Multiple 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. $\blacksquare$

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

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

$\blacksquare$

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

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

$\blacksquare$

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

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

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 $0 \leq \rho \leq a$, $0 \leq \phi \leq \pi$, $0 \leq \theta \leq 2\pi$.

$$\iiint_E (x^2 + y^2 + z^2)\, dV = \int_0^{2\pi} \int_0^{\pi} \int_0^a \rho^2 \cdot \rho^2 \sin\phi\, d\rho\, d\phi\, d\theta$$ $$= \left(\int_0^a \rho^4\, d\rho\right)\left(\int_0^{\pi} \sin\phi\, d\phi\right)\left(\int_0^{2\pi} d\theta\right)$$ $$= \frac{a^5}{5} \cdot 2 \cdot 2\pi = \frac{4\pi a^5}{5}$$

$\blacksquare$

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.

A multiple integral extends the idea of area under a curve to higher dimensions. A double integral computes the signed volume under a surface $z = f(x,y)$ 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 $x$ and integrating over $y$, 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.

• Partial Derivatives: The Jacobian determinant in change of variables is computed from partial derivatives of the coordinate transformation.
• Vector Calculus: Surface integrals and flux integrals extend multiple integrals to curved surfaces in three dimensions.
• Optimization: Multiple integrals are used to compute volumes and averages needed in constrained optimisation problems.
• Curves and Surfaces: Surface area and surface integrals are computed using parametrisations and the cross product of partial derivatives.
• Quantum Mechanics
• Graph Theory
• Classical Mechanics
• Electromagnetism

• Curves and Surfaces
• Optimization
• Partial Derivatives
• Problem Set