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Multiple Integrals

2.1 Double Integrals

The double integral of ff over a rectangle R=[a,b]×[c,d]R = [a,b] \times [c,d] is defined as the limit of Riemann sums:

Rf(x,y)dA=limP0i,jf(xij,yij)ΔAij\iint_R f(x,y)\, dA = \lim_{\lVert P \rVert \to 0} \sum_{i,j} f(x_{ij}^*, y_{ij}^*) \Delta A_{ij}

Theorem 2.1 (Fubini”s Theorem). If ff is continuous on R=[a,b]×[c,d]R = [a,b] \times [c,d]Then

Rf(x,y)dA=ab(cdf(x,y)dy)dx=cd(abf(x,y)dx)dy\iint_R f(x,y)\, dA = \int_a^b \left(\int_c^d f(x,y)\, dy\right) dx = \int_c^d \left(\int_a^b f(x,y)\, dx\right) dy

Proof (sketch). For a continuous function ff on the compact rectangle RRDefine

F(x)=cdf(x,y)dyF(x) = \int_c^d f(x,y)\, dy

Since ff is continuous, FF is continuous on [a,b][a,b]. For each partition P=(x0,,xm)P = \\{(x_0, \ldots, x_m)\\} of [a,b][a,b]Define Riemann sums for the outer integral:

S(P)=i=1mF(xi)Δxi=i=1mcdf(xi,y)dyΔxiS(P) = \sum_{i=1}^m F(x_i^*)\, \Delta x_i = \sum_{i=1}^m \int_c^d f(x_i^*, y)\, dy\, \Delta x_i

By Fubini’s theorem for Riemann integrals (proven via uniform continuity of ff on the compact set RR), As P0\lVert P \rVert \to 0 these sums converge to both RfdA\iint_R f\, dA and abF(x)dx\int_a^b F(x)\, dx. The Reversal of integration order follows by symmetry. \blacksquare

2.2 General Regions

For a general region DD in R2\mathbb{R}^2:

  • Type I region: D=(x,y):axb,g1(x)yg2(x)D = \\{(x,y) : a \leq x \leq b,\, g_1(x) \leq y \leq g_2(x)\\}

DfdA=abg1(x)g2(x)f(x,y)dydx\iint_D f\, dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y)\, dy\, dx

  • Type II region: D=(x,y):cyd,h1(y)xh2(y)D = \\{(x,y) : c \leq y \leq d,\, h_1(y) \leq x \leq h_2(y)\\}

DfdA=cdh1(y)h2(y)f(x,y)dxdy\iint_D f\, dA = \int_c^d \int_{h_1(y)}^{h_2(y)} f(x,y)\, dx\, dy

Problem. Evaluate DxydA\iint_D xy\, dA where DD is the region bounded by y=x2y = x^2 and y=x+2y = x + 2.

Solution

The curves intersect when x2=x+2x^2 = x + 2I.e., x2x2=0x^2 - x - 2 = 0So (x2)(x+1)=0(x-2)(x+1) = 0Giving x=1x = -1 and x=2x = 2. As a Type I region, D=(x,y):1x2,x2yx+2D = \\{(x,y) : -1 \leq x \leq 2,\, x^2 \leq y \leq x+2\\}.

DxydA=12x2x+2xydydx=12x[y22]x2x+2dx\iint_D xy\, dA = \int_{-1}^{2} \int_{x^2}^{x+2} xy\, dy\, dx = \int_{-1}^{2} x \left[\frac{y^2}{2}\right]_{x^2}^{x+2}\, dx

=12x2[(x+2)2x4]dx=1212[x(x+2)2x5]dx= \int_{-1}^{2} \frac{x}{2}\left[(x+2)^2 - x^4\right]\, dx = \frac{1}{2} \int_{-1}^{2} \left[x(x+2)^2 - x^5\right]\, dx

=1212[x3+4x2+4xx5]dx= \frac{1}{2} \int_{-1}^{2} \left[x^3 + 4x^2 + 4x - x^5\right]\, dx

=12[x44+4x33+2x2x66]12= \frac{1}{2}\left[\frac{x^4}{4} + \frac{4x^3}{3} + 2x^2 - \frac{x^6}{6}\right]_{-1}^{2}

