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method of Lagrange multipliers : a method of solving an optimization problem subject to one or more constraints
https://openstax.org/books/calculus-volume-3/pages/4-key-terms
mixed partial derivatives : second-order or higher partial derivatives, in which at least two of the differentiations are with respect to different variables
https://openstax.org/books/calculus-volume-3/pages/4-key-terms
objective function : the function that is to be maximized or minimized in an optimization problem
https://openstax.org/books/calculus-volume-3/pages/4-key-terms
open set : a setSSthat contains none of its boundary points
https://openstax.org/books/calculus-volume-3/pages/4-key-terms
optimization problem : calculation of a maximum or minimum value of a function of several variables, often using Lagrange multipliers
https://openstax.org/books/calculus-volume-3/pages/4-key-terms
partial derivative : a derivative of a function of more than one independent variable in which all the variables but one are held constant
https://openstax.org/books/calculus-volume-3/pages/4-key-terms
partial differential equation : an equation that involves an unknown function of more than one independent variable and one or more of its partial derivatives
https://openstax.org/books/calculus-volume-3/pages/4-key-terms
region : an open, connected, nonempty subset ofℝ2ℝ2
https://openstax.org/books/calculus-volume-3/pages/4-key-terms
saddle point : given the functionz=f(x,y),z=f(x,y),the point(x0,y0,f(x0,y0))(x0,y0,f(x0,y0))is a saddle point if bothfx(x0,y0)=0fx(x0,y0)=0andfy(x0,y0)=0,fy(x0,y0)=0,butffdoes not have a local extremum at(x0,y0)(x0,y0)
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surface : the graph of a function of two variables,z=f(x,y)z=f(x,y)
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tangent plane : given a functionf(x,y)f(x,y)that is differentiable at a point(x0,y0),(x0,y0),the equation of the tangent plane to the surfacez=f(x,y)z=f(x,y)is given byz=f(x0,y0)+fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)z=f(x0,y0)+fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)
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total differential : the total differential of the functionf(x,y)f(x,y)at(x0,y0)(x0,y0)is given by the formuladz=fx(x0,y0)dx+fy(x0,y0)dydz=fx(x0,y0)dx+fy(x0,y0)dy
https://openstax.org/books/calculus-volume-3/pages/4-key-terms
tree diagram : illustrates and derives formulas for the generalized chain rule, in which each independent variable is accounted for
https://openstax.org/books/calculus-volume-3/pages/4-key-terms
vertical trace : the set of ordered triples(c,y,z)(c,y,z)that solves the equationf(c,y)=zf(c,y)=zfor a given constantx=cx=cor the set of ordered triples(x,d,z)(x,d,z)that solves the equationf(x,d)=zf(x,d)=zfor a given constanty=dy=d
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δδball : all points inℝ3ℝ3lying at a distance of less thanδδfrom(x0,y0,z0)(x0,y0,z0)
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δδdisk : an open disk of radiusδδcentered at point(a,b)(a,b)
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We can use a double Riemann sum to approximate the volume of a solid bounded above by a function of two variables over a rectangular region. By taking the limit, this becomes a double integral representing the volume of the solid.
https://openstax.org/books/calculus-volume-3/pages/5-key-concepts
Properties of double integral are useful to simplify computation and find bounds on their values.
https://openstax.org/books/calculus-volume-3/pages/5-key-concepts
We can use Fubini’s theorem to write and evaluate a double integral as an iterated integral.
https://openstax.org/books/calculus-volume-3/pages/5-key-concepts
Double integrals are used to calculate the area of a region, the volume under a surface, and the average value of a function of two variables over a rectangular region.
https://openstax.org/books/calculus-volume-3/pages/5-key-concepts
A general bounded regionDDon the plane is a region that can be enclosed inside a rectangular region. We can use this idea to define a double integral over a general bounded region.
https://openstax.org/books/calculus-volume-3/pages/5-key-concepts
To evaluate an iterated integral of a function over a general nonrectangular region, we sketch the region and express it as a Type I or as a Type II region or as a union of several Type I or Type II regions that overlap only on their boundaries.
https://openstax.org/books/calculus-volume-3/pages/5-key-concepts
We can use double integrals to find volumes, areas, and average values of a function over general regions, similarly to calculations over rectangular regions.
