x. Show that f is a potential function for F on U, i.e., ∂f ∂x = F1 and ∂f ∂y = F2. In particular, F is a conservative vector field. (Hint: Given a point c in U, choose an open ball B = B(c, r) that is contained in U. Define functions f1 and f2 as in parts (a) and (b). For x in B, how is f(x) related to f(c) + f1(x) and f(c) + f2(x)?) | Multivariable_Calculus_Shimamoto_Page_246_Chunk3460 |
Chapter 10 Surface integrals We next study how to integrate vector fields over surfaces. One important caveat: our discussion applies only to surfaces in R3. For surfaces in Rn with n > 3, the expressions that one integrates are more complicated than vector fields. It is easy to get lost in the weeds with the details of surface integrals, so, in the first section of the chapter, we just try to build some intuition about what the integral measures and how to compute that measurement. The formal definition of the integral appears in the second section. Calculating surface integrals is not necessarily difficult but it can be messy, so having a range of options is especially welcome. In addition to the definition and the intuitive approach, sometimes a surface integral can be converted to a different type of integral altogether. For example, one of the theorems of the chapter, Stokes’s theorem, relates surface integrals to line integrals, and another, Gauss’s theorem, relates them to triple integrals. It would be a mistake to think of these results primarily as computational tools, however. Their theoretical consequences are at least as important, and we try to give a taste of some of the new lines of reasoning about integrals that they open up. The chapter closes with a couple of sections in which we tie up some loose ends. This also puts us in position to wrap things up in the next, and final, chapter. 10.1 What the surface integral measures If S is a surface in R3, we learned in Section 5.5 how to integrate a continuous real-valued function f : S →R over S. By definition: ZZ S f dS = ZZ D f(σ(s, t)) ∂σ ∂s × ∂σ ∂t ds dt, (10.1) where σ: D →R3 is a smooth parametrization of S defined on a subset D of the st-plane, as in Figure 10.1. This is called the integral with respect to surface area. As before, we assume now and in the future that D is the closure of a bounded open set in R2, that is, a bounded open set together with all its boundary points, such as a closed disk or rectangle (see page 134). As a matter of full disclosure, we did not show that the value of the integral is independent of the particular parametrization σ, a point that we address at the end of the chapter. Now, let F: U →R3 be a vector field on an open set U in R3 that contains S. The idea is that the integral of F over S measures the amount that F flows through S. This quantity is called the flux. For example, if F points in a direction normal to the surface, then it is flowing through effectively, whereas, if it points in a tangent direction, it is not flowing through at all. This stands 235 | Multivariable_Calculus_Shimamoto_Page_247_Chunk3461 |
236 CHAPTER 10. SURFACE INTEGRALS Figure 10.1: A parametrization σ of a surface S in R3 in contrast with the interpretation of the line integral, where we were interested in the degree to which the vector field flowed along a curve. To begin, we need to designate a direction of flow through S that is considered to be positive. Definition. An orientation of a smooth surface S in R3 is a continuous vector field n: S →R3 such that, for every x in S, n(x) is a unit normal vector to S at x. See Figure 10.2. An oriented surface is a surface S together with a specified orientation n. Figure 10.2: A surface S with orienting unit normal vector field n Example 10.1. Suppose that S is a level set defined by f(x, y, z) = c for some smooth real-valued function f of three variables. If x ∈S, then, by Proposition 4.17, ∇f(x) is a normal vector to S at x, so the formula n(x) = 1 ∥∇f(x)∥∇f(x) describes an orientation of S, provided that ∇f(x) ̸= 0 for all x in S. For instance, let S be the sphere x2 + y2 + z2 = a2 of radius a. Here, f(x, y, z) = x2 + y2 + z2, so ∇f = (2x, 2y, 2z) = 2(x, y, z). As a result, an orientation of the sphere is given by: n(x, y, z) = 1 2 p x2 + y2 + z2 · 2(x, y, z) = 1 √ a2 (x, y, z) = 1 a(x, y, z) for all points (x, y, z) on the sphere. This vector points away from the origin, so we say that it orients S with the outward normal. See Figure 10.3, left. We could equally well orient the sphere with the inward unit normal. This would be given by n(x, y, z) = −1 a(x, y, z). Or let S be the circular cylinder x2 + y2 = a2. This is a level set of f(x, y, z) = x2 + y2. Then ∇f = (2x, 2y, 0) = 2(x, y, 0), and an orientation is: n(x, y, z) = 1 2 p x2 + y2 · 2(x, y, 0) = 1 a(x, y, 0) for all (x, y, z) on the cylinder. Once again, this is the outward normal (Figure 10.3, right). | Multivariable_Calculus_Shimamoto_Page_248_Chunk3462 |
10.1. WHAT THE SURFACE INTEGRAL MEASURES 237 Figure 10.3: A sphere and cylinder oriented by the outward normal Figure 10.4: A mammal with orientation9 In general, let S be an oriented surface with orienting normal n. To measure the flow of a vector field F = (F1, F2, F3) through S, we find the scalar component Fnorm of F in the normal direction at every point of S and integrate it over S. This component is given by Fnorm = ∥F∥cos θ, where θ is the angle between F and n, as shown in Figure 10.5. Figure 10.5: The component of a vector field in the normal direction at a point x Hence: Fnorm = ∥F∥cos θ = ∥F∥∥n∥cos θ = F · n. The integral of F over S, denoted by RR S F · dS, can be expressed as: ZZ S F · dS = ZZ S Fnorm dS = ZZ S F · n dS. (10.2) 9Photo credit: https://pixabay.com/photos/nature-animal-porcupine-mammal-3588682/ by analogicus under Pixabay license. | Multivariable_Calculus_Shimamoto_Page_249_Chunk3463 |
238 CHAPTER 10. SURFACE INTEGRALS As the integral of the real-valued function F · n, the expression on the right is an integral with respect to surface area of the type considered previously in (10.1). We take equation (10.2) as a tentative definition of the integral of F and use it to work through some examples. Example 10.2. Let S be the sphere x2 +y2 +z2 = a2 of radius a, oriented by the outward normal. Find RR S F · dS if: (a) F(x, y, z) = − 1 (x2+y2+z2)3/2 (x, y, z) (i.e., the inverse square field), (b) F(x, y, z) = (0, 0, z), (c) F(x, y, z) = (0, 0, −1). As shown in Example 10.1, the outward unit normal to S is given by n = 1 a(x, y, z). (a) The inverse square field, illustrated in Figure 10.6, points directly inwards towards the origin, Figure 10.6: The inverse square field so F · n is negative at every point of S. In other words, if the positive direction is taken to be outward, then F is flowing in the negative direction. Thus we expect the integral to be negative. To calculate it precisely, note that on S, F(x, y, z) = − 1 (a2)3/2 (x, y, z) = −1 a3 (x, y, z), so: F · n = −1 a3 (x, y, z) · 1 a(x, y, z) = −1 a4 | Multivariable_Calculus_Shimamoto_Page_250_Chunk3464 |
