A
double integral
over a disk or a wedge is a nightmare in x, y — the limits are
ugly square roots like y = \pm\sqrt{a^2 - x^2}. Circular regions
beg for polar coordinates, where a disk is just a rectangle:
x = r\cos\theta, \qquad y = r\sin\theta, \qquad 0 \le r \le a, \quad 0 \le \theta \le 2\pi.
But you cannot simply swap dA = dx\, dy for
dr\, d\theta. There is an extra factor of
r — and seeing exactly where it comes from is the whole point.
The area element: where the extra r comes from
In Cartesian coordinates a tiny tile is a rectangle of sides
dx and dy, area
dA = dx\, dy. The polar grid is different: it is made of circular
arcs and radial spokes, so a tiny "polar rectangle" is a curved patch. Let us measure it.
Step 1 — name the two sides of the polar patch. Hold a corner at radius
r and angle \theta, then nudge each
coordinate. Increasing r by dr moves
you outward along a spoke — a straight radial side of length
\text{radial side} = dr.
Step 2 — measure the arc side. Increasing
\theta by d\theta sweeps along a
circular arc of radius r. Arc length is radius times angle:
\text{arc side} = r\, d\theta.
Step 3 — multiply the two sides. The patch is, to first order, a tiny
rectangle with perpendicular sides dr and
r\, d\theta, so its area is their product:
dA = (dr)\,(r\, d\theta) = r\, dr\, d\theta.
That stray r is the Jacobian of the polar map —
it accounts for the fact that patches far from the origin (large
r) are fatter than patches near it, because the arc side
r\, d\theta grows with r. The
change-of-variables theorem
will derive this same r from a determinant.
The polar conversion in full
Replace x, y by r\cos\theta, r\sin\theta
in the integrand, and replace dA by
r\, dr\, d\theta:
\iint_R f(x, y)\, dA = \iint_S f(r\cos\theta,\, r\sin\theta)\; r\, dr\, d\theta.
Worked example — \displaystyle \iint_R e^{-(x^2+y^2)}\, dA over the disk of radius a
Step 1 — convert the integrand. Since
x^2 + y^2 = r^2, the awkward exponential becomes a function of
r alone:
e^{-(x^2+y^2)} = e^{-r^2}.
Step 2 — set up the polar integral. The disk is the polar rectangle
0 \le r \le a, 0 \le \theta \le 2\pi,
and we must include the r\, dr\, d\theta:
\iint_R e^{-(x^2+y^2)}\, dA = \int_0^{2\pi} \!\int_0^a e^{-r^2}\; r\, dr\, d\theta.
Step 3 — the inner integral, where the extra r earns its
keep. Substitute u = r^2, so
du = 2r\, dr — the leftover r is
exactly what the substitution needs:
\int_0^a e^{-r^2}\, r\, dr = \frac{1}{2}\int_0^{a^2} e^{-u}\, du = \frac{1}{2}\left(1 - e^{-a^2}\right).
Step 4 — the outer integral. The inner result is constant in
\theta, so the \theta integral just
multiplies by 2\pi:
\int_0^{2\pi} \frac{1}{2}\left(1 - e^{-a^2}\right) d\theta = \pi\left(1 - e^{-a^2}\right).
Step 5 — let the disk fill the plane. Send
a \to \infty; the term e^{-a^2} \to 0:
\iint_{\mathbb{R}^2} e^{-(x^2+y^2)}\, dA = \pi.
A clean \pi from an integral that has no elementary
antiderivative in Cartesian form. This single fact unlocks the famous Gaussian integral.
Let R be a region described in polar coordinates by
\alpha \le \theta \le \beta and
r_1(\theta) \le r \le r_2(\theta), and let
f be continuous on R. Then:
-
The substitution is
x = r\cos\theta, y = r\sin\theta,
with x^2 + y^2 = r^2.
-
The area element carries an extra
r: dA = r\, dr\, d\theta
(the Jacobian of the polar map).
-
The integral becomes
\iint_R f\, dA = \int_\alpha^\beta \!\int_{r_1(\theta)}^{r_2(\theta)} f(r\cos\theta, r\sin\theta)\; r\, dr\, d\theta.
-
A full disk of radius a is the polar
rectangle 0 \le r \le a,
0 \le \theta \le 2\pi.
The bell-curve integral \int_{-\infty}^{\infty} e^{-x^2}\, dx
resists every one-variable trick — e^{-x^2} has no elementary
antiderivative. The escape, due to Gauss's
trick, is to go up a dimension. Call the unknown value I and
square it, writing the two copies in different dummy variables:
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.
That right-hand side is exactly the polar integral we just evaluated to
\pi. Hence
I^2 = \pi \quad\Longrightarrow\quad I = \int_{-\infty}^{\infty} e^{-x^2}\, dx = \sqrt{\pi}.
A one-dimensional integral that defeats single-variable calculus falls instantly once you
let it breathe in two dimensions and switch to polar. The whole of statistics —
normalising the normal distribution — rests on this \sqrt{\pi}.