https://openstax.org/books/calculus-volume-3/pages/5-key-concepts
We can use Fubini’s theorem for improper integrals to evaluate some types of improper integrals.
https://openstax.org/books/calculus-volume-3/pages/5-key-concepts
To apply a double integral to a situation with circular symmetry, it is often convenient to use a double integral in polar coordinates. We can apply these double integrals over a polar rectangular region or a general polar region, using an iterated integral similar to those used with rectangular double integrals.
https://openstax.org/books/calculus-volume-3/pages/5-key-concepts
The areadAdAin polar coordinates becomesrdrdθ.rdrdθ.
https://openstax.org/books/calculus-volume-3/pages/5-key-concepts
Usex=rcosθ,x=rcosθ,y=rsinθ,y=rsinθ,anddA=rdrdθdA=rdrdθto convert an integral in rectangular coordinates to an integral in polar coordinates.
https://openstax.org/books/calculus-volume-3/pages/5-key-concepts
User2=x2+y2r2=x2+y2andθ=tan−1(yx)θ=tan−1(yx)to convert an integral in polar coordinates to an integral in rectangular coordinates, if needed.
https://openstax.org/books/calculus-volume-3/pages/5-key-concepts
To find the volume in polar coordinates bounded above by a surfacez=f(r,θ)z=f(r,θ)over a region on thexyxy-plane, use a double integral in polar coordinates.
https://openstax.org/books/calculus-volume-3/pages/5-key-concepts
To compute a triple integral we use Fubini’s theorem, which states that iff(x,y,z)f(x,y,z)is continuous on a rectangular boxB=[a,b]×[c,d]×[e,f],B=[a,b]×[c,d]×[e,f],then∭Bf(x,y,z)dV=∫ef∫cd∫abf(x,y,z)dxdydz∭Bf(x,y,z)dV=∫ef∫cd∫abf(x,y,z)dxdydzand is also equal to any of the other five possible orderi...
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To compute the volume of a general solid bounded regionEEwe use the triple integralV(E)=∭E1dV.V(E)=∭E1dV.
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Interchanging the order of the iterated integrals does not change the answer. As a matter of fact, interchanging the order of integration can help simplify the computation.
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To compute the average value of a function over a general three-dimensional region, we usefave=1V(E)∭Ef(x,y,z)dV.fave=1V(E)∭Ef(x,y,z)dV.
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To evaluate a triple integral in cylindrical coordinates, use the iterated integral∫θ=αθ=β∫r=g1(θ)r=g2(θ)∫z=u1(r,θ)z=u2(r,θ)f(r,θ,z)rdzdrdθ.∫θ=αθ=β∫r=g1(θ)r=g2(θ)∫z=u1(r,θ)z=u2(r,θ)f(r,θ,z)rdzdrdθ.
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To evaluate a triple integral in spherical coordinates, use the iterated integral∫θ=αθ=β∫ρ=g1(θ)ρ=g2(θ)∫φ=u1(r,θ)φ=u2(r,θ)f(ρ,θ,φ)ρ2sinφdφdρdθ.∫θ=αθ=β∫ρ=g1(θ)ρ=g2(θ)∫φ=u1(r,θ)φ=u2(r,θ)f(ρ,θ,φ)ρ2sinφdφdρdθ.
https://openstax.org/books/calculus-volume-3/pages/5-key-concepts
Finding the mass, center of mass, moments, and moments of inertia in double integrals:
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For a laminaRRwith a density functionρ(x,y)ρ(x,y)at any point(x,y)(x,y)in the plane, the mass ism=∬Rρ(x,y)dA.m=∬Rρ(x,y)dA.
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The moments about thex-axisx-axisandy-axisy-axisareMx=∬Ryρ(x,y)dAandMy=∬Rxρ(x,y)dA.Mx=∬Ryρ(x,y)dAandMy=∬Rxρ(x,y)dA.
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The center of mass is given byx−=Mym,y−=Mxm.x−=Mym,y−=Mxm.
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The center of mass becomes the centroid of the plane when the density is constant.
