Analysis

Multiple integrals

Integration over regions in two or more dimensions, computing volumes and higher-dimensional totals.

IntuitionFrom slices of area to columns of volume

A single-variable integral $\int_a^b f(x)\,dx$ chops an interval into thin segments of width $\Delta x$ and sums the areas of rectangles. A double integral $\iint_D f(x,y)\,dA$ extends the exact same idea to a two-dimensional domain $D$ : we tile $D$ with tiny rectangles of area $\Delta A = \Delta x\,\Delta y$ , erect a thin column of height $f(x_i,y_j)$ over each tile, and let the grid refine. Adding up the columns gives the signed volume under the surface $z = f(x,y)$ , or the total mass of a plate whose surface density is $f(x,y)$ .

The surface

z = f(x,y)

over a planar domain

D

: the double integral

\iint_D f(x,y)\,dA

accumulates the volume of vertical columns under the surface, and Fubini's theorem computes it by integrating slice by slice.

UndergraduateFormal definitions: Double and triple integrals, Fubini reduction, and Jacobians

Definition: Double and triple Riemann integrals

For a bounded function $f : D \subseteq \mathbb{R}^2 \to \mathbb{R}$ on a bounded measurable domain $D$ , the double integral $\iint_D f(x,y)\,dA$ is the limit of Riemann sums $\sum_{i,j} f(x_i^*, y_j^*)\,\Delta x_i\,\Delta y_j$ as the mesh size of the partition approaches $0$ . Analogously, over a solid region $E \subseteq \mathbb{R}^3$ , the triple integral $\iiint_E f(x,y,z)\,dV$ sums over small boxes of volume $\Delta V = \Delta x\,\Delta y\,\Delta z$ .

\iint_D f(x,y)\,dA = \int_a^b \left(\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\right) dx = \int_c^d \left(\int_{h_1(y)}^{h_2(y)} f(x,y)\,dx\right) dy

When a region is circular, cylindrical, or spherical, a smooth coordinate transformation $\Phi(u,v) = (x(u,v), y(u,v))$ simplifies the boundary limits. Under $\Phi$ , a tiny $du \times dv$ rectangle is stretched and sheared into a parallelogram whose area is scaled by the absolute value of the Jacobian determinant $|\det D\Phi| = \left|\frac{\partial(x,y)}{\partial(u,v)}\right|$ . In polar coordinates $(x,y) = (r\cos\theta, r\sin\theta)$ this factor is $r$ , and in spherical coordinates $(x,y,z) = (\rho\sin\varphi\cos\theta, \rho\sin\varphi\sin\theta, \rho\cos\varphi)$ it is $\rho^2\sin\varphi$ .

\iint_{\Phi(U)} f(x,y)\,dx\,dy = \iint_U f(\Phi(u,v))\,|\det D\Phi(u,v)|\,du\,dv, \qquad \int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}

Coordinate systems and Jacobian volume elements in 2D and 3D
Coordinate system	Transformation	Differential element
Cartesian 2D	$(x,y)$	$dA = dx\,dy$
Polar 2D	$x = r\cos\theta,\; y = r\sin\theta$	$dA = r\,dr\,d\theta$
Cylindrical 3D	$x = r\cos\theta,\; y = r\sin\theta,\; z = z$	$dV = r\,dr\,d\theta\,dz$
Spherical 3D	$x = \rho\sin\varphi\cos\theta,\; y = \rho\sin\varphi\sin\theta,\; z = \rho\cos\varphi$	$dV = \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$

UndergraduateCore theorems: Fubini's theorem and Jacobian change of variables

Theorem: Fubini's theorem for iterated integrals

If $f(x,y)$ is continuous on the rectangle $R = [a,b] \times [c,d]$ , then the double integral equals both iterated single integrals: $\iint_R f(x,y)\,dA = \int_a^b \left(\int_c^d f(x,y)\,dy\right) dx = \int_c^d \left(\int_a^b f(x,y)\,dx\right) dy$ . More generally, on a vertically simple region $D = \{(x,y) : a \le x \le b,\; g_1(x) \le y \le g_2(x)\}$ , we have $\iint_D f(x,y)\,dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx$ .

