Multiple Integrals

2.1 Double Integrals

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

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

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

$\iint_R f(x,y)\, dA = \int_a^b \left(\int_c^d f(x,y)\, dy\right) dx = \int_c^d \left(\int_a^b f(x,y)\, dx\right) dy$

Proof (sketch). For a continuous function $f$ on the compact rectangle $R$ Define

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

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

$S(P) = \sum_{i=1}^m F(x_i^*)\, \Delta x_i = \sum_{i=1}^m \int_c^d f(x_i^*, y)\, dy\, \Delta x_i$

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

2.2 General Regions

For a general region $D$ in $\mathbb{R}^2$ :

Type I region: $D = \\{(x,y) : a \leq x \leq b,\, g_1(x) \leq y \leq g_2(x)\\}$

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

Type II region: $D = \\{(x,y) : c \leq y \leq d,\, h_1(y) \leq x \leq h_2(y)\\}$

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

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

Solution

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

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

$= \int_{-1}^{2} \frac{x}{2}\left[(x+2)^2 - x^4\right]\, dx = \frac{1}{2} \int_{-1}^{2} \left[x(x+2)^2 - x^5\right]\, dx$

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

$= \frac{1}{2}\left[\frac{x^4}{4} + \frac{4x^3}{3} + 2x^2 - \frac{x^6}{6}\right]_{-1}^{2}$

$= \frac{1}{2}\left[\left(4 + \frac{32}{3} + 8 - \frac{64}{6}\right) - \left(\frac{1}{4} - \frac{4}{3} + 2 - \frac{1}{6}\right)\right]$

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

$\blacksquare$

Problem. Evaluate $\iint_D x\, dA$ where $D$ is the region bounded by $y = x$ , $y = 2x$ And $x + y = 2$ .

Solution

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

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

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

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

Wait, this is getting messy. Let me use Type II instead. For each $y$ , $x$ ranges from $y/2$ to $y$ (for $0 \leq y \leq 4/3$ ) and from $y/2$ to $2 - y$ (for $4/3 \leq y \leq 1$ ). Actually, the simplest approach is to split $D$ at $y = 4/3$ .

For $0 \leq y \leq 1$ : $y/2 \leq x \leq y$ (between $y = x$ and $y = 2x$ But only up to $x + y = 2$ ). Actually $y = 2x$ gives $x = y/2$ And $y = x$ gives $x = y$ . But $x + y = 2$ gives $x = 2 - y$ . For $y \leq 1$ : both $y \leq 2 - y$ (since $y \leq 1$ ) and $y/2 \leq y$ So the right boundary is $y$ . But we also need $x + y \leq 2$ I.e., $x \leq 2 - y$ . For $y \leq 1$ : $y \leq 2 - y$ So the constraint $x \leq y$ is tighter.

For $0 \leq y \leq 1$ : $y/2 \leq x \leq y$ .

$\iint_D x\, dA = \int_0^1 \int_{y/2}^y x\, dx\, dy = \int_0^1 \left[\frac{x^2}{2}\right]_{y/2}^y\, dy = \int_0^1 \frac{y^2}{2} - \frac{y^2}{8}\, dy = \int_0^1 \frac{3y^2}{8}\, dy = \frac{3}{8} \cdot \frac{1}{3} = \frac{1}{8}$

$\blacksquare$

2.3 Triple Integrals

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

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

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

Solution

The region $E$ can be described as $\\{(x,y,z) : 0 \leq x \leq 1,\, 0 \leq y \leq 1-x,\, 0 \leq z \leq 1-x-y\\}$ .

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

$= \int_0^1 \int_0^{1-x} \left[\frac{z^2}{2}\right]_0^{1-x-y}\, dy\, dx = \int_0^1 \int_0^{1-x} \frac{(1-x-y)^2}{2}\, dy\, dx$

Substituting $u = 1 - x - y$ , $du = -dy$ :

$= \int_0^1 \frac{(1-x)^3}{6}\, dx = \frac{1}{6}\left[-\frac{(1-x)^4}{4}\right]_0^1 = \frac{1}{6} \cdot \frac{1}{4} = \frac{1}{24}$

$\blacksquare$

2.4 Change of Variables

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

$\int_{T(D)} f(\mathbf{u})\, d\mathbf{u} = \int_D f(T(\mathbf{x}))\, \lvert J_T(\mathbf{x})\rvert\, d\mathbf{x}$

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

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

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

The area of this parallelogram is $\lvert \mathbf{a} \times \mathbf{b} \rvert$ Which equals

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

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

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

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

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

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

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

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

2.5 Coordinate System Worked Examples

Problem. Evaluate $\iint_D e^{-(x^2+y^2)}\, dA$ where $D$ is the entire $\mathbb{R}^2$ plane.

Solution

Use polar coordinates. The region $D'$ is $0 \leq r \lt \infty$ , $0 \leq \theta \leq 2\pi$ .