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The moments of inertia about thex−axis,x−axis,y−axis,y−axis,and the origin areIx=∬Ry2ρ(x,y)dA,Iy=∬Rx2ρ(x,y)dA,andI0=Ix+Iy=∬R(x2+y2)ρ(x,y)dA.Ix=∬Ry2ρ(x,y)dA,Iy=∬Rx2ρ(x,y)dA,andI0=Ix+Iy=∬R(x2+y2)ρ(x,y)dA.
https://openstax.org/books/calculus-volume-3/pages/5-key-concepts
Finding the mass, center of mass, moments, and moments of inertia in triple integrals:
https://openstax.org/books/calculus-volume-3/pages/5-key-concepts
For a solid objectQQwith a density functionρ(x,y,z)ρ(x,y,z)at any point(x,y,z)(x,y,z)in space, the mass ism=∭Qρ(x,y,z)dV.m=∭Qρ(x,y,z)dV.
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The moments about thexy-plane,xy-plane,thexz-plane,xz-plane,and theyz-planeyz-planeareMxy=∭Qzρ(x,y,z)dV,Mxz=∭Qyρ(x,y,z)dV,Myz=∭Qxρ(x,y,z)dV.Mxy=∭Qzρ(x,y,z)dV,Mxz=∭Qyρ(x,y,z)dV,Myz=∭Qxρ(x,y,z)dV.
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The center of mass is given byx−=Myzm,y−=Mxzm,z−=Mxym.x−=Myzm,y−=Mxzm,z−=Mxym.
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The center of mass becomes the centroid of the solid when the density is constant.
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The moments of inertia about theyz-plane,yz-plane,thexz-plane,xz-plane,and thexy-planexy-planeareIx=∭Q(y2+z2)ρ(x,y,z)dV,Iy=∭Q(x2+z2)ρ(x,y,z)dV,Iz=∭Q(x2+y2)ρ(x,y,z)dV.Ix=∭Q(y2+z2)ρ(x,y,z)dV,Iy=∭Q(x2+z2)ρ(x,y,z)dV,Iz=∭Q(x2+y2)ρ(x,y,z)dV.
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A transformationTTis a function that transforms a regionGGin one plane (space) into a regionRRin another plane (space) by a change of variables.
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A transformationT:G→RT:G→Rdefined asT(u,v)=(x,y)T(u,v)=(x,y)(orT(u,v,w)=(x,y,z))(orT(u,v,w)=(x,y,z))is said to be a one-to-one transformation if no two points map to the same image point.
https://openstax.org/books/calculus-volume-3/pages/5-key-concepts
Ifffis continuous onR,R,then∬Rf(x,y)dA=∬Sf(g(u,v),h(u,v))|∂(x,y)∂(u,v)|dudv.∬Rf(x,y)dA=∬Sf(g(u,v),h(u,v))|∂(x,y)∂(u,v)|dudv.
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IfFFis continuous onR,R,then∭RF(x,y,z)dV=∭GF(g(u,v,w),h(u,v,w),k(u,v,w))|∂(x,y,z)∂(u,v,w)|dudvdw=∭GH(u,v,w)|J(u,v,w)|dudvdw.∭RF(x,y,z)dV=∭GF(g(u,v,w),h(u,v,w),k(u,v,w))|∂(x,y,z)∂(u,v,w)|dudvdw=∭GH(u,v,w)|J(u,v,w)|dudvdw.