Why is it true?

Computing a volume by summing tiny boxes in a grid gives the same answer whether you first add each column along $y$ to get cross-sectional slice areas $A(x)$ and then integrate $A(x)$ along $x$ , or slice perpendicular to the $y$ -axis first.

Proof

Partition $[a,b]$ into $m$ subintervals $[x_{i-1}, x_i]$ of width $\Delta x$ and $[c,d]$ into $n$ subintervals $[y_{j-1}, y_j]$ of width $\Delta y$ . For each fixed $x$ , define the cross-sectional integral $A(x) = \int_c^d f(x,y)\,dy$ . By the Mean Value Theorem for integrals, on each strip $[y_{j-1}, y_j]$ there is $y_{ij}^* \in [y_{j-1}, y_j]$ such that $\int_{y_{j-1}}^{y_j} f(x_i, y)\,dy = f(x_i, y_{ij}^*)\,\Delta y$ .

Summing over $j = 1, \dots, n$ gives $A(x_i) = \sum_{j=1}^n f(x_i, y_{ij}^*)\,\Delta y$ . Multiplying by $\Delta x$ and summing over $i = 1, \dots, m$ yields $\sum_{i=1}^m A(x_i)\,\Delta x = \sum_{i=1}^m \sum_{j=1}^n f(x_i, y_{ij}^*)\,\Delta x\,\Delta y$ . Because $f$ is uniformly continuous on the compact rectangle $R$ , letting $\Delta x, \Delta y \to 0$ makes the left-hand side converge to $\int_a^b A(x)\,dx = \int_a^b \left(\int_c^d f(x,y)\,dy\right) dx$ while the right-hand side converges to $\iint_R f(x,y)\,dA$ . Repeating the argument with $x$ and $y$ swapped establishes equality with the other order of integration.

Theorem: Change of variables theorem and the Jacobian determinant

Let $\Phi : U \to \Phi(U) \subseteq \mathbb{R}^2$ be a $C^1$ diffeomorphism with Jacobian matrix $D\Phi(u,v) = \begin{pmatrix} x_u & x_v \\ y_u & y_v \end{pmatrix}$ . For any integrable function $f$ on $\Phi(U)$ , we have $\iint_{\Phi(U)} f(x,y)\,dx\,dy = \iint_U f(x(u,v), y(u,v))\,|\det D\Phi(u,v)|\,du\,dv$ , where $\det D\Phi = x_u y_v - x_v y_u$ .

Why is it true?

Just as $dx = g'(u)\,du$ rescales length in one-variable substitution, $|\det D\Phi(u,v)|$ measures the local area-stretching ratio when a tiny $(u,v)$ -rectangle is mapped to an $(x,y)$ -parallelogram.

Proof

Consider a small rectangle $R_{ij} = [u_i, u_i + \Delta u] \times [v_j, v_j + \Delta v]$ in $U$ . By first-order Taylor expansion around $(u_i, v_j)$ , the edges $( \Delta u, 0 )$ and $( 0, \Delta v )$ map approximately to the tangent vectors $\mathbf{a} = \Phi_u(u_i,v_j)\,\Delta u = (x_u, y_u)\,\Delta u$ and $\mathbf{b} = \Phi_v(u_i,v_j)\,\Delta v = (x_v, y_v)\,\Delta v$ . The area of the parallelogram spanned by $\mathbf{a}$ and $\mathbf{b}$ in $\mathbb{R}^2$ equals $|x_u y_v - x_v y_u|\,\Delta u\,\Delta v = |\det D\Phi(u_i, v_j)|\,\Delta u\,\Delta v$ up to higher-order error $o(\Delta u\,\Delta v)$ .

Substituting $\Delta A_{ij} \approx |\det D\Phi(u_i, v_j)|\,\Delta u\,\Delta v$ into the Riemann sum $\sum_{i,j} f(\Phi(u_i, v_j))\,\Delta A_{ij}$ yields $\sum_{i,j} f(\Phi(u_i, v_j))\,|\det D\Phi(u_i, v_j)|\,\Delta u\,\Delta v$ . In particular, for polar coordinates $x = r\cos\theta$ , $y = r\sin\theta$ , we have $\det D\Phi = (\cos\theta)(r\cos\theta) - (-r\sin\theta)(\sin\theta) = r(\cos^2\theta + \sin^2\theta) = r$ , giving $dx\,dy = r\,dr\,d\theta$ .

