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Multiple Integrals | Mathematics

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

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 = 0 So (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 = 2x And 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 = 2x But only up to x+y=2x + y = 2). Actually y=2xy = 2x gives x=y/2x = y/2 And 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 y So 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 - y So 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

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

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

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) = 0 So 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^2 And 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

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 = 2x And 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 Intuition: What Are Multiple Integrals?

Section titled “2.7 Intuition: What Are Multiple Integrals?”

A multiple integral extends the idea of area under a curve to higher dimensions. A double integral computes the signed volume under a surface z=f(x,y)z = f(x,y) over a region in the plane, while a triple integral computes the integral of a function over a three-dimensional region. Fubini’s theorem says you can evaluate these integrals one variable at a time, converting a double integral into iterated single integrals, which is the computational workhorse.

The key to setting up multiple integrals is describing the region correctly. Type I and Type II regions in the plane correspond to fixing xx and integrating over yy, or vice versa. When the region has a complicated shape, you may need to split the integral or change the order of integration. Changes of variables, such as polar or spherical coordinates, simplify the integrand or the region at the cost of introducing a Jacobian determinant that accounts for how the coordinate transformation stretches or compresses space. The Jacobian is the multivariable analogue of the chain rule: it measures how much a small volume element is distorted by the transformation.

flowchart TD
A[2_Multiple Integrals] --> B[Key Concepts]
A --> C[Core Principles]
A --> D[Practical Applications]
B --> E[Fundamental definitions]
C --> F[Design patterns]
D --> G[Real-world usage]