=12[(4+323+8646)(1443+216)]= \frac{1}{2}\left[\left(4 + \frac{32}{3} + 8 - \frac{64}{6}\right) - \left(\frac{1}{4} - \frac{4}{3} + 2 - \frac{1}{6}\right)\right]

=12[363912]=12[1234]=458= \frac{1}{2}\left[\frac{36}{3} - \frac{9}{12}\right] = \frac{1}{2}\left[12 - \frac{3}{4}\right] = \frac{45}{8}

\blacksquare

Problem. Evaluate DxdA\iint_D x\, dA where DD is the region bounded by y=xy = x, y=2xy = 2xAnd x+y=2x + y = 2.

Solution

First, find the intersections. The lines y=xy = x and y=2xy = 2x intersect at (0,0)(0, 0). The line x+y=2x + y = 2 intersects y=xy = x at (1,1)(1, 1) and y=2xy = 2x at (2/3,4/3)(2/3, 4/3).

As a Type I region, we must split: for 0x2/30 \leq x \leq 2/3, xy2xx \leq y \leq 2x; for 2/3x12/3 \leq x \leq 1, xy2xx \leq y \leq 2 - x.

DxdA=02/3x2xxdydx+2/31x2xxdydx\iint_D x\, dA = \int_0^{2/3} \int_x^{2x} x\, dy\, dx + \int_{2/3}^1 \int_x^{2-x} x\, dy\, dx

=02/3x(xx)dx...= \int_0^{2/3} x(x - x)\, dx...

Wait, this is getting messy. Let me use Type II instead. For each yy, xx ranges from y/2y/2 to yy (for 0y4/30 \leq y \leq 4/3) and from y/2y/2 to 2y2 - y (for 4/3y14/3 \leq y \leq 1). Actually, the simplest approach is to split DD at y=4/3y = 4/3.

For 0y10 \leq y \leq 1: y/2xyy/2 \leq x \leq y (between y=xy = x and y=2xy = 2xBut only up to x+y=2x + y = 2). Actually y=2xy = 2x gives x=y/2x = y/2And y=xy = x gives x=yx = y. But x+y=2x + y = 2 gives x=2yx = 2 - y. For y1y \leq 1: both y2yy \leq 2 - y (since y1y \leq 1) and y/2yy/2 \leq ySo the right boundary is yy. But we also need x+y2x + y \leq 2I.e., x2yx \leq 2 - y. For y1y \leq 1: y2yy \leq 2 - ySo the constraint xyx \leq y is tighter.

For 0y10 \leq y \leq 1: y/2xyy/2 \leq x \leq y.

DxdA=01y/2yxdxdy=01[x22]y/2ydy=01y22y28dy=013y28dy=3813=18\iint_D x\, dA = \int_0^1 \int_{y/2}^y x\, dx\, dy = \int_0^1 \left[\frac{x^2}{2}\right]_{y/2}^y\, dy = \int_0^1 \frac{y^2}{2} - \frac{y^2}{8}\, dy = \int_0^1 \frac{3y^2}{8}\, dy = \frac{3}{8} \cdot \frac{1}{3} = \frac{1}{8}

\blacksquare

2.3 Triple Integrals

Triple integrals extend to R3\mathbb{R}^3:

Ef(x,y,z)dV=D(g1(x,y)g2(x,y)f(x,y,z)dz)dA\iiint_E f(x,y,z)\, dV = \iint_D \left(\int_{g_1(x,y)}^{g_2(x,y)} f(x,y,z)\, dz\right) dA

Problem. Evaluate EzdV\iiint_E z\, dV where EE is the tetrahedron in the first octant bounded by The coordinate planes and x+y+z=1x + y + z = 1.

Solution

The region EE can be described as (x,y,z):0x1,0y1x,0z1xy\\{(x,y,z) : 0 \leq x \leq 1,\, 0 \leq y \leq 1-x,\, 0 \leq z \leq 1-x-y\\}.