UndergraduateReal-World Applications and Worked Examples

Multiple integrals appear across physics, probability, and engineering: in mechanics, triple integrals compute the total mass $M = \iiint_E \rho(x,y,z)\,dV$ , center of mass, and moment of inertia $I_z = \iiint_E (x^2+y^2)\rho\,dV$ of rotating machinery; in probability and statistics, normalizing the normal distribution relies on the Gaussian integral $\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}$ , proved by squaring into a double integral in polar coordinates; and in astrophysics and electromagnetism, spherical integrals $\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$ integrate gravitational and charge distributions over stars and planets.

Example: Reversing the order of integration when the inner integral has no elementary antiderivative

Evaluate the iterated integral $I = \int_0^1 \int_y^1 e^{x^2}\,dx\,dy$ by sketching the region of integration and reversing the order of integration via Fubini's theorem.

Solution

The inner integral $\int_y^1 e^{x^2}\,dx$ cannot be evaluated directly in closed form because $e^{x^2}$ has no elementary antiderivative. However, the inequalities $0 \le y \le 1$ and $y \le x \le 1$ describe the triangle $D = \{(x,y) : 0 \le x \le 1,\; 0 \le y \le x\}$ .

By Fubini's theorem, slicing vertically first rewrites the integral as $I = \int_0^1 \left(\int_0^x e^{x^2}\,dy\right) dx = \int_0^1 x e^{x^2}\,dx$ . Now substitute $u = x^2$ , $du = 2x\,dx$ to obtain $I = \left[\frac{1}{2}e^{x^2}\right]_0^1 = \frac{e - 1}{2}$ .

Example: Evaluating the Gaussian integral via polar coordinates

Prove the Gaussian integral formula $I = \int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}$ by expressing $I^2$ as a double integral over $\mathbb{R}^2$ and converting to polar coordinates.

Solution

Write the product of two independent copies of $I$ using dummy variables $x$ and $y$ : $I^2 = \left(\int_{-\infty}^{\infty} e^{-x^2}\,dx\right)\left(\int_{-\infty}^{\infty} e^{-y^2}\,dy\right) = \iint_{\mathbb{R}^2} e^{-(x^2+y^2)}\,dx\,dy$ .

Switch to polar coordinates $x = r\cos\theta$ , $y = r\sin\theta$ with $0 \le r < \infty$ , $0 \le \theta \le 2\pi$ , and Jacobian factor $dx\,dy = r\,dr\,d\theta$ . Then $I^2 = \int_0^{2\pi} \int_0^{\infty} e^{-r^2} r\,dr\,d\theta = 2\pi \left[-\frac{1}{2}e^{-r^2}\right]_0^{\infty} = 2\pi \cdot \frac{1}{2} = \pi$ . Since $e^{-x^2} > 0$ , taking the positive square root yields $I = \sqrt{\pi}$ .

Evaluate the double integral $\iint_R (2x + 4y)\,dA$ over the rectangle $R = [0, 2] \times [0, 1]$ .

When reversing the order of integration in $\int_0^2 \int_{x^2}^4 f(x,y)\,dy\,dx$ , which iterated integral is obtained?

In 3D spherical coordinates $(x,y,z) = (\rho\sin\varphi\cos\theta, \rho\sin\varphi\sin\theta, \rho\cos\varphi)$ , what is the volume element $dV$ ?

Using the Gaussian integral $\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}$ , what is the value of the 2D probability integral $\iint_{\mathbb{R}^2} e^{-(x^2+y^2)/2}\,dx\,dy$ ?

References

Jerrold E. Marsden, Anthony J. Tromba (2012). Vector Calculus
Tom M. Apostol (1974). Mathematical Analysis
Tom M. Apostol (1969). Calculus, Vol. 2: Multi-Variable Calculus and Linear Algebra with Applications