10.1. WHAT THE SURFACE INTEGRAL MEASURES 239 Figure 10.7: The vector field F(x, y, z) = (0, 0, z) flowing through a sphere This last integral is precisely one of the examples we calculated using a parametrization when we studied integrals with respect to surface area (Example 5.19 again). The answer is RR S z2 dS = 4 3πa4. (We also gave a symmetry argument that made use of the facts that RR S x2 dS = RR S y2 dS = RR S z2 dS and RR S(x2 + y2 + z2) dS = RR S a2 dS = a2 · (4πa2) = 4πa4.) Thus: ZZ S F · dS = 1 a · 4 3πa4 = 4 3πa3. We shall take another look at this example later as well. (c) Here, F(x, y, z) = (0, 0, −1) is a constant vector field consisting of unit vectors that point straight down everywhere (Figure 10.8). Roughly speaking, F points inward on the upper hemisphere, i.e., the component in the outward direction is negative, and outward on the lower hemisphere. From the symmetry of the situation, we might expect these two contributions to cancel out exactly, giving an integral of 0. Informally, the flow in equals the flow out, so the net flow is 0. Figure 10.8: The vector field F(x, y, z) = (0, 0, −1) flowing through a sphere To verify this, note that F · n = (0, 0, −1) · 1 a(x, y, z) = −1 az. Thus: ZZ S F · dS = ZZ S −1 az dS = −1 a ZZ S z dS. By symmetry, this last integral is indeed 0. More precisely, S is symmetric in the xy-plane, i.e., (x, y, −z) is in S whenever (x, y, z) is, and the integrand f(x, y, z) = z satisfies f(x, y, −z) = −z = −f(x, y, z), so the contributions of the northern and southern hemispheres cancel. Thus RR S F · dS = 0. | Multivariable_Calculus_Shimamoto_Page_251_Chunk3465 |
240 CHAPTER 10. SURFACE INTEGRALS In this last example, if we had integrated instead only over the northern hemisphere, the same line of thought would lead us to expect that the integral is negative. As a test of your intuition about what the integral represents, you might think about whether this is related to the following example. Example 10.3. Let S be the disk of radius a in the xy-plane described by x2 + y2 ≤a2, z = 0, oriented by the upward normal, and let F(x, y, z) = (0, 0, −1) as in Example 10.2(c) above. See Figure 10.9. Find RR S F · dS. Figure 10.9: The vector field F(x, y, z) = (0, 0, −1) flowing through a disk in the xy-plane The upward unit normal to S is n = k = (0, 0, 1), so F·n = (0, 0, −1)·(0, 0, 1) = −1. Therefore: ZZ S F · dS = ZZ S −1 dS = −Area (S) = −πa2. Continuing with the comments right before the example, the result turned out be negative, but is the flow of F = (0, 0, −1) through the disk really related to the flow through the northern hemisphere? We shall be able to give a precise answer based on general principles once we learn a little more about surface integrals (see Exercises 4.4 and 5.5 at the end of the chapter). 10.2 The definition of the surface integral We now present the actual definition of the vector field surface integral RR S F · dS. This follows a strategy we have used before: to define this new type of integral, we parametrize the surface, substitute for x, y, and z in terms of the two parameters, and then pull back to a double integral in the parameter plane. Of course, we need to figure out what the pullback should be. Let S be an oriented surface in R3 with orienting normal n, and let σ: D →R3 be a smooth parametrization of S with domain D in R2: σ(s, t) = (x(s, t), y(s, t), z(s, t)), (s, t) ∈D. We first describe how σ can be used to give an independent orientation of S. Recall from Section 5.5 that the vectors ∂σ ∂s = | Multivariable_Calculus_Shimamoto_Page_252_Chunk3466 |
10.2. THE DEFINITION OF THE SURFACE INTEGRAL 241 is a unit normal, i.e., an orientation of S, provided that ∂σ ∂s × ∂σ ∂t ̸= 0. Since the orientation n that is given for S is also a unit normal, it must be true that: n = ± ∂σ ∂s × ∂σ ∂t ∥∂σ ∂s × ∂σ ∂t ∥. (10.3) We say that σ preserves orientation if the sign is + and that it reverses orientation if −. Now, to formulate the definition of the surface integral, recall that the idea was that RR S F · dS should equal RR S F·n dS. Using (10.3) to substitute for n, this becomes ± RR S F· ∂σ ∂s × ∂σ ∂t ∥∂σ ∂s × ∂σ ∂t ∥dS. Then the definition of the integral with respect to surface area (10.1) with f = F · ∂σ ∂s × ∂σ ∂t ∥∂σ ∂s × ∂σ ∂t ∥gives: ZZ S F · n dS = ± ZZ D F(σ(s, t)) · ∂σ ∂s × ∂σ ∂t ∥∂σ ∂s × ∂σ ∂t ∥ ∂σ ∂s × ∂σ ∂t ds dt = ± ZZ D F(σ(s, t)) · | Multivariable_Calculus_Shimamoto_Page_253_Chunk3467 |
242 CHAPTER 10. SURFACE INTEGRALS Figure 10.10: A parametrization of the sphere of radius a Substituting in terms of the parameters gives F(σ(ϕ, θ)) = (0, 0, z(ϕ, θ)) = (0, 0, a cos ϕ). In addition: ∂σ ∂ϕ × ∂σ ∂θ = det i j k a cos ϕ cos θ a cos ϕ sin θ −a sin ϕ −a sin ϕ sin θ a sin ϕ cos θ 0 = ... (we calculated this before in Example 5.19) = a2 sin ϕ | Multivariable_Calculus_Shimamoto_Page_254_Chunk3468 |
10.2. THE DEFINITION OF THE SURFACE INTEGRAL 243 To use the definition of the integral (10.4), we integrate the function: F(σ(ϕ, θ)) · ∂σ ∂ϕ × ∂σ ∂θ = (0, 0, a cos ϕ) · | Multivariable_Calculus_Shimamoto_Page_255_Chunk3469 |
244 CHAPTER 10. SURFACE INTEGRALS Figure 10.12: Rain (left), the cone z = p x2 + y2, x2 + y2 ≤4 (middle), and the cap of the cone, z = 2, x2 + y2 ≤4 (right) The z-component is positive, so ∂σ ∂x × ∂σ ∂y points upward. Thus σ is orientation-preserving. In addition: F(σ(x, y)) · | Multivariable_Calculus_Shimamoto_Page_256_Chunk3470 |
10.3. STOKES’S THEOREM 245 10.3 Stokes’s theorem There are two types of points on a surface. At some points, one can approach along the surface from any direction and remain on the surface, at least for a little while, after continuing on through the point. These are sometimes called interior points. Other points lie on the edge of the surface in the sense that, after approaching the point from certain directions, one falls off the surface immediately after continuing through. These are called boundary points. The set of all boundary points is called the boundary of the surface. For instance, a hemisphere has a boundary consisting of the equatorial circle, and a cylinder has a boundary consisting of two pieces, namely, the circles at each end. On the other hand, a sphere and a torus have no boundary points. See Figure 10.13. In general, the boundary of a surface is either empty or consists of one or more simple closed curves. Figure 10.13: The boundary of a hemisphere is a circle and that of a cylinder is two circles. The boundaries of a sphere and torus are empty. Stokes’s theorem relates a line integral R B F · ds around the boundary B of a surface in R3 to a vector field integral over the surface itself. It can be seen as an extension of Green’s theorem from double integrals to surface integrals. The theorem is not obvious, and we shall try to show where it comes from, though in a very special case similar to the one we used to explain Green’s theorem. Namely, we assume that S is an oriented surface with a smooth orientation-preserving parametri- zation σ(s, t) = | Multivariable_Calculus_Shimamoto_Page_257_Chunk3471 |