EzdV=0101x01xyzdzdydx\iiint_E z\, dV = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} z\, dz\, dy\, dx

=0101x[z22]01xydydx=0101x(1xy)22dydx= \int_0^1 \int_0^{1-x} \left[\frac{z^2}{2}\right]_0^{1-x-y}\, dy\, dx = \int_0^1 \int_0^{1-x} \frac{(1-x-y)^2}{2}\, dy\, dx

Substituting u=1xyu = 1 - x - y, du=dydu = -dy:

=01(1x)36dx=16[(1x)44]01=1614=124= \int_0^1 \frac{(1-x)^3}{6}\, dx = \frac{1}{6}\left[-\frac{(1-x)^4}{4}\right]_0^1 = \frac{1}{6} \cdot \frac{1}{4} = \frac{1}{24}

\blacksquare

2.4 Change of Variables

Theorem 2.2 (Change of Variables). Let T:DRnRnT : D \subseteq \mathbb{R}^n \to \mathbb{R}^n be a C1C^1 diffeomorphism with Jacobian determinant JTJ_T. Then

T(D)f(u)du=Df(T(x))JT(x)dx\int_{T(D)} f(\mathbf{u})\, d\mathbf{u} = \int_D f(T(\mathbf{x}))\, \lvert J_T(\mathbf{x})\rvert\, d\mathbf{x}

Derivation of the Jacobian factor (for n=2n = 2). Let T(x,y)=(u(x,y),v(x,y))T(x, y) = (u(x,y),\, v(x,y)) be a C1C^1 Diffeomorphism. Partition DD into small rectangles RijR_{ij} of area ΔxΔy\Delta x\, \Delta y. The image T(Rij)T(R_{ij}) is approximately a parallelogram spanned by the vectors

a=T(x+Δx,y)T(x,y)(uxΔx,vxΔx)\mathbf{a} = T(x + \Delta x, y) - T(x, y) \approx \left(\frac{\partial u}{\partial x}\Delta x,\, \frac{\partial v}{\partial x}\Delta x\right)

b=T(x,y+Δy)T(x,y)(uyΔy,vyΔy)\mathbf{b} = T(x, y + \Delta y) - T(x, y) \approx \left(\frac{\partial u}{\partial y}\Delta y,\, \frac{\partial v}{\partial y}\Delta y\right)

The area of this parallelogram is a×b\lvert \mathbf{a} \times \mathbf{b} \rvertWhich equals

uxvyuyvxΔxΔy=JTΔxΔy\left\lvert \frac{\partial u}{\partial x}\frac{\partial v}{\partial y} - \frac{\partial u}{\partial y}\frac{\partial v}{\partial x} \right\rvert \Delta x\, \Delta y = \lvert J_T \rvert\, \Delta x\, \Delta y

Summing over all subrectangles and taking the limit gives the change of variables formula. \blacksquare

Polar coordinates: x=rcosθx = r\cos\theta, y=rsinθy = r\sin\theta, J=r\lvert J \rvert = r.

Df(x,y)dA=Df(rcosθ,rsinθ)rdrdθ\iint_D f(x,y)\, dA = \iint_{D'} f(r\cos\theta, r\sin\theta)\, r\, dr\, d\theta

Cylindrical coordinates: x=rcosθx = r\cos\theta, y=rsinθy = r\sin\theta, z=zz = z, J=r\lvert J \rvert = r.

Ef(x,y,z)dV=Ef(rcosθ,rsinθ,z)rdrdθdz\iiint_E f(x,y,z)\, dV = \iiint_{E'} f(r\cos\theta, r\sin\theta, z)\, r\, dr\, d\theta\, dz

Spherical coordinates: x=ρsinϕcosθx = \rho\sin\phi\cos\theta, y=ρsinϕsinθy = \rho\sin\phi\sin\theta, z=ρcosϕz = \rho\cos\phi J=ρ2sinϕ\lvert J \rvert = \rho^2 \sin\phi.

Ef(x,y,z)dV=Ef(ρsinϕcosθ,ρsinϕsinθ,ρcosϕ)ρ2sinϕdρdϕdθ\iiint_E f(x,y,z)\, dV = \iiint_{E'} f(\rho\sin\phi\cos\theta, \rho\sin\phi\sin\theta, \rho\cos\phi)\, \rho^2 \sin\phi\, d\rho\, d\phi\, d\theta

2.5 Coordinate System Worked Examples

Problem. Evaluate De(x2+y2)dA\iint_D e^{-(x^2+y^2)}\, dA where DD is the entire R2\mathbb{R}^2 plane.

