Multiple Integrals and Their Calculation

Why Multiple Integrals Are Needed

A one-dimensional integral finds the area under a curve. Multiple integrals generalize this to several dimensions: the double integral computes the volume under a surface, the mass of a flat plate with variable density, the probability of hitting a region for a two-dimensional distribution. Triple integrals — volumes, centers of mass, moments of inertia of three-dimensional bodies. Physics, mechanics, probability theory — all are unthinkable without multiple integrals.

Historically, multiple integrals arose with Leibniz and Newton in solving mechanics problems, but a rigorous justification appeared only with Cauchy and Riemann in the 19th century. The key obstacle — the region of integration can be an arbitrary shape, not a rectangle. Fubini’s theorem, which we will discuss below, is the central tool, reducing computation to successive one-dimensional integrals.

Double Integral: Definition and Geometric Meaning

Let $f(x,y)$ be defined and bounded on a closed region $D \subset \mathbb{R}^2$. Divide $D$ into small pieces with areas $\Delta A_k$, choose a point $(\xi_k, \eta_k)$ in each, and form the integral sum $\sum f(\xi_k, \eta_k) \Delta A_k$. If this limit exists as the partition becomes finer and does not depend on the method of partitioning or the choice of points, it is called a double integral:

$ \iint_D f(x,y), dA = \lim \sum f(\xi_k, \eta_k) \Delta A_k. $

Geometric meaning: if $f(x,y) \geq 0$, then the double integral equals the volume under the surface $z = f(x,y)$ above region $D$. For $f = 1$ we get the area of the region: $S(D) = \iint_D dA$.

Physical meaning: if $\rho(x,y)$ is surface density (kg/m²), then the mass of the body $M = \iint_D \rho(x,y) , dA$. Coordinates of the center of mass: $\overline{x} = (1/M) \iint_D x,\rho, dA$, $\overline{y} = (1/M) \iint_D y,\rho, dA$. Moment of inertia relative to the origin: $I_0 = \iint_D (x^2 + y^2), \rho, dA$.

Fubini’s Theorem: Reduction to Iterated Integral

If the region $D$ is “regular” (x-simple or y-simple) and $f$ is continuous on $D$, then the double integral reduces to successive ordinary integrals.

For x-simple region $D = {(x, y): a \leq x \leq b,, \varphi_1(x) \leq y \leq \varphi_2(x)}$:

$ \iint_D f, dA = \int_a^b \left[\int_{\varphi_1(x)}^{\varphi_2(x)} f(x,y) , dy\right] dx. $

Meaning: fix $x$ (a vertical “layer”), integrate over $y$ within the layer, then sum all layers over $x$.

Expanded example. Compute $\iint_D (x^2 + y^2), dA$, where $D$ is the triangle with vertices $(0,0)$, $(1,0)$, $(1,1)$.

Description of the region: $0 \leq x \leq 1$, $0 \leq y \leq x$.

$ \int_0^1 \left[\int_0^x (x^2 + y^2), dy\right] dx = \int_0^1 \left[x^2y + \frac{y^3}{3}\right]_0^x dx = \int_0^1 (x^3 + \frac{x^3}{3}) dx = \int_0^1 \frac{4x^3}{3} dx = \left[\frac{x^4}{3}\right]_0^1 = \frac{1}{3}. $

Changing the order of integration. Sometimes an integral in one order cannot be solved analytically, but is easily computed in another. The same triangle as a y-simple region: $0 \leq y \leq 1$, $y \leq x \leq 1$.