246 CHAPTER 10. SURFACE INTEGRALS right and the vertical sides A2 and A4 are oriented upward. Taking orientation into account, we have A = A1 ∪A2 ∪(−A3) ∪(−A4). Applying σ then breaks B up into four curves B1, B2, B3, B4, where Bj = σ(Aj) for j = 1, 2, 3, 4. B inherits the orientation B = B1 ∪B2 ∪(−B3) ∪(−B4). Let F = (F1, F2, F3) be a smooth vector field defined on an open set in R3 that contains S. We consider the line integral of F around B, the boundary of S: Z B F · ds = Z B1 + Z B2 − Z B3 − Z B4 F · ds. We use differential form notation: R B F · ds = R B F1 dx + F2 dy + F3 dz. To cut down on the amount of notation in any one place, we find R B F1 dx, R B F2 dy, R B F3 dz separately and add the results at the end. We can parametrize each of the four pieces of B by applying σ to the corresponding side of A, fixing one of s or t and using the other as parameter. For B1: α1(s) = σ(s, c), a ≤s ≤b. For B2: α2(t) = σ(b, t), c ≤t ≤d. For B3: α3(s) = σ(s, d), a ≤s ≤b. For B4: α4(t) = σ(a, t), c ≤t ≤d. So for R B F1 dx, we integrate over each of B1 through B4 and take the sum with appropri- ate signs. For example, for B1, the parametrization is α1(s) = σ(s, c) = (x(s, c), y(s, c), z(s, c)), where t = c is held fixed. Hence the derivatives with respect to the parameter are deriva- tives with respect to s, partial derivatives where appropriate. For instance, α′ 1(s) = ∂σ ∂s (s, c) = (∂x ∂s(s, c), ∂y ∂s(s, c), ∂z ∂s(s, c)). As a result, we obtain: Z B1 F1 dx = Z b a F1(σ(s, c)) ∂x ∂s (s, c) ds. Similarly, over B2 with parameter t: Z B2 F1 dx = Z d c F1(σ(b, t)) ∂x ∂t (b, t) dt. Continuing in this way with B3 and B4 and combining the results gives: Z B F1 dx = Z b a F1(σ(s, c)) ∂x ∂s (s, c) ds + Z d c F1(σ(b, t)) ∂x ∂t (b, t) dt − Z b a F1(σ(s, d)) ∂x ∂s (s, d) ds − Z d c F1(σ(a, t)) ∂x ∂t (a, t) dt. (10.6) Upon closer examination, each of the terms in this last expression is a line integral over one of the sides of the boundary of R back in the st-plane. For instance, the first term is R A1(F1 ◦σ) ∂x ∂s ds and the second is R A2(F1 ◦σ) ∂x ∂t dt, line integrals over A1 and A2, respectively. In fact, since t is constant on A1 and s is constant on A2, we can give these terms a more uniform appearance by writing them as: Z A1 (F1 ◦σ) ∂x ∂s ds + (F1 ◦σ) ∂x ∂t dt and Z A2 (F1 ◦σ) ∂x ∂s ds + (F1 ◦σ) ∂x ∂t dt, respectively. The extra summands are superfluous to the integrals, since the derivatives of the extra coordinates are zero on the corresponding segment. After converting the remaining two terms to | Multivariable_Calculus_Shimamoto_Page_258_Chunk3472 |
10.3. STOKES’S THEOREM 247 line integrals over A3 and A4, equation (10.6) becomes a line integral over all of A: Z B F1 dx = Z A (F1 ◦σ) ∂x ∂s ds + (F1 ◦σ) ∂x ∂t dt. But now that we are back to a line integral in the plane, Green’s theorem applies, and, as A = ∂R, we can replace the line integral over A with a double integral over the rectangle R. Thus: Z B F1 dx = ZZ R ∂ ∂s | Multivariable_Calculus_Shimamoto_Page_259_Chunk3473 |
248 CHAPTER 10. SURFACE INTEGRALS Adding the three calculations and suitably regrouping the terms results in the formula: Z B F1 dx + F2 dy + F3 dz = ZZ R | Multivariable_Calculus_Shimamoto_Page_260_Chunk3474 |
10.3. STOKES’S THEOREM 249 Figure 10.16: Three oriented surfaces and their oriented boundaries For example, let S be a sphere that is oriented by the outward normal. Then the boundary of the northern hemisphere is the equatorial circle oriented counterclockwise, while the boundary of the southern hemisphere is the same circle but oriented clockwise. If S is a pair of pants oriented by the outward normal, then ∂S consists of three closed curves: the waist oriented one way—let’s call it clockwise—and the bottoms of the two legs oriented counterclockwise. See Figure 10.16. For a piecewise smooth surface S = S1 ∪S2 ∪· · · ∪Sk, we require that the smooth pieces S1, S2, . . . , Sk are oriented compatibly in the sense that the oriented boundaries of adjacent pieces are traversed in opposite directions wherever they intersect. These intersections are not considered to be part of the boundary of the whole surface S. This is illustrated in Figure 10.17. Figure 10.17: Compatible orienting normals n1, n2 for adjacent surfaces S1, S2, respectively, with corresponding orientations of ∂S1, ∂S2 For the extension of (10.10) beyond surfaces whose parameter domains are rectangles, one approach might be to mimic the strategy we outlined informally for Green’s theorem. Namely, first verify it in the case where the parameter domain is a union of rectangles that overlap at most along parts of their boundaries and then argue that the domain of any parametrization can be approximated arbitrarily well by such a union. We won’t give the details. With these preliminaries out of the way, we can state the theorem. Theorem 10.6 (Stokes’s theorem). Let S be a piecewise smooth oriented surface in R3 that is bounded as a subset of R3, and let F be a smooth vector field defined on an open set containing S. Then: Z ∂S F · ds = ZZ S (∇× F) · dS. In the form stated, Stokes’s theorem converts a line integral to a surface integral. Since surface integrals are usually messier than line integrals, this is not the usual way in which the theorem is applied, at least as far as calculating specific examples goes. (See Exercises 3.3–3.6 for some exceptions, however.) Sometimes though, the theorem can be used to go the other way, converting a surface integral to a line integral. This possibility has implications for the general properties of surface integrals. | Multivariable_Calculus_Shimamoto_Page_261_Chunk3475 |
250 CHAPTER 10. SURFACE INTEGRALS Example 10.7. Let F be the “falling rain” vector field F(x, y, z) = (0, 1, −3), and let S be the cone z = p x2 + y2, x2 + y2 ≤4, oriented by the upward normal. In Example 10.5, we integrated F over S using a parametrization. Since we evaluated the integral before, there seems no harm in giving away a spoiler: the answer is RR S F · dS = −12π. Could this surface integral have been computed using Stokes’s theorem? The answer is yes provided that there is a vector field G such that F = ∇× G, for then Stokes’s theorem gives RR S F · dS = RR S(∇× G) · dS = R ∂S G · ds. In other words, we could find the surface integral of F over S by computing the line integral of G around the boundary. We try to find such a G = (G1, G2, G3) by setting F = ∇× G and guessing: F = (0, 1, −3) = det i j k ∂ ∂x ∂ ∂y ∂ ∂z G1 G2 G3 . This leads to the system: 0 = ∂G3 ∂y −∂G2 ∂z 1 = ∂G1 ∂z −∂G3 ∂x −3 = ∂G2 ∂x −∂G1 ∂y . From the second equation, we guess G1 = z, from the third G2 = −3x, and from the first G3 = 0. One can check that these guesses are actually consistent with all three equations. Thus G(x, y, z) = (z, −3x, 0) works! Our virtuous lifestyle has paid off. This gives: ZZ S F · dS = Z ∂S G · ds = Z ∂S z dx −3x dy + 0 dz = Z ∂S z dx −3x dy. We calculate the line integral by parametrizing C = ∂S, which is a circle of radius 2 in the plane z = 2, oriented counterclockwise (see Figure 10.12, middle, or, if you’re willing to peek ahead, Figure 10.18, left). The circle can be parametrized by α(t) = (2 cos t, 2 sin t, 2), 0 ≤t ≤2π. Thus: ZZ S F · dS = Z C z dx −3x dy = Z 2π 0 | Multivariable_Calculus_Shimamoto_Page_262_Chunk3476 |
10.4. CURL FIELDS 251 Figure 10.18: An oriented cone S and two other oriented surfaces eS, all three having the same oriented boundary could be the disk in the plane z = 2 that caps off the cone, oriented by the upward normal, which was also considered in Example 10.5. See Figure 10.18. If we integrate the falling rain vector field F over the new surface eS, we obtain: ZZ eS F · dS = ZZ eS (∇× G) · dS (where G = (z, −3x, 0) as before) = Z ∂eS G · ds (by Stokes’s theorem) = Z C G · ds = −12π (by the calculation in the previous example). In other words, the integral of F is −12π over all oriented surfaces whose boundary is C. This idea can be formulated more generally. Definition. Let U be an open set in R3. A vector field F: U →R3 is called a curl field on U if there exists a smooth vector field G: U →R3 such that F = ∇× G. For instance, we saw in Example 10.7 that F = (0, 1, −3) is a curl field on R3 with “anti-curl” G = (z, −3x, 0). One perspective on curl fields is the following general working principle: curl fields are to surface integrals as conservative fields are to line integrals. We consider a couple of illustrations of this. Example 10.8. For line integrals: The integral of a conservative field over an oriented curve depends only on the endpoints of the curve. For surface integrals: The integral of a curl field over an oriented surface depends only on the oriented boundary of the surface. More precisely: Theorem 10.9. Let S1 and S2 be piecewise smooth oriented surfaces in an open set U in R3 that have the same oriented boundary, i.e., ∂S1 = ∂S2. If F is a curl field on U, then: ZZ S1 F · dS = ZZ S2 F · dS. Proof. Suppose that F = ∇× G. Then by Stokes’s theorem, both sides of the equation are equal to R C G · ds, where C = ∂S1 = ∂S2. | Multivariable_Calculus_Shimamoto_Page_263_Chunk3477 |