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∬ R f ( x , y ) d A = lim m , n → ∞ ∑ i = 1 m ∑ j = 1 n f ( x i j * , y i j * ) Δ A ∬ R f ( x , y ) d A = lim m , n → ∞ ∑ i = 1 m ∑ j = 1 n f ( x i j * , y i j * ) Δ A
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∫ a b ∫ c d f ( x , y ) d x d y = ∫ a b [ ∫ c d f ( x , y ) d y ] d x ∫ a b ∫ c d f ( x , y ) d x d y = ∫ a b [ ∫ c d f ( x , y ) d y ] d x or ∫ c d ∫ b a f ( x , y ) d x d y = ∫ c d [ ∫ a b f ( x , y ) d x ] d y ∫ c d ∫ b a f ( x , y ) d x d y = ∫ c d [ ∫ a b f ( x , y ) d x ] d y
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f ave = 1 Area R ∬ R f ( x , y ) d x d y f ave = 1 Area R ∬ R f ( x , y ) d x d y
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∬ D f ( x , y ) d A = ∬ D f ( x , y ) d y d x = ∫ a b [ ∫ g 1 ( x ) g 2 ( x ) f ( x , y ) d y ] d x ∬ D f ( x , y ) d A = ∬ D f ( x , y ) d y d x = ∫ a b [ ∫ g 1 ( x ) g 2 ( x ) f ( x , y ) d y ] d x
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∬ D f ( x , y ) d A = ∬ D f ( x , y ) d x d y = ∫ c d [ ∫ h 1 ( y ) h 2 ( y ) f ( x , y ) d x ] d y ∬ D f ( x , y ) d A = ∬ D f ( x , y ) d x d y = ∫ c d [ ∫ h 1 ( y ) h 2 ( y ) f ( x , y ) d x ] d y
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∬ R f ( r , θ ) d A = lim m , n → ∞ ∑ i = 1 m ∑ j = 1 n f ( r i j * , θ i j * ) Δ A = lim m , n → ∞ ∑ i = 1 m ∑ j = 1 n f ( r i j * , θ i j * ) r i j * Δ r Δ θ ∬ R f ( r , θ ) d A = lim m , n → ∞ ∑ i = 1 m ∑ j = 1 n f ( r i j * , θ i j * ) Δ A = lim m , n → ∞ ∑ i = 1 m ∑ j ...
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∬ D f ( r , θ ) r d r d θ = ∫ θ = α θ = β ∫ r = h 1 ( θ ) r = h 2 ( θ ) f ( r , θ ) r d r d θ ∬ D f ( r , θ ) r d r d θ = ∫ θ = α θ = β ∫ r = h 1 ( θ ) r = h 2 ( θ ) f ( r , θ ) r d r d θ
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lim l , m , n → ∞ ∑ i = 1 l ∑ j = 1 m ∑ k = 1 n f ( x i j k * , y i j k * , z i j k * ) Δ x Δ y Δ z = ∭ B f ( x , y , z ) d V lim l , m , n → ∞ ∑ i = 1 l ∑ j = 1 m ∑ k = 1 n f ( x i j k * , y i j k * , z i j k * ) Δ x Δ y Δ z = ∭ B f ( x , y , z ) d V
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∭ B g ( x , y , z ) d V = ∭ B g ( r cos θ , r sin θ , z ) r d r d θ d z = ∭ B f ( r , θ , z ) r d r d θ d z ∭ B g ( x , y , z ) d V = ∭ B g ( r cos θ , r sin θ , z ) r d r d θ d z = ∭ B f ( r , θ , z ) r d r d θ d z
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∭ B f ( ρ , θ , φ ) ρ 2 sin φ d ρ d φ d θ = ∫ φ = γ φ = ψ ∫ θ = α θ = β ∫ ρ = a ρ = b f ( ρ , θ , φ ) ρ 2 sin φ d ρ d φ d θ ∭ B f ( ρ , θ , φ ) ρ 2 sin φ d ρ d φ d θ = ∫ φ = γ φ = ψ ∫ θ = α θ = β ∫ ρ = a ρ = b f ( ρ , θ , φ ) ρ 2 sin φ d ρ d φ d θ
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m = lim k , l → ∞ ∑ i = 1 k ∑ j = 1 l m i j = lim k , l → ∞ ∑ i = 1 k ∑ j = 1 l ρ ( x i j * , y i j * ) Δ A = ∬ R ρ ( x , y ) d A m = lim k , l → ∞ ∑ i = 1 k ∑ j = 1 l m i j = lim k , l → ∞ ∑ i = 1 k ∑ j = 1 l ρ ( x i j * , y i j * ) Δ A = ∬ R ρ ( x , y ) d A
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M x = lim k , l → ∞ ∑ i = 1 k ∑ j = 1 l ( y i j * ) m i j = lim k , l → ∞ ∑ i = 1 k ∑ j = 1 l ( y i j * ) ρ ( x i j * , y i j * ) Δ A = ∬ R y ρ ( x , y ) d A M x = lim k , l → ∞ ∑ i = 1 k ∑ j = 1 l ( y i j * ) m i j = lim k , l → ∞ ∑ i = 1 k ∑ j = 1 l ( y i j * ) ρ ( x i j * , y i ...