$ \int_0^1 \left[\int_y^1 (x^2 + y^2), dx\right] dy = \int_0^1 \left[\frac{x^3}{3} + y^2x\right]_y^1 dy = \int_0^1 \left(\frac{1}{3} + y^2 - \frac{y^3}{3} - y^3\right) dy = \frac{1}{3} + \frac{1}{3} - \frac{1}{12} - \frac{1}{4} = \frac{1}{3}. , \checkmark $

Change of Variables and the Jacobian

When making the substitution $(x, y) = \Phi(u, v)$ the formula is:

$ \iint_D f(x,y), dx, dy = \iint_{D'} f(x(u,v), y(u,v)), |J|, du, dv, $

where Jacobian $J = \det\left(\frac{\partial(x,y)}{\partial(u,v)}\right) = \frac{\partial x}{\partial u} \cdot \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \cdot \frac{\partial y}{\partial u}$ — the “stretching” factor for area under the mapping. The absolute value $|J|$ ensures the positivity of area.

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

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

The $r$ factor before $dr, d\theta$ is the “cost” of the polar area element: a small sector at distance $r$ from the origin has area $r \cdot \Delta r \cdot \Delta \theta$.

Classical example: Gaussian integral $\int_{-\infty}^{\infty} e^{-x^2}, dx = \sqrt{\pi}$.

Let us compute $I^2 = \left(\int_{-\infty}^{\infty} e^{-x^2}, dx\right)\left(\int_{-\infty}^{\infty} e^{-y^2}, dy\right) = \iint_{\mathbb{R}^2} e^{-(x^2 + y^2)}, dA$.

In polar coordinates:

$ I^2 = \int_0^{2\pi} \int_0^{\infty} e^{-r^2}, r, dr, d\theta = 2\pi \cdot \left[-\frac{e^{-r^2}}{2}\right]_0^{\infty} = 2\pi \cdot \frac{1}{2} = \pi. $

Therefore, $I = \sqrt{\pi}$ — a result fundamental to probability theory and statistics.

Triple Integrals and Curvilinear Coordinates

The triple integral $\iiint_V f(x, y, z), dV$ is defined analogously. For calculation, use:

Cylindrical coordinates: $x = r \cos\theta$, $y = r \sin\theta$, $z = z$; $dV = r, dr, d\theta, dz$. Convenient for bodies with axis symmetry (cylinders, cones).

Spherical coordinates: $x = \rho\sin\varphi\cos\theta$, $y = \rho\sin\varphi\sin\theta$, $z = \rho\cos\varphi$; $dV = \rho^2 \sin\varphi, d\rho, d\varphi, d\theta$. Convenient for bodies with central symmetry (spheres, spherical parts).

Volume of a sphere of radius $R$:

$ \iiint_{\rho \leq R} dV = \int_0^{2\pi} \int_0^\pi \int_0^R \rho^2 \sin\varphi, d\rho, d\varphi, d\theta = 2\pi \cdot 2 \cdot \frac{R^3}{3} = \frac{4\pi R^3}{3}. , \checkmark $

Moment of inertia of a sphere relative to the $z$-axis with uniform density $\rho_0$:

$ I_z = \rho_0 \iiint_{\rho \leq R} (x^2 + y^2), dV = \rho_0 \int_0^{2\pi} \int_0^\pi \int_0^R \rho^2 \sin^2\varphi \cdot \rho^2 \sin\varphi, d\rho, d\varphi, d\theta = \frac{2MR^2}{5}, $

where $M$ is the mass of the sphere. This is the formula used in mechanics to calculate rotational kinetic energy.

Probabilistic Applications

Double and triple integrals are the language of multidimensional probability distributions. If $(X, Y)$ is a continuous random vector with density $p(x, y)$, then:

$ P\big((X, Y) \in D\big) = \iint_D p(x, y), dA. $

Marginal density of $X$: $p_X(x) = \int_{-\infty}^{\infty} p(x, y), dy$ — integration “over the unnecessary” variable. Joint normal distribution, Dirichlet distributions, computation of normalization constants — all these are problems concerning multiple integrals.

Question for reflection: the problem of calculating a multiple integral over a complicated region is often reduced to changing the order of integration or changing coordinates. Try to compute $\iint_D y e^{x^2}, dA$ over the triangle $D: 0 \leq y \leq x \leq 1$, first in the order $dy, dx$, and then by changing the order.