252 CHAPTER 10. SURFACE INTEGRALS Example 10.10. For line integrals: The integral of a conservative field over a piecewise smooth oriented closed curve is 0. For surface integrals: Definition. Let S be a piecewise smooth surface in Rn such that every pair of points in S can be joined by a piecewise smooth curve in S. Then S is called closed if ∂S = ∅(the empty set). For example, spheres and tori are closed surfaces, but hemispheres and cylinders are not. Theorem 10.11. If F is a curl field on an open set U in R3, then RR S F · dS = 0 for any piecewise smooth oriented closed surface S in U. Proof. Let S be a piecewise smooth oriented closed surface in U. The simplest argument would be to say that, if F = ∇× G, then, by Stokes’s theorem, RR S F · dS = R ∂S G · ds = 0 since ∂S = ∅. If integrating over the empty set makes you nervous, then perhaps a more aboveboard alternative would be to cut S into two nonempty pieces S1 and S2 such that S = S1∪S2 and S1 and S2 intersect only along their common boundary. For instance, S1 could be a small disklike piece of S, and S2 could be what’s left. See Figure 10.19. Then: ZZ S F · dS = ZZ S1 F · dS + ZZ S2 F · dS = Z ∂S1 G · ds + Z ∂S2 G · ds. (10.12) Figure 10.19: Dividing S into two pieces, S = S1 ∪S2 S1 and S2 inherit the orienting unit normal from S, so, in order to keep them on the left, their boundaries must be traversed in opposite directions. This is consistent with our requirement for orientations of piecewise smooth surfaces. In other words, ∂S1 and ∂S2 are the same curve but with opposite orientations. Hence the line integrals in equation (10.12) cancel each other out. Lastly, we look for an analogue for curl fields of the mixed partials theorem for conservative fields. In R3, the mixed partials theorem takes the form that, if F is conservative, then ∇× F = 0. Now, let F be a curl field, say F = ∇× G, so (F1, F2, F3) = det i j k ∂ ∂x ∂ ∂y ∂ ∂z G1 G2 G3 . Hence: F1 = ∂G3 ∂y −∂G2 ∂z F2 = ∂G1 ∂z −∂G3 ∂x F3 = ∂G2 ∂x −∂G1 ∂y . | Multivariable_Calculus_Shimamoto_Page_264_Chunk3478 |
10.5. GAUSS’S THEOREM 253 Taking certain partial derivatives of F1, F2, F3 introduces the second-order mixed partials of G1, G2, G3: ∂F1 ∂x = ∂2G3 ∂x ∂y −∂2G2 ∂x ∂z ∂F2 ∂y = ∂2G1 ∂y ∂z −∂2G3 ∂y ∂x ∂F3 ∂z = ∂2G2 ∂z ∂x −∂2G1 ∂z ∂y. The terms on the right side are mixed partial pairs that appear with opposite signs. Thus, by the equality of mixed partials, taking the sum gives: ∂F1 ∂x + ∂F2 ∂y + ∂F3 ∂z = 0. (10.13) The sum can also be written in terms of the ∇operator as ∂F1 ∂x + ∂F2 ∂y + ∂F3 ∂z = ( ∂ ∂x, ∂ ∂y, ∂ ∂z) · (F1, F2, F3) = ∇· F, where the product is dot product. This quantity has a name. Definition. If F = (F1, F2, F3) is a smooth vector field on an open set in R3, then: ∇· F = ∂F1 ∂x + ∂F2 ∂y + ∂F3 ∂z is called the divergence of F. Note that, while the curl of a vector field is again a vector field, the divergence is a real-valued function. From (10.13), we have proven the following. Theorem 10.12. Let U be an open set in R3, and let F be a smooth vector field on U. If F is a curl field on U, then: ∇· F = 0 at all points of U. Equivalently, given a smooth vector field F on U, if ∇· F ̸= 0 at some point, then F is not a curl field. Example 10.13. The vector field F(x, y, z) = (2xy + z2, 2yz + x2, 2xz + y2) is conservative on R3. It has f(x, y, z) = x2y + y2z + z2x as a potential function. Is F also a curl field on R3? We could write down the conditions for being a curl field and try to guess a vector field G such that F = ∇× G the way we did before. It is easy, however, to compute the divergence: ∇· F = ∂ ∂x | Multivariable_Calculus_Shimamoto_Page_265_Chunk3479 |
254 CHAPTER 10. SURFACE INTEGRALS Figure 10.20: The rectangular box W = [a, b]×[c, d]×[e, f] in R3 and the orientation of its boundary surface S We integrate F over S. To do so, we consider each face separately and add the results: ZZ S F · dS = ZZ top + ZZ bottom + ZZ front + ZZ back + ZZ left + ZZ right F · dS. (10.14) For instance, the top may be parametrized using x and y as parameters with z = f fixed: σ(x, y) = (x, y, f), a ≤x ≤b, c ≤y ≤d. Then: ∂σ ∂x × ∂σ ∂y = (1, 0, 0) × (0, 1, 0) = i × j = k = (0, 0, 1). This points upward, hence away from the box W, so σ is orientation-preserving. Therefore: ZZ top F · dS = ZZ [a,b]×[c,d] F(σ(x, y)) · ∂σ ∂x × ∂σ ∂y dx dy = ZZ [a,b]×[c,d] (F1, F2, F3) · (0, 0, 1) dx dy = ZZ [a,b]×[c,d] F3(σ(x, y)) dx dy = ZZ [a,b]×[c,d] F3(x, y, f) dx dy. For the face on the bottom, the similar parametrization σ(x, y) = (x, y, e) works, only now σ is orientation-reversing since the orientation of S on the bottom points downward in order to point away from W. This gives: ZZ bottom F · dS = − ZZ [a,b]×[c,d] F3(x, y, e) dx dy. | Multivariable_Calculus_Shimamoto_Page_266_Chunk3480 |
10.5. GAUSS’S THEOREM 255 We add these results and apply the fundamental theorem of calculus: ZZ top F · dS + ZZ bottom F · dS = ZZ [a,b]×[c,d] | Multivariable_Calculus_Shimamoto_Page_267_Chunk3481 |
256 CHAPTER 10. SURFACE INTEGRALS Figure 10.21: A solid region W in R3 and its oriented boundary ∂W In the case that F is a curl field, the left side of Gauss’s theorem is 0. This is because ∂W consists of closed surfaces and the integral of a curl field over a closed surface is 0 (Theorem 10.11). The right side is 0, too, because curl fields have divergence 0 (Theorem 10.12). Thus, for curl fields, Gauss’s theorem reflects properties already known to us. To illustrate Gauss’s theorem, we begin by taking a new look at two surface integrals we evaluated earlier. Example 10.15. Let Sa be the sphere x2 + y2 + z2 = a2 of radius a, oriented by the outward normal. Find RR Sa F · dS if: (a) F(x, y, z) = (0, 0, −1), (b) F(x, y, z) = (0, 0, z). Figure 10.22: The solid ball Wa of radius a and its oriented boundary, the sphere Sa (a) This was Example 10.2(c). To answer it using Gauss’s theorem, let Wa be the solid ball x2 + y2 + z2 ≤a2 so that Sa = ∂Wa, as in Figure 10.22. Since ∇· F = ∂ ∂x(0) + ∂ ∂y(0) + ∂ ∂z(−1) = 0, Gauss’s theorem says that: ZZ Sa F · dS = ZZZ Wa ∇· F dx dy dz = ZZZ Wa 0 dx dy dz = 0. (b) This is the third time we have evaluated this integral (see Examples 10.2(b) and 10.4). Here, ∇· F = 0 + 0 + 1 = 1, so with Wa as above: ZZ Sa F · dS = ZZZ Wa 1 dx dy dz = Vol (Wa) = 4 3πa3, | Multivariable_Calculus_Shimamoto_Page_268_Chunk3482 |
