Double Integrals, Cartesian and Polar

• Evaluate a double integral as an iterated integral over a rectangle or over a region bounded by curves, and change the order of integration by redescribing the region.
• State the hypotheses of Fubini's theorem, and recognize a case where the two iterated integrals disagree because a hypothesis fails.
• State the change-of-variables theorem with its conditions, compute a Jacobian, and derive the factor $r$ of polar coordinates.
• Evaluate $\int_{-\infty}^{\infty} e^{-x^2}\, dx$ by switching to polar coordinates, and use it to show that the Normal density integrates to 1.

A double integral adds up a quantity spread over a region of the plane: the volume under a surface $z = f(x, y)$, or the mass of a plate whose density varies from point to point. As in the-definite-integral-and-riemann-sums, it is defined as a limit of sums, and it is computed by reducing it to two single integrals.

The lesson the-normal-distribution writes down the density $\frac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/(2\sigma^2)}$ and says only that the constant in front makes the area equal 1. That constant is not obvious, because $e^{-x^2}$ has no antiderivative that can be written with elementary functions, so the fundamental theorem of calculus cannot evaluate the integral directly. The way around it uses two dimensions: the square of the integral is a double integral over the plane, and in polar coordinates it becomes easy. The same idea gives the Box-Muller method for turning uniform random numbers into Normal ones.

Double integrals over rectangles

Split a rectangle $R = [a, b] \times [c, d]$ into small subrectangles of area $\Delta A$, pick a sample point in each, and form $\sum f(x_i^*, y_j^*)\, \Delta A$. When the sums converge to the same number as the subrectangles shrink, the limit is $\iint_R f\, dA$. It exists when $f$ is continuous on $R$, and also when $f$ is bounded and is discontinuous only along finitely many smooth curves. Where $f \ge 0$ it is the volume under the graph.

Iterated integrals and Fubini's theorem

Fubini's theorem (continuous case). If $f$ is continuous on the closed rectangle $R = [a, b] \times [c, d]$, then

The hypotheses are a closed bounded rectangle and continuity. The version for general integrable functions (Lebesgue integral) replaces continuity by absolute integrability: if $\iint |f|\, dA < \infty$, both iterated integrals exist and equal $\iint f\, dA$. Tonelli's theorem covers $f \ge 0$: both iterated integrals always equal the double integral, as values in $[0, \infty]$, and no finiteness is needed.

When the hypothesis fails, the conclusion can fail. Take $f(x, y) = \dfrac{x^2 - y^2}{(x^2 + y^2)^2}$ on $(0, 1)^2$, which blows up at the origin. Since $\frac{\partial}{\partial x}\frac{x}{x^2 + y^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2}$, integrating in $x$ first gives $-\frac{1}{1 + y^2}$, and then

Both values were confirmed numerically. The function is not absolutely integrable: $\iint |f|\, dA$ over $[\varepsilon, 1]^2$ grows by about $2.3$ each time $\varepsilon$ shrinks tenfold (it approaches $\ln 10$ per step), which is logarithmic divergence. Without absolute integrability, the order can matter.

General regions and changing the order

A region is of type I if it is $a \le x \le b$, $g_1(x) \le y \le g_2(x)$ with $g_1, g_2$ continuous, and of type II if the roles of $x$ and $y$ are swapped. For $f$ continuous on such a region $D$,

Swapping the order is not swapping the limits. The region has to be redescribed from the other direction, so a sketch comes first: read off the extreme values of the new outer variable, then the bounds of the inner one as functions of it.

Change of variables and the Jacobian

Suppose $T(u, v) = (x(u, v), y(u, v))$ maps a region $S$ in the $uv$-plane onto a region $D$ in the $xy$-plane. The Jacobian is the determinant

Change of variables. If $T$ is continuously differentiable ($C^1$), one-to-one on the interior of $S$, and has a nonzero Jacobian on the interior of $S$, and if $f$ is continuous on $D$, then

The reason is the local linear approximation from partial-derivatives-and-the-multivariable-chain-rule: a small $\Delta u \times \Delta v$ rectangle is sent to nearly a parallelogram of area $\lvert\det\rvert\, \Delta u\, \Delta v$. For the linear map $x = (u + v)/2$, $y = (u - v)/2$ the determinant is $-1/2$, so areas are halved and the factor is $1/2$.

For polar coordinates, $x = r\cos\theta$ and $y = r\sin\theta$ give

so $dA = r\, dr\, d\theta$, the factor previewed in polar-coordinates. The map from $[0, R] \times [0, 2\pi]$ to the disk of radius $R$ breaks the hypotheses on the boundary: the whole segment $r = 0$ lands on the origin, the rays $\theta = 0$ and $\theta = 2\pi$ coincide, and $r = 0$ makes the Jacobian vanish. These sets have zero area, and the theorem needs only the interior, so for continuous $f$ the integral is unchanged.

A hypothesis that does matter is one-to-one. Let $\theta$ run over $[0, 4\pi]$ and the unit disk is covered twice: $\int_0^{4\pi}\int_0^1 r\, dr\, d\theta = 2\pi$, twice the true area $\pi$.

Unbounded regions

An integral over the whole plane is a limit, as in improper-integrals: integrate over a disk or square that grows, then let it grow without bound. For $f \ge 0$ the result does not depend on which growing family is used. For a function that changes sign, absolute integrability is the safe requirement.