Solution

Use polar coordinates. The region DD' is 0r<0 \leq r \lt \infty, 0θ2π0 \leq \theta \leq 2\pi.

De(x2+y2)dA=02π0er2rdrdθ\iint_D e^{-(x^2+y^2)}\, dA = \int_0^{2\pi} \int_0^{\infty} e^{-r^2}\, r\, dr\, d\theta

The inner integral: 0rer2dr=[12er2]0=12\int_0^{\infty} r e^{-r^2}\, dr = \left[-\frac{1}{2}e^{-r^2}\right]_0^{\infty} = \frac{1}{2}.

=02π12dθ=π= \int_0^{2\pi} \frac{1}{2}\, d\theta = \pi

\blacksquare

Remark. This is the classic Gaussian integral computation, yielding ex2dx=π\int_{-\infty}^{\infty} e^{-x^2}\, dx = \sqrt{\pi}.

Problem. Evaluate EzdV\iiint_E z\, dV where EE is the solid bounded above by the sphere x2+y2+z2=2x^2 + y^2 + z^2 = 2 and below by the paraboloid z=x2+y2z = x^2 + y^2.

Solution

The surfaces intersect when x2+y2+(x2+y2)2=2x^2 + y^2 + (x^2 + y^2)^2 = 2. Let r2=x2+y2r^2 = x^2 + y^2. Then r2+r4=2r^2 + r^4 = 2I.e., (r2+2)(r21)=0(r^2 + 2)(r^2 - 1) = 0So r=1r = 1 (positive root). Use Cylindrical coordinates. The region EE' is

0r1,0θ2π,r2z2r20 \leq r \leq 1, \quad 0 \leq \theta \leq 2\pi, \quad r^2 \leq z \leq \sqrt{2 - r^2}

EzdV=02π01r22r2zrdzdrdθ\iiint_E z\, dV = \int_0^{2\pi} \int_0^1 \int_{r^2}^{\sqrt{2-r^2}} z\, r\, dz\, dr\, d\theta

=02π01r2[(2r2)r4]drdθ=02π01r2(2r2r4)drdθ= \int_0^{2\pi} \int_0^1 \frac{r}{2}\left[(2 - r^2) - r^4\right]\, dr\, d\theta = \int_0^{2\pi} \int_0^1 \frac{r}{2}(2 - r^2 - r^4)\, dr\, d\theta

=02π12[r2r44r66]01dθ=02π12712dθ=7π12= \int_0^{2\pi} \frac{1}{2}\left[r^2 - \frac{r^4}{4} - \frac{r^6}{6}\right]_0^1\, d\theta = \int_0^{2\pi} \frac{1}{2} \cdot \frac{7}{12}\, d\theta = \frac{7\pi}{12}

\blacksquare

Problem. Evaluate E(x2+y2+z2)dV\iiint_E (x^2 + y^2 + z^2)\, dV where EE is the solid ball x2+y2+z2a2x^2 + y^2 + z^2 \leq a^2.

Solution

Use spherical coordinates. In spherical: x2+y2+z2=ρ2x^2 + y^2 + z^2 = \rho^2And EE' is 0ρa0 \leq \rho \leq a, 0ϕπ0 \leq \phi \leq \pi, 0θ2π0 \leq \theta \leq 2\pi.

E(x2+y2+z2)dV=02π0π0aρ2ρ2sinϕdρdϕdθ\iiint_E (x^2 + y^2 + z^2)\, dV = \int_0^{2\pi} \int_0^{\pi} \int_0^a \rho^2 \cdot \rho^2 \sin\phi\, d\rho\, d\phi\, d\theta

=(0aρ4dρ)(0πsinϕdϕ)(02πdθ)= \left(\int_0^a \rho^4\, d\rho\right)\left(\int_0^{\pi} \sin\phi\, d\phi\right)\left(\int_0^{2\pi} d\theta\right)

=a5522π=4πa55= \frac{a^5}{5} \cdot 2 \cdot 2\pi = \frac{4\pi a^5}{5}

\blacksquare

2.6 Worked Example

Problem. Compute D(x2+y2)dA\iint_D (x^2 + y^2)\, dA where DD is the region bounded by x2+y2=4x^2 + y^2 = 4.