10.6. THE INVERSE SQUARE FIELD 257 where we have used the formula for the volume of a three-dimensional ball from Example 7.8 in Chapter 7. It seems odd to say that turning a problem into a triple integral makes it simpler, but, with the help of Gauss’s theorem, finding the preceding two surface integrals became essentially trivial. Example 10.16. Let F(x, y, z) = (x, y, z). This vector field radiates directly away from the origin at every point, and the arrows get longer the further out you go. Let W be a solid region in R3 as in Gauss’s theorem. Then ∇· F = 1 + 1 + 1 = 3, so RR ∂W F · dS = RRR W 3 dx dy dz = 3 Vol (W), that is: Vol (W) = 1 3 ZZ ∂W (x, y, z) · dS. This is another instance where information about a set can be gleaned from a calculation just involving its boundary. (Question: Thinking of the surface integral as the integral of the normal component and keeping in mind that ∂W is oriented by the outward normal, does it make sense that RR ∂W (x, y, z) · dS is always positive? Note that the origin need not lie in W, so there may be places where the orienting normal n on the boundary points towards the origin and the normal component F · n = (x, y, z) · n is negative.) 10.6 The inverse square field Throughout this section, G denotes the inverse square field G: R3 −{(0, 0, 0)} →R3, given by: G(x, y, z) = − x (x2 + y2 + z2)3/2 , − y (x2 + y2 + z2)3/2 , − z (x2 + y2 + z2)3/2 . (10.16) We study G using Gauss’s theorem, starting with a couple of preliminary facts. Fact 1. Let Sa be the sphere x2 + y2 + z2 = a2, oriented by the outward normal. Then: ZZ Sa G · dS = −4π, independent of the radius a. Justification. We calculated and even emphasized this earlier in Example 10.2(a), which was the first surface integral of a vector field that we found. Note that this implies that G is not a curl field, since a curl field would integrate to 0 over the closed surface Sa. Fact 2. ∇· G = 0. Justification. This is a brute force calculation. For instance: ∂G1 ∂x = ∂ ∂x − x (x2 + y2 + z2)3/2 = −(x2 + y2 + z2)3/2 · 1 −x · 3 2(x2 + y2 + z2)1/2 · 2x (x2 + y2 + z2)3 = −(x2 + y2 + z2) −3x2 (x2 + y2 + z2)5/2 = 2x2 −y2 −z2 (x2 + y2 + z2)5/2 . By symmetric calculations: ∂G2 ∂y = 2y2 −x2 −z2 (x2 + y2 + z2)5/2 and ∂G3 ∂z = 2z2 −x2 −y2 (x2 + y2 + z2)5/2 . | Multivariable_Calculus_Shimamoto_Page_269_Chunk3483 |
258 CHAPTER 10. SURFACE INTEGRALS Taking the sum gives ∇· G = (2x2−y2−z2)+(2y2−x2−z2)+(2z2−x2−y2) (x2+y2+z2)5/2 = 0, as claimed. This fact has an interesting implication, too, for we know that curl fields have divergence 0. This shows that the converse need not hold, since ∇· G = 0 yet G is not a curl field by Fact 1. Now, let S be a piecewise smooth closed surface in R3 −{(0, 0, 0)}, oriented by the outward normal. Is that enough for us to be able to say something about RR S G · dS? Strikingly, the answer is yes. There are two cases. Case 1. Suppose that the origin 0 lies in the exterior of S (Figure 10.23). Figure 10.23: The origin lies exterior to S. Let W be the three-dimensional solid inside of S so that S = ∂W. By assumption, 0 /∈W, so G is defined throughout W and we may apply Gauss’s theorem. This gives: ZZ S G · dS = ZZZ W ∇· G dx dy dz = ZZZ W 0 dx dy dz (by Fact 2) = 0. Case 2. Suppose that 0 lies in the interior of S. The preceding argument is no longer valid, since filling in the interior of S would include the origin, which is excluded from the domain of G. Instead, we choose a very small sphere Sϵ of radius ϵ centered at the origin, oriented by the outward normal, and we let W be the region inside S but outside Sϵ. See Figure 10.24. Figure 10.24: The origin lies in the interior of S. Then Gauss’s theorem applies to W, giving: ZZ ∂W G · dS = ZZZ W ∇· G dx dy dz = 0, (10.17) | Multivariable_Calculus_Shimamoto_Page_270_Chunk3484 |
10.7. A MORE SUBSTITUTION-FRIENDLY NOTATION FOR SURFACE INTEGRALS 259 as before. This time, however, the boundary of W consists of S and Sϵ, oriented by the normal pointing away from W. This is the outward normal on S but points inward on Sϵ. In other words, ∂W = S ∪(−Sϵ), and therefore RR ∂W G · dS = RR S G · dS − RR Sϵ G · dS. By equation (10.17), the difference is 0, from which it follows that RR S G · dS = RR Sϵ G · dS. And, by Fact 1, the integral of G over any sphere centered at the origin is −4π. Hence: ZZ S G · dS = −4π. After a slight rearrangement, we can summarize the discussion as follows. For a piecewise smooth closed surface S that doesn’t pass through the origin and is oriented by the outward normal: −1 4π ZZ S G · dS = ( 0 if 0 lies in the exterior of S, 1 if 0 lies in the interior of S. As we noted back in Chapter 8, the inverse square field describes the gravitational, or electro- static, force due to an object, or charged particle, located at the origin. If several objects of equal mass, or equal charge, are placed at the points p1, p2, . . . , pk in R3, then the total force at a point x due to these objects is described by the vector field: F(x) = k X i=1 G(x −pi), (10.18) where G is the same as above (10.16). It is not hard to verify that again ∇· F = 0. A modification of the argument just given shows that, if S is a piecewise smooth closed surface contained in the set R3 −{p1, p2, . . . , pk} and oriented by the outward normal, then: −1 4π ZZ S F · dS counts the number of the points p1, p2, . . . , pk that lie in the interior of S. This is an instance of what is known in physics as Gauss’s law. In effect, the total mass, or charge, contained inside of S determines, and is determined by, the integral of F over S. An explicit formula or parametrization for S is not needed to obtain this information. 10.7 A more substitution-friendly notation for surface integrals We now fill in a few details that we have been postponing. Some of the arguments are technical, but they also unveil themes that unify the theory behind what we have been doing. In the case of line integrals, we have typically replaced the notation R C F · ds in favor of the more suggestive R C F1 dx1 + F2 dx2 + · · · + Fn dxn. The latter is useful in that it tells us more or less what we need to substitute in order to express the integral in terms of the parameter used to trace out C. We now introduce a corresponding notation for surface integrals. Let F = (F1, F2, F3) be a continuous vector field on an open set U in R3, and let S be a smooth oriented surface in U. To integrate F over S, we want to turn the calculation into a double integral in the two parameters of a parametrization. So let σ: D →R3 be a smooth parametrization of S, where σ(s, t) = (x(s, t), y(s, t), z(s, t)). For any function of (x, y, z), such as the components F1, F2, and F3 of F, the parametrization tells us how to substitute for x, y, and z, resulting in a function of s and t. | Multivariable_Calculus_Shimamoto_Page_271_Chunk3485 |
260 CHAPTER 10. SURFACE INTEGRALS For simplicity, let’s assume that σ is orientation-preserving. By definition: ZZ S F · dS = ZZ D F(σ(s, t)) · ∂σ ∂s × ∂σ ∂t ds dt = ZZ D | Multivariable_Calculus_Shimamoto_Page_272_Chunk3486 |