Example 1: choosing the order on a rectangle

Evaluate $\iint_R x e^{xy}\, dA$ on $R = [0, 1] \times [0, 2]$. The integrand is continuous on a closed rectangle, so either order is allowed. Integrating $y$ first is easy:

Integrating $x$ first needs integration by parts and gives $\frac{(y - 1)e^y + 1}{y^2}$, a harder second step. Symbolic evaluation in both orders gives $(e^2 - 3)/2$, and numerical quadrature gives $2.19453$.

Example 2: an integral that only works after swapping

Evaluate $\int_0^1\!\int_x^1 e^{y^2}\, dy\, dx$. The inner integral is $\frac{\sqrt\pi}{2}(\operatorname{erfi}(1) - \operatorname{erfi}(x))$, not elementary. The region is $0 \le x \le y \le 1$, a triangle. Described the other way, $0 \le y \le 1$ and $0 \le x \le y$, so

Numerical quadrature of the original form returns $0.85914$.

Example 3: volume under a paraboloid

Find the volume between $z = 4 - x^2 - y^2$ and the plane $z = 0$. The surface meets the plane at $r = 2$, so $D$ is the disk $r \le 2$ and

A numerical double integral over the disk in $x, y$ gives $25.13274$. Dropping the Jacobian, $\int_0^{2\pi}\int_0^2 (4 - r^2)\, dr\, d\theta$, gives $32\pi/3 \approx 33.51$, which is wrong.

Example 4: the Gaussian integral

Let $I = \int_{-\infty}^{\infty} e^{-x^2}\, dx$. Then

The step from a product to a double integral is Fubini in the Tonelli form: the integrand is positive, so no finiteness is assumed. Evaluate the double integral as the limit of integrals over the disk $D_R$ of radius $R$ (allowed because the integrand is nonnegative). On $D_R$ polar coordinates apply, with $e^{-(x^2+y^2)} = e^{-r^2}$ continuous:

The extra factor $r$ is what makes the inner integral elementary: $r\, e^{-r^2}$ has the antiderivative $-\tfrac{1}{2}e^{-r^2}$. So $I^2 = \pi$, and since $I > 0$, $I = \sqrt\pi \approx 1.77245$, which numerical quadrature confirms. Squares exhaust the plane too, and give the same limit:

text
R   disk: pi(1 - e^-R^2)   square [-R,R]^2: pi erf(R)^2
1   1.9859                 2.2310
2   3.0841                 3.1123
3   3.1412                 3.1415     (pi = 3.1416)

Back to the Normal density. Substituting $u = (x - \mu)/(\sigma\sqrt 2)$, so $dx = \sigma\sqrt 2\, du$,

Dividing by $\sigma\sqrt{2\pi}$ produces total area 1, which is the constant in the-normal-distribution. For $\mu = 175$ and $\sigma = 7$ a numerical integral gives $17.5464 = 7\sqrt{2\pi}$. This derives the normalizing constant; why the bell shape arises in the first place is the Central Limit Theorem's job.

Example 5: Box-Muller

If $X, Y$ are independent standard Normals, their joint density is $\frac{1}{2\pi}e^{-(x^2 + y^2)/2}$. In polar form the density becomes $\frac{1}{2\pi}e^{-r^2/2}\, r\, dr\, d\theta$: the angle is uniform on $[0, 2\pi)$, and $R$ has distribution function $1 - e^{-r^2/2}$. Inverting that gives $R = \sqrt{-2\ln U_1}$ and $\Theta = 2\pi U_2$ for uniform $U_1, U_2$, and $X = R\cos\Theta$, $Y = R\sin\Theta$. In a run of $10^6$ pairs (seed 7) the sample variances were $0.9986$ and $0.9980$, the correlation was $-0.00003$, and the fraction with $\lvert X\rvert \le 1$ was $0.6826$ against the exact $0.6827$.

• "Fubini always lets you swap the order." It needs continuity on a closed rectangle (or absolute integrability, or $f \ge 0$). The example with $-\pi/4$ and $+\pi/4$ shows the order changing the answer.
• "Swapping the order means swapping the limits." The new bounds come from the region's picture, not from exchanging numbers.
• "Polar coordinates only change $dx\, dy$ into $dr\, d\theta$." The area element is $r\, dr\, d\theta$, and in Example 3 the missing $r$ changed the answer from $8\pi$ to $32\pi/3$. Also check that the angle range covers the region once. A circle not centered at the origin, such as $x^2 + y^2 \le 2x$, becomes $0 \le r \le 2\cos\theta$ with $-\pi/2 \le \theta \le \pi/2$.
• "The scaling factor is the Jacobian determinant itself." The determinant can be negative because it records orientation, but the factor is its absolute value. The linear map above has determinant $-1/2$ and factor $1/2$.

A double integral is a limit of sums, evaluated as an iterated integral when Fubini applies: continuity on a closed rectangle, or absolute integrability, or a nonnegative integrand (Tonelli). Without those, the two orders can disagree, as $\pm\pi/4$ shows. On general regions the order is changed by redescribing the region. A change of variables needs a $C^1$ map that is one-to-one on the interior and has a nonzero Jacobian there, and it multiplies the integrand by $\lvert\partial(x, y)/\partial(u, v)\rvert$; for polar coordinates that is $r$. With it, $\iint e^{-(x^2 + y^2)}\, dA = \pi$ gives $\int e^{-x^2}\, dx = \sqrt\pi$, the normalizing constant $\sigma\sqrt{2\pi}$ of the Normal density, and the Box-Muller sampler.