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M y = lim k , l → ∞ ∑ i = 1 k ∑ j = 1 l ( x i j * ) m i j = lim k , l → ∞ ∑ i = 1 k ∑ j = 1 l ( x i j * ) ρ ( x i j * , y i j * ) Δ A = ∬ R x ρ ( x , y ) d A M y = lim k , l → ∞ ∑ i = 1 k ∑ j = 1 l ( x i j * ) m i j = lim k , l → ∞ ∑ i = 1 k ∑ j = 1 l ( x i j * ) ρ ( x i j * , y i ...
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x − = M y m = ∬ R x ρ ( x , y ) d A ∬ R ρ ( x , y ) d A x − = M y m = ∬ R x ρ ( x , y ) d A ∬ R ρ ( x , y ) d A and y − = M x m = ∬ R y ρ ( x , y ) d A ∬ R ρ ( x , y ) d A y − = M x m = ∬ R y ρ ( x , y ) d A ∬ R ρ ( x , y ) d A
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double integral : of the functionf(x,y)f(x,y)over the regionRRin thexyxy-plane is defined as the limit of a double Riemann sum,∬Rf(x,y)dA=limm,n→∞∑i=1m∑j=1nf(xij*,yij*)ΔA.∬Rf(x,y)dA=limm,n→∞∑i=1m∑j=1nf(xij*,yij*)ΔA.
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double Riemann sum : of the functionf(x,y)f(x,y)over a rectangular regionRRis∑i=1m∑j=1nf(xij*,yij*)ΔA∑i=1m∑j=1nf(xij*,yij*)ΔAwhereRRis divided into smaller subrectanglesRijRijand(xij*,yij*)(xij*,yij*)is an arbitrary point inRijRij
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Fubini’s theorem : iff(x,y)f(x,y)is a function of two variables that is continuous over a rectangular regionR={(x,y)∈ℝ2|a≤x≤b,c≤y≤d},R={(x,y)∈ℝ2|a≤x≤b,c≤y≤d},then the double integral offfover the region equals an iterated integral,∬Rf(x,y)dydx=∫ab∫cdf(x,y)dxdy=∫cd∫abf(x,y)dxdy∬Rf(x...
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improper double integral : a double integral over an unbounded region or of an unbounded function
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iterated integral : for a functionf(x,y)f(x,y)over the regionRRis∫ab∫cdf(x,y)dxdy=∫ab[∫cdf(x,y)dy]dx,∫ab∫cdf(x,y)dxdy=∫ab[∫cdf(x,y)dy]dx,∫cd∫baf(x,y)dxdy=∫cd[∫abf(x,y)dx]dy,∫cd∫baf(x,y)dxdy=∫cd[∫abf(x,y)dx]dy,wherea,b,c,a,b,c,andddare any real numbers andR=[a,b]×[c,d]R=[a,b]×[c,d]
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Jacobian : the JacobianJ(u,v)J(u,v)in two variables is a2×22×2determinant:J(u,v)=|∂x∂u∂y∂u∂x∂v∂y∂v|;J(u,v)=|∂x∂u∂y∂u∂x∂v∂y∂v|;the JacobianJ(u,v,w)J(u,v,w)in three variables is a3×33×3determinant:J(u,v,w)=|∂x∂u∂y∂u∂z∂u∂x∂v∂y∂v∂z∂v∂x∂w∂y∂w∂z∂w|J(u,v,...