10.8. INDEPENDENCE OF PARAMETRIZATION 261 which also have the form of certain changes of variables. The integrand η = F1 dy∧dz+F2 dz∧dx+F3 dx∧dy of (10.20) is called a differential 2-form. By comparison, the expressions ω = F1 dx + F2 dy + F3 dz that we integrated over curves in R3 are called differential 1-forms. To integrate the 2-form η over S, we substitute for x, y, z in terms of s, t in F1, F2, F3 and replace the wedge products like dy ∧dz with expressions that come formally from the change of variables theorem. This converts the integral over S into a double integral over the parameter domain. The integrand of this double integral is called the pullback of η, denoted by σ∗(η). From (10.19), it can be written: σ∗(η) = (F1 ◦σ) · det ∂y ∂s ∂y ∂t ∂z ∂s ∂z ∂t + (F2 ◦σ) · det ∂z ∂s ∂z ∂t ∂x ∂s ∂x ∂t + (F3 ◦σ) · det ∂x ∂s ∂x ∂t ∂y ∂s ∂y ∂t ds ∧dt. With this notation, the definition of the surface integral takes the form: ZZ σ(D) η = ZZ D σ∗(η), (10.23) where, on the right, an integral of the form RR D f(s, t) ds ∧dt is defined to be the usual double integral RR D f(s, t) ds dt. Equation (10.23) highlights that the definition has the same form as the change of variables theorem, continuing an emerging pattern (see (7.3) and (9.6)). We say more in the next chapter about properties of differential forms, but our discussion so far can be used to motivate some elementary manipulations. The substitution formulas (10.21) and (10.22) for dy ∧dz and the like suggest a connection between wedge products and determinants. To reinforce this, we adopt the following rules for working with differential forms: • dz ∧dy = −dy ∧dz, dx ∧dz = −dz ∧dx, dy ∧dx = −dx ∧dy, • dx ∧dx = 0, dy ∧dy = 0, dz ∧dz = 0. The first collection mirrors the property that interchanging two rows changes the sign of a deter- minant, and the second mirrors the property that two equal rows imply a determinant of 0. We postpone until the next chapter the discussion of how these properties may be used. 10.8 Independence of parametrization Finally, we address the long-standing question of the extent to which the official definition of the surface integral, RR S F·dS = ± RR D F(σ(s, t))· | Multivariable_Calculus_Shimamoto_Page_273_Chunk3487 |
262 CHAPTER 10. SURFACE INTEGRALS integral after it has been expanded out in terms of coordinates as suggested by the differential form notation, that is: ZZ σ F · dS = ZZ D F1(σ(s, t)) · det ∂y ∂s ∂y ∂t ∂z ∂s ∂z ∂t + F2(σ(s, t)) · det ∂z ∂s ∂z ∂t ∂x ∂s ∂x ∂t + F3(σ(s, t)) · det ∂x ∂s ∂x ∂t ∂y ∂s ∂y ∂t ds dt, (10.24) as in equation (10.19), except that the 2 by 2 matrices have been transposed. The expression for RR τ F · dS is similar, integrating over D∗and using u’s and v’s instead of s’s and t’s. If x ∈S, then x = σ(s, t) for some (s, t) in D and x = τ(u, v) for some (u, v) in D∗. We write this as (s, t) = T(u, v). This defines a function T : D∗→D, where, given (u, v) in D∗, T(u, v) is the point (s, t) in D such that σ(s, t) = τ(u, v). Then, by construction: τ(u, v) = σ(s, t) = σ(T(u, v)). (10.25) See Figure 10.26. (If σ or τ is not one-to-one, which is permitted only on the boundaries of their Figure 10.26: Two parametrizations σ and τ of the same surface S respective domains, there is some ambiguity on how to define T there, but, as we have observed before, the integral can tolerate a certain amount of misbehavior on the boundary.) By equation (10.25) and the chain rule, Dτ(u, v) = Dσ(T(u, v))·DT(u, v) = Dσ(s, t)·DT(u, v), or: ∂x ∂u ∂x ∂v ∂y ∂u ∂y ∂v ∂z ∂u ∂z ∂v = ∂x ∂s ∂x ∂t ∂y ∂s ∂y ∂t ∂z ∂s ∂z ∂t ∂s ∂u ∂s ∂v ∂t ∂u ∂t ∂v . (10.26) We focus on the 2 by 2 blocks within the first two matrices of this equation. For instance, isolating the portions of the product involving the second and third rows, as indicated in red above, gives: ∂y ∂u ∂y ∂v ∂z ∂u ∂z ∂v = ∂y ∂s ∂y ∂t ∂z ∂s ∂z ∂t ∂s ∂u ∂s ∂v ∂t ∂u ∂t ∂v = ∂y ∂s ∂y ∂t ∂z ∂s ∂z ∂t · DT(u, v). Since the determinant of a product is the product of the determinants—that is, det(AB) = (det A)(det B)—we obtain: det ∂y ∂u ∂y ∂v ∂z ∂u ∂z ∂v = det ∂y ∂s ∂y ∂t ∂z ∂s ∂z ∂t · det DT(u, v). (10.27) | Multivariable_Calculus_Shimamoto_Page_274_Chunk3488 |
10.8. INDEPENDENCE OF PARAMETRIZATION 263 Referring to (10.24), the first determinant in the preceding equation appears in the definition of RR τ F · dS, while the second appears in the definition of RR σ F · dS. By choosing different pairs of rows in the chain rule (10.26), the same reasoning applies to the other two determinants that appear in (10.24). In each case, the determinants for τ and σ differ by a factor of det DT(u, v). In fact, these 2 by 2 determinants are precisely the components of the cross products ∂τ ∂u × ∂τ ∂v and ∂σ ∂s × ∂σ ∂t , so this calculation shows that: ∂τ ∂u × ∂τ ∂v = | Multivariable_Calculus_Shimamoto_Page_275_Chunk3489 |
264 CHAPTER 10. SURFACE INTEGRALS where the sign depends on whether τ is orientation-preserving or orientation-reversing. In particular, in order to compute RR S F · dS = RR σ F · dS, it does not matter which orientation- preserving parametrization σ is used. For integrals of real-valued functions with respect to surface area, the corresponding assertion is that RR σ f dS = RR τ f dS for any two smooth parametrizations σ, τ of S. The proof combines the ideas just given above with those given in the analogous situation for integrals over curves with respect to arclength. We won’t go through the details. This may be one of those rare occasions where it is all right to say that the authorities have weighed in and to let them have their way. One last remark: We commented earlier that the definition of the vector field surface integral can be regarded as pulling back the integral of the 2–form η = F1 dy∧dz+F2 dz∧dx+F3 dx∧dy to a double integral in the parameter plane (see equation (10.23)). Using that language, our calculation of the independence of parametrization can be summarized in one line: ZZ D∗τ ∗(η) = Z D∗(σ ◦T)∗(η) = ZZ D∗T ∗(σ∗(η)) = ZZ D σ∗(η). (10.30) In other words, using either σ or τ to pull back η gives the same value. Don’t panic! The first equation follows since T was defined so that τ = σ ◦T. The second equation uses the fact that (σ ◦T)∗(η) = T ∗(σ∗(η)). This is essentially the chain rule-based calculation that brought out the factor of det DT(u, v) in going from (10.28) to (10.29), though we would need to develop the properties of pullbacks more carefully to explain this. (To get a taste of these properties, see Exercises 4.14–4.18 in Chapter 11.) Finally, the step going from D∗ to D uses the pullback form of the change of variables theorem, which is a refinement that takes orientation into account and eliminates the absolute value around the determinant (see Exercise 4.20 in Chapter 11). Though there are gaps that need to be filled in, the intended lesson here is that differential forms, properly developed, provide a structure in which the arguments can be presented with remarkable conciseness. Of course, the proper development may involve lengthy calculations similar to the ones we presented, but their validity is guided by and rooted in basic principles: the chain rule and change of variables. 10.9 Exercises for Chapter 10 Section 1 What the surface integral measures 1.1. Let S be the hemisphere x2 + y2 + z2 = a2, z ≥0, oriented by the outward normal. (a) Find RR S F · dS if F(x, y, z) = (x, y, z). (b) Find RR S F · dS if F(x, y, z) = (y −z, z −x, x −y). 1.2. Let F(x, y, z) = (z, x, y), and let S be the triangular surface in R3 with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1), oriented by the upward normal. Find RR S F · dS. 1.3. Let F be a continuous vector field on R3 such that F(x, y, z) is always a scalar multiple of (0, 0, 1), i.e., F(x, y, z) = | Multivariable_Calculus_Shimamoto_Page_276_Chunk3490 |
10.9. EXERCISES FOR CHAPTER 10 265 1.5. Let F: R3 →R3 be a vector field such that F(−x) = −F(x) for all x in R3, and let S be the sphere x2 + y2 + z2 = a2, oriented by the outward normal. Is it necessarily true that RR S F · dS = 0? Explain. Section 2 The definition of the surface integral 2.1. Let S be the portion of the plane x + y + z = 3 for which 0 ≤x ≤1 and 0 ≤y ≤1. Let F(x, y, z) = (2y, x, z). If S is oriented by the upward normal, find RR S F · dS. 2.2. Let F(x, y, z) = (xz, yz, z2). (a) Let S be the cylinder x2 + y2 = 4, 1 ≤z ≤3, oriented by the outward normal. Find RR S F · dS. (b) Suppose that you “close up” the cylinder S in part (a) by adding the disks x2 + y2 ≤4, z = 3, and x2 + y2 ≤4, z = 1, at the top and bottom, respectively, where the disk at the top is oriented by the upward normal and the one at the bottom is oriented by the downward normal. Let S1 be the resulting surface: cylinder, top, and bottom. Find RR S1 F · dS. 2.3. Let S be the helicoid parametrized by: σ(r, θ) = (r cos θ, r sin θ, θ), 0 ≤r ≤1, 0 ≤θ ≤4π, oriented by the upward normal. Find RR S F · dS if F(x, y, z) = (y, −x, z2). 2.4. Let F(x, y, z) = | Multivariable_Calculus_Shimamoto_Page_277_Chunk3491 |