Solution. Use polar coordinates. The region DD' is 0r20 \leq r \leq 2, 0θ2π0 \leq \theta \leq 2\pi.

D(x2+y2)dA=02π02r2rdrdθ=02π02r3drdθ\iint_D (x^2 + y^2)\, dA = \int_0^{2\pi} \int_0^2 r^2 \cdot r\, dr\, d\theta = \int_0^{2\pi} \int_0^2 r^3\, dr\, d\theta

=02π[r44]02dθ=02π4dθ=8π= \int_0^{2\pi} \left[\frac{r^4}{4}\right]_0^2 d\theta = \int_0^{2\pi} 4\, d\theta = 8\pi

\blacksquare

Problem. Evaluate DyxdA\iint_D \frac{y}{x}\, dA where DD is bounded by y=xy = x, y=2xy = 2xAnd x=1x = 1.

Solution

The region D=(x,y):0x1,xy2xD = \\{(x,y) : 0 \leq x \leq 1,\, x \leq y \leq 2x\\}.

DyxdA=01x2xyxdydx=011x[y22]x2xdx\iint_D \frac{y}{x}\, dA = \int_0^1 \int_x^{2x} \frac{y}{x}\, dy\, dx = \int_0^1 \frac{1}{x}\left[\frac{y^2}{2}\right]_x^{2x}\, dx

=011x[4x22x22]dx=011x3x22dx=3201xdx=34= \int_0^1 \frac{1}{x}\left[\frac{4x^2}{2} - \frac{x^2}{2}\right]\, dx = \int_0^1 \frac{1}{x} \cdot \frac{3x^2}{2}\, dx = \frac{3}{2}\int_0^1 x\, dx = \frac{3}{4}

\blacksquare

Problem. Swap the order of integration and evaluate: 01x21xey2dydx\int_0^1 \int_{x^2}^1 x e^{y^2}\, dy\, dx.

Solution

The region is 0x10 \leq x \leq 1, x2y1x^2 \leq y \leq 1Which is the same as 0y10 \leq y \leq 1 0xy0 \leq x \leq \sqrt{y}.

01x21xey2dydx=010yxey2dxdy=01ey2[x22]0ydy\int_0^1 \int_{x^2}^1 x e^{y^2}\, dy\, dx = \int_0^1 \int_0^{\sqrt{y}} x e^{y^2}\, dx\, dy = \int_0^1 e^{y^2}\left[\frac{x^2}{2}\right]_0^{\sqrt{y}}\, dy

=01y2ey2dy= \int_0^1 \frac{y}{2} e^{y^2}\, dy

Let u=y2u = y^2, du=2ydydu = 2y\, dy:

=1401eudu=14(e1)= \frac{1}{4}\int_0^1 e^u\, du = \frac{1}{4}(e - 1)

\blacksquare

Remark. This integral cannot be evaluated in the original order because ey2e^{y^2} has no elementary Antiderivative with respect to yy. Swapping the order was essential.

2.7 Common Pitfalls

:::caution Common Pitfalls

  • Order of integration limits. When setting up abg1(x)g2(x)fdydx\int_a^b \int_{g_1(x)}^{g_2(x)} f\, dy\, dxVerify that g1(x)g2(x)g_1(x) \leq g_2(x) for all x[a,b]x \in [a, b]. If the region is described as “between two curves,” determine which curve is above the other.
  • Forgetting the Jacobian. In a change of variables, the Jacobian determinant J\lvert J \rvert must be included. For polar coordinates, this factor is rr; omitting it is one of the most common errors.
  • Spherical coordinate conventions. Different texts use different conventions for ϕ\phi and θ\theta. Here, ϕ[0,π]\phi \in [0, \pi] is the polar angle (from the positive zz-axis) and θ[0,2π]\theta \in [0, 2\pi] is the azimuthal angle.
  • Region description. When swapping integration order, carefully redraw the region and re-derive the bounds. The new bounds may require splitting the integral into multiple pieces.

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