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one-to-one transformation : a transformationT:G→RT:G→Rdefined asT(u,v)=(x,y)T(u,v)=(x,y)is said to be one-to-one if no two points map to the same image point
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planar transformation : a functionTTthat transforms a regionGGin one plane into a regionRRin another plane by a change of variables
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polar rectangle : the region enclosed between the circlesr=ar=aandr=br=band the anglesθ=αθ=αandθ=β;θ=β;it is described asR={(r,θ)|a≤r≤b,α≤θ≤β}R={(r,θ)|a≤r≤b,α≤θ≤β}
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radius of gyration : the distance between the rotational axis of the object and the point where the entire mass of the object can be concentrated and have the same moment of inertia
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transformation : a function that transforms a regionGGin one plane into a regionRRin another plane by a change of variables
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triple integral : the triple integral of a continuous functionf(x,y,z)f(x,y,z)over a rectangular solid boxBBis the limit of a Riemann sum for a function of three variables, if this limit exists
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triple integral in cylindrical coordinates : the limit of a triple Riemann sum, provided the following limit exists:liml,m,n→∞∑i=1l∑j=1m∑k=1nf(rijk*,θijk*,zijk*)rijk*ΔrΔθΔzliml,m,n→∞∑i=1l∑j=1m∑k=1nf(rijk*,θijk*,zijk*)rijk*ΔrΔθΔz
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triple integral in spherical coordinates : the limit of a triple Riemann sum, provided the following limit exists:liml,m,n→∞∑i=1l∑j=1m∑k=1nf(ρijk*,θijk*,φijk*)(ρijk*)2sinφΔρΔθΔφliml,m,n→∞∑i=1l∑j=1m∑k=1nf(ρijk*,θijk*,φijk*)(ρijk*)2sinφΔρΔθΔφ
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Type I : a regionDDin thexyxy-plane is Type I if it lies between two vertical lines and the graphs of two continuous functionsg1(x)g1(x)andg2(x)g2(x)
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Type II : a regionDDin thexyxy-plane is Type II if it lies between two horizontal lines and the graphs of two continuous functionsh1(y)andh2(y)h1(y)andh2(y)
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A vector field assigns a vectorF(x,y)F(x,y)to each point(x,y)(x,y)in a subsetDofℝ2orℝ3.ℝ2orℝ3.F(x,y,z)F(x,y,z)to each point(x,y,z)(x,y,z)in a subsetDofℝ3.ℝ3.
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Vector fields can describe the distribution of vector quantities such as forces or velocities over a region of the plane or of space. They are in common use in such areas as physics, engineering, meteorology, oceanography.
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We can sketch a vector field by examining its defining equation to determine relative magnitudes in various locations and then drawing enough vectors to determine a pattern.
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A vector fieldFFis called conservative if there exists a scalar functionffsuch that∇f=F.∇f=F.
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Line integrals generalize the notion of a single-variable integral to higher dimensions. The domain of integration in a single-variable integral is a line segment along thex-axis, but the domain of integration in a line integral is a curve in a plane or in space.
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IfCis a curve, then the length ofCis∫Cds.∫Cds.
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There are two kinds of line integral: scalar line integrals and vector line integrals. Scalar line integrals can be used to calculate the mass of a wire; vector line integrals can be used to calculate the work done on a particle traveling through a field.
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Scalar line integrals can be calculated usingEquation 6.8; vector line integrals can be calculated usingEquation 6.9.
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Two key concepts expressed in terms of line integrals are flux and circulation. Flux measures the rate that a field crosses a given line; circulation measures the tendency of a field to move in the same direction as a given closed curve.
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The theorems in this section require curves that are closed, simple, or both, and regions that are connected or simply connected.
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The line integral of a conservative vector field can be calculated using the Fundamental Theorem for Line Integrals. This theorem is a generalization of the Fundamental Theorem of Calculus in higher dimensions. Using this theorem usually makes the calculation of the line integral easier.
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Conservative fields are independent of path. The line integral of a conservative field depends only on the value of the potential function at the endpoints of the domain curve.
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Given vector fieldF, we can test whetherFis conservative by using the cross-partial property. IfFhas the cross-partial property and the domain is simply connected, thenFis conservative (and thus has a potential function). IfFis conservative, we can find a potential function by using the Problem-Solving Strategy.
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The circulation of a conservative vector field on a simply connected domain over a closed curve is zero.
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Green’s theorem relates the integral over a connected region to an integral over the boundary of the region. Green’s theorem is a version of the Fundamental Theorem of Calculus in one higher dimension.
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Green’s Theorem comes in two forms: a circulation form and a flux form. In the circulation form, the integrand isF·T.F·T.In the flux form, the integrand isF·N.F·N.
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Green’s theorem can be used to transform a difficult line integral into an easier double integral, or to transform a difficult double integral into an easier line integral.
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A vector field is source free if it has a stream function. The flux of a source-free vector field across a closed curve is zero, just as the circulation of a conservative vector field across a closed curve is zero.
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The divergence of a vector field is a scalar function. Divergence measures the “outflowing-ness” of a vector field. Ifvis the velocity field of a fluid, then the divergence ofvat a point is the outflow of the fluid less the inflow at the point.
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