$\iint_D e^{-(x^2+y^2)}\, dA = \int_0^{2\pi} \int_0^{\infty} e^{-r^2}\, r\, dr\, d\theta$

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

$= \int_0^{2\pi} \frac{1}{2}\, d\theta = \pi$

$\blacksquare$

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

Problem. Evaluate $\iiint_E z\, dV$ where $E$ is the solid bounded above by the sphere $x^2 + y^2 + z^2 = 2$ and below by the paraboloid $z = x^2 + y^2$ .

Solution

The surfaces intersect when $x^2 + y^2 + (x^2 + y^2)^2 = 2$ . Let $r^2 = x^2 + y^2$ . Then $r^2 + r^4 = 2$ I.e., $(r^2 + 2)(r^2 - 1) = 0$ So $r = 1$ (positive root). Use Cylindrical coordinates. The region $E'$ is

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

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

$= \int_0^{2\pi} \int_0^1 \frac{r}{2}\left[(2 - r^2) - r^4\right]\, dr\, d\theta = \int_0^{2\pi} \int_0^1 \frac{r}{2}(2 - r^2 - r^4)\, dr\, d\theta$

$= \int_0^{2\pi} \frac{1}{2}\left[r^2 - \frac{r^4}{4} - \frac{r^6}{6}\right]_0^1\, d\theta = \int_0^{2\pi} \frac{1}{2} \cdot \frac{7}{12}\, d\theta = \frac{7\pi}{12}$

$\blacksquare$

Problem. Evaluate $\iiint_E (x^2 + y^2 + z^2)\, dV$ where $E$ is the solid ball $x^2 + y^2 + z^2 \leq a^2$ .

Solution

Use spherical coordinates. In spherical: $x^2 + y^2 + z^2 = \rho^2$ And $E'$ is $0 \leq \rho \leq a$ , $0 \leq \phi \leq \pi$ , $0 \leq \theta \leq 2\pi$ .

$\iiint_E (x^2 + y^2 + z^2)\, dV = \int_0^{2\pi} \int_0^{\pi} \int_0^a \rho^2 \cdot \rho^2 \sin\phi\, d\rho\, d\phi\, d\theta$

$= \left(\int_0^a \rho^4\, d\rho\right)\left(\int_0^{\pi} \sin\phi\, d\phi\right)\left(\int_0^{2\pi} d\theta\right)$

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

$\blacksquare$

2.6 Worked Example

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

Solution. Use polar coordinates. The region $D'$ is $0 \leq r \leq 2$ , $0 \leq \theta \leq 2\pi$ .

$\iint_D (x^2 + y^2)\, dA = \int_0^{2\pi} \int_0^2 r^2 \cdot r\, dr\, d\theta = \int_0^{2\pi} \int_0^2 r^3\, dr\, d\theta$

$= \int_0^{2\pi} \left[\frac{r^4}{4}\right]_0^2 d\theta = \int_0^{2\pi} 4\, d\theta = 8\pi$

$\blacksquare$

Problem. Evaluate $\iint_D \frac{y}{x}\, dA$ where $D$ is bounded by $y = x$ , $y = 2x$ And $x = 1$ .

Solution

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

$\iint_D \frac{y}{x}\, dA = \int_0^1 \int_x^{2x} \frac{y}{x}\, dy\, dx = \int_0^1 \frac{1}{x}\left[\frac{y^2}{2}\right]_x^{2x}\, dx$

$= \int_0^1 \frac{1}{x}\left[\frac{4x^2}{2} - \frac{x^2}{2}\right]\, dx = \int_0^1 \frac{1}{x} \cdot \frac{3x^2}{2}\, dx = \frac{3}{2}\int_0^1 x\, dx = \frac{3}{4}$

$\blacksquare$

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

Solution

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

$\int_0^1 \int_{x^2}^1 x e^{y^2}\, dy\, dx = \int_0^1 \int_0^{\sqrt{y}} x e^{y^2}\, dx\, dy = \int_0^1 e^{y^2}\left[\frac{x^2}{2}\right]_0^{\sqrt{y}}\, dy$

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

Let $u = y^2$ , $du = 2y\, dy$ :

$= \frac{1}{4}\int_0^1 e^u\, du = \frac{1}{4}(e - 1)$

$\blacksquare$

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

2.7 Common Pitfalls

:::caution Common Pitfalls

Order of integration limits. When setting up $\int_a^b \int_{g_1(x)}^{g_2(x)} f\, dy\, dx$ Verify that $g_1(x) \leq g_2(x)$ for all $x \in [a, b]$ . 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 $\lvert J \rvert$ must be included. For polar coordinates, this factor is $r$ ; omitting it is one of the most common errors.
Spherical coordinate conventions. Different texts use different conventions for $\phi$ and $\theta$ . Here, $\phi \in [0, \pi]$ is the polar angle (from the positive $z$ -axis) and $\theta \in [0, 2\pi]$ 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.

:::

Mark this page as reviewed