266 CHAPTER 10. SURFACE INTEGRALS Section 3 Stokes’s theorem 3.1. Consider the vector field F(x, y, z) = (2yz, y sin z, 1 + cos z). (a) Find a vector field G whose curl is F. (b) Let S be the half-ellipsoid 4x2 + 4y2 + z2 = 4, z ≥0, oriented by the upward normal. Use Stokes’s theorem to find RR S F · dS. (c) Find RR eS F · dS if eS is the portion of the surface z = 1 −x2 −y2 above the xy-plane, oriented by the upward normal. (Hint: Take advantage of what you’ve already done.) 3.2. Consider the vector field F(x, y, z) = (ex+y −xey+z, ey+z −ex+y + yez, −ez). (a) Is F a conservative vector field? Explain. (b) Find a vector field G = (G1, G2, G3) such that G2 = 0 and the curl of G is F. (c) Find RR S F · dS if S is the hemisphere x2 + y2 + z2 = 4, z ≥0, oriented by the outward normal. (d) Find RR S F · dS if S is the hemisphere x2 + y2 + z2 = 4, z ≤0, oriented by the outward normal. (e) Find RR S F · dS if S is the cylinder x2 + y2 = 4, 0 ≤z ≤4, oriented by the outward normal. Exercises 3.3–3.6 illustrate situations where, by bringing in a surface integral, Stokes’s theorem can be used to obtain information about line integrals that would be hard to find directly. 3.3. Let F(x, y, z) = (xz + yz, 2xz, 1 2y2). (a) Find ∇× F. (b) Let C be the curve of intersection in R3 of the cylinder x2 + y2 = 1 and the saddle surface z = x2 −y2, oriented counterclockwise as viewed from high above the xy-plane, looking down. Use Stokes’s theorem to find the line integral R C F · ds. (Hint: Find a surface whose boundary is C.) 3.4. Let P be a plane in R3 described by the equation ax + by + cz = d. Let C be a piecewise smooth oriented simple closed curve that lies in P, and let S be the subset of P that is enclosed by C, i.e., the region of P “inside” C. Show that the area of S is equal to the absolute value of: 1 2 √ a2 + b2 + c2 Z C (bz −cy) dx + (cx −az) dy + (ay −bx) dz. 3.5. (a) Find an example of a smooth vector field F = (F1, F2, F3) defined on an open set U of R3 such that: • its mixed partials are equal, i.e., ∂F1 ∂y = ∂F2 ∂x , ∂F1 ∂z = ∂F3 ∂x , and ∂F2 ∂z = ∂F3 ∂y , and • there exists a piecewise smooth oriented simple closed curve C in U such that R C F1 dx + F2 dy + F3 dz ̸= 0. (b) On the other hand, show that, if F is any smooth vector field on an open set U in R3 whose mixed partials are equal, then R C F1 dx + F2 dy + F3 dz = 0 for any piecewise smooth oriented simple closed curve C that is the boundary of an oriented surface S contained in U. | Multivariable_Calculus_Shimamoto_Page_278_Chunk3492 |
10.9. EXERCISES FOR CHAPTER 10 267 (c) What does part (b) tell you about the curve C you found in part (a)? 3.6. Let C1 and C2 be piecewise smooth simple closed curves contained in the cylinder x2 +y2 = 1 that do not intersect one another, both oriented counterclockwise when viewed from high above the xy-plane, looking down, as illustrated in Figure 10.28. If F is the vector field F(x, y, z) = (z −2y3, 2yz + x5, y2 + x), show that R C1 F · ds = R C2 F · ds. Figure 10.28: Two nonintersecting counterclockwise curves C1 and C2 in the cylinder x2 + y2 = 1 Section 4 Curl fields In Exercises 4.1–4.3, determine whether the given vector field F is a curl field. If it is, find a vector field G whose curl is F. If not, explain why not. 4.1. F(x, y, z) = (x, y, z) 4.2. F(x, y, z) = (y, z, x) 4.3. F(x, y, z) = (x + 2y + 3z, x4 + y5 + z6, x7y8z9) 4.4. This exercise ties up some loose ends left dangling after Example 10.3 back in Section 10.1. Let S1 be the hemisphere x2 + y2 + z2 = a2, z ≥0, oriented by the outward normal, and let S2 be the disk x2 + y2 ≤a2, z = 0, oriented by the upward normal. Let F be the vector field F(x, y, z) = (0, 0, −1). (a) Show that F is a curl field by finding a vector field G whose curl is F. (b) In Example 10.3, we showed that it is fairly easy to calculate that RR S2 F · dS = −πa2. Without any further calculation, explain why RR S1 F · dS = −πa2 as well. 4.5. Let F and G be vector fields on R3. (a) True or false: If G = F + C for some constant vector C, then F and G have the same curl. Either prove that the statement is true, or find an example in which it is false. (b) True or false: If F and G have the same curl, then G = F + C for some constant vector C. Either prove that the statement is true, or find an example in which it is false. 4.6. Does there exist a vector field F on R3, other than the constant vector field F = 0, with the property that: • R C F · ds = 0 for every piecewise smooth oriented closed curve C in R3 and • RR S F · dS = 0 for every piecewise smooth oriented closed surface S in R3? If so, find such a vector field, and justify why it works. If no such vector field exists, explain why not. | Multivariable_Calculus_Shimamoto_Page_279_Chunk3493 |
268 CHAPTER 10. SURFACE INTEGRALS Section 5 Gauss’s theorem 5.1. Let F(x, y, z) = (x3, y3, z3). Use Gauss’s theorem to find RR S F · dS if S is the sphere x2 + y2 + z2 = 4, oriented by the outward normal. 5.2. Find RR S F · dS if F(x, y, z) = (x3y3z4, x2y4z4, x2y3z5) and S is the boundary of the cube W = [0, 1] × [0, 1] × [0, 1], oriented by the normal pointing away from W. 5.3. Let S be the silo-shaped closed surface consisting of: • the cylinder x2 + y2 = 4, 0 ≤z ≤3, • a hemispherical cap of radius 2 centered at (0, 0, 3), i.e., x2 + y2 + (z −3)2 = 4, z ≥3, and • the disk x2 + y2 ≤4, z = 0, at the base, all oriented by the normal pointing away from the silo. Find RR S F · dS if F(x, y, z) = (3xz2, 2y, x + y2 −z3). 5.4. One way to produce a torus is to take the circle C in the xz-plane with center (a, 0, 0) and radius b and rotate it about the z-axis. We assume that a > b. The surface that is swept out is a torus. The circle C can be parametrized by α(ψ) = (a + b cos ψ, 0, b sin ψ), 0 ≤ψ ≤2π. The distance of α(ψ) to the z-axis is a+b cos ψ, so rotating α(ψ) counterclockwise by θ about the z-axis brings it to the point | Multivariable_Calculus_Shimamoto_Page_280_Chunk3494 |
