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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... | 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) s... | 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 show... | 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... | 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 o... | 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 origin... | 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, s... | 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 bou... | 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 si... | 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 ... | 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 . Lastl... | 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−x... | 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 i... | 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... | 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) e... | 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,... | 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. W... | 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... | 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 th... | 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 arro... | 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)... | 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... | 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 :... | 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... | 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 t... | 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... | 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... | 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... | 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 ... | 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, o... | 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 ... | 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 ... | 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... | 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, ... | 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, o... | 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 comp... | 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 ... | 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 ... | 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 + ... | 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 ... | 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 win... | 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... | 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 l... | 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 parametr... | 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... | 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 ob... | 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 Corol... | 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... | 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 tha... | 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 ... | 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 o... | 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 p... | 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... | 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. Examp... | 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 3... | 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 fol... | 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 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 calculate... | 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 o... | 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 = ... | 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 i... | 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... | 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 i... | 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... | 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. T... | 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 opposit... | 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. (1... | 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, ... | 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 essent... | 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. ... | 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 ... | 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 i... | 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 ... | 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 ... | 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 surfa... | 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 ... | 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... | 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... | 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... | 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 ... | 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 ... | 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... | 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 sm... | 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 coordinat... | Multivariable_Calculus_Shimamoto_Page_286_Chunk3500 |
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