10.9. EXERCISES FOR CHAPTER 10 269 (b) Is σ orientation-preserving or orientation-reversing? (c) Find the surface area of the torus. (d) Let F(x, y, z) = (x, y, z). Use the parametrization of S and the definition of the surface integral to find RR S F · dS. Then, use your answer and the result of Example 10.16 to find the volume of the solid torus, that is, the three-dimensional region enclosed by the torus. (e) Find the integral of the vector field F(x, y, z) = (0, 0, 1) over S. Use whatever method seems best. What does your answer say about the flow of F through S? 5.5. We describe an alternative approach to Exercise 4.4. Let F(x, y, z) = (0, 0, −1). The main point of the aforementioned exercise is that: ZZ S1 F · dS = ZZ S2 F · dS, (10.31) where S1 is the hemisphere x2 + y2 + z2 = a2, z ≥0, oriented by the outward normal, and S2 is the disk x2 + y2 ≤a2, z = 0, oriented by the upward normal. The argument was based on Stokes’s theorem. Show that the same conclusion (10.31) can be reached using Gauss’s theorem. 5.6. A bounded solid W in R3 is contained in the region a ≤z ≤b, where a and b are constant and a < b. The boundary of W is the union of three parts, as shown in Figure 10.30, left: • a top surface T that is a subset of the plane z = b, • a bottom surface B that is a subset of the plane z = a, and • a surface S in between that connects the bottom and the top. Let F(x, y, z) = (0, 0, 1). If the third piece above, S, is oriented by the normal pointing away from W, express RR S F · dS in terms of Area (T) and Area (B). T S B x2+ y2 = 9, z = n– x² + y² = 4, z = 8 n Figure 10.30: A solid region W lying between the planes z = a and z = b whose boundary consists of a surface S, a top T, and a bottom B (left) and a lumpy shopping bag of volume 75 (right) 5.7. A shopping bag is modeled as a lumpy surface S having an open circular top. Suppose that the bag is placed in R3 in such a way that the boundary of the opening is the circle x2+y2 = 4 in the plane z = 8. See Figure 10.30, right. Assume that, when filled exactly to the brim, the bag has a volume of 75. Let F(x, y, z) = (x, y, z). If S is oriented by the outward normal, use Gauss’s theorem to find RR S F · dS. (Note that S is not a closed surface.) | Multivariable_Calculus_Shimamoto_Page_281_Chunk3495 |
270 CHAPTER 10. SURFACE INTEGRALS Let U be an open set in R3. A smooth real-valued function h: U →R is called harmonic on U if: ∂2h ∂x2 + ∂2h ∂y2 + ∂2h ∂z2 = 0 at all points of U. Exercises 5.8–5.10 concern harmonic functions. The results are the analogues of Exercises 4.12–4.15 in Chapter 9, which were about harmonic functions of two variables. 5.8. Show that h(x, y, z) = e−5x sin(3y + 4z) is a harmonic function on R3. 5.9. Let W be a bounded subset of an open set U in R3 whose boundary ∂W consists of a finite number of piecewise smooth closed surfaces. Assume that every pair of points in W can be connected by a piecewise smooth curve in W. If h is harmonic on U and if h = 0 at all points of ∂W, prove that h = 0 on all of W. (Hint: Consider the vector field F = h ∇h = h(∂h ∂x, ∂h ∂y , ∂h ∂z ).) 5.10. Let U and W be as in the preceding exercise. Show that, if h1 and h2 are harmonic on U and if h1 = h2 at all points of ∂W, then h1 = h2 on all of W. Thus a harmonic function is determined on W by its values on the boundary. Section 6 The inverse square field 6.1. Let Sa denote the sphere x2 + y2 + z2 = a2 of radius a, oriented by the outward normal. If we did not already know the answer, it might be tempting to calculate the integral of the inverse square field G(x, y, z) = | Multivariable_Calculus_Shimamoto_Page_282_Chunk3496 |
10.9. EXERCISES FOR CHAPTER 10 271 where G is our usual inverse square field (10.16). Let S be a piecewise smooth closed surface in R3−{p1, p2, . . . , pk}, oriented by the outward normal. What can you say about −1 4π RR S F·dS in this case? Section 7 A more substitution-friendly notation for surface integrals 7.1. Let S be the unit sphere x2 + y2 + z2 = 1, oriented by the outward normal. Consider the surface integral: ZZ S xy dy ∧dz + yz dz ∧dx + xz dx ∧dy. (a) What is the vector field that is being integrated? (b) Evaluate the integral. 7.2. Suppose that a surface S in R3 is the graph z = f(x, y) of a smooth real-valued function f : D →R whose domain D is a bounded subset of R2. Let S be oriented by the upward normal. Show that: ZZ S 1 dx ∧dy = Area (D). In other words, the integral is the area of the projection of S on the xy-plane. Similar interpretations hold for RR S 1 dy ∧dz and RR S 1 dz ∧dx under analogous hypotheses. Section 8 Independence of parametrization 8.1. Let S be the portion of the cone z = p x2 + y2 in R3 where 0 ≤z ≤2. In Section 5.4, we showed that S could be parametrized in more than one way. One possibility is based on using polar/cylindrical coordinates as parameters (Example 5.14): σ: [0, 2] × [0, 2π] →R3, σ(s, t) = (s cos t, s sin t, s). Another is based on spherical coordinates (Example 5.17): τ : [0, 2 √ 2] × [0, 2π] →R3, τ(u, v) = | Multivariable_Calculus_Shimamoto_Page_283_Chunk3497 |
272 CHAPTER 10. SURFACE INTEGRALS | Multivariable_Calculus_Shimamoto_Page_284_Chunk3498 |
Chapter 11 Working with differential forms This final chapter is a whirlwind survey of how to work with differential forms. We focus particularly on what differential forms have to say about some of the main theorems we have learned recently. In what follows, all functions, including vector fields, are assumed to be smooth. Four theorems, version 1 1. Conservative vector fields. Let C be a piecewise smooth oriented curve in Rn from p to q. If F = ∇f is a conservative vector field on an open set containing C, then: Z C ∇f · ds = f(q) −f(p). (C) 2. Green’s theorem. Let D be a bounded subset of R2. If F = (F1, F2) is a vector field on an open set containing D, then: ZZ D | Multivariable_Calculus_Shimamoto_Page_285_Chunk3499 |
274 CHAPTER 11. WORKING WITH DIFFERENTIAL FORMS 11.1 Integrals of differential forms Roughly speaking, a differential k-form is an expression that can be integrated over an oriented k-dimensional domain. For instance, in Rn, here are the cases at opposite ends of the range of dimensions. • n-forms. In Rn with coordinates x1, x2, . . . , xn, an n-form is an expression: f dx1 ∧dx2 ∧· · · ∧dxn, where f is a real-valued function of x = (x1, x2, . . . , xn). It could also be written f ∧dx1 ∧ dx2 ∧· · · ∧dxn, though f is treated commonly as a scalar factor. If W is an n-dimensional bounded subset of Rn, which by default we say has the “positive” orientation, then the integral R W f dx1 ∧dx2 ∧· · · ∧dxn is defined to be the usual Riemann integral: Z W f dx1 ∧dx2 ∧· · · ∧dxn = ZZ · · · Z W f(x1, x2, . . . , xn) dx1 dx2 · · · dxn. (11.1) As indicated on the left of equation (11.1), we won’t use multiple integral signs with differential forms. The dimension of the integral can be inferred from the type of form that is being integrated. • 0-forms. A 0-form is simply another name for a real-valued function f. 0-forms are integrated over zero-dimensional domains, i.e., individual points, by evaluating the function at that point and where we orient a point by attaching a + or −sign. So R +{p} f = f(p) and R −{p} f = −f(p). If C is an oriented curve from p to q, we define its oriented boundary to be the oriented points ∂C = (+{q}) ∪(−{p}). We restrict our attention here to differential forms on subsets of R3 (or R or R2, depending on the context), real-valued functions f of three variables (or one or two), and vector fields on subsets of R3 (or R or R2), since these are the cases of interest in the four big theorems listed above. Table 11.1 below gives the notation for 0, 1, 2, and 3-forms and their associated integrals. The integrals that the notation stands for, shown in the last column, are the kinds of integrals we have been studying: 1-forms integrate as line integrals, 2-forms as surface integrals, and 3-forms as triple integrals. Type of Domain of Notation for integral Meaning form integration using forms of integral 0-form oriented points R +{p} f or R −{p} f = f(p) or −f(p) 1-form oriented curves R C F1 dx + F2 dy + F3 dz = R C F · ds 2-form oriented surfaces R S F1 dy ∧dz + F2 dz ∧dx +F3 dx ∧dy = RR S F · dS 3-form 3-dim solids R W f dx ∧dy ∧dz = RRR W f dx dy dz Table 11.1: Integrals of differential forms In this notation, the four theorems (C), (Gr), (S), and (Ga) translate as integrals of forms as follows. | Multivariable_Calculus_Shimamoto_Page_286_Chunk3500 |
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