Double Integrals in Polar Coordinates

Disks, rings and wedges are awkward in rectangular coordinates: the limits involve $\sqrt{r^2 - x^2}$. In polar coordinates they are rectangles: $a \le r \le b$, $\alpha \le \theta \le \beta$. You met polar area in Calculus 2; now the integrand can be any function.

Idea clave

$$\iint_D f(x, y)\,dA = \int_\alpha^\beta\int_{r_1(\theta)}^{r_2(\theta)} f(r\cos\theta, r\sin\theta)\,r\,dr\,d\theta.$$ The area element is $dA = r\,dr\,d\theta$, with the extra factor $r$.

Por qué funciona

Chop the region with circles ($r$ constant) and rays ($\theta$ constant). A small piece is a "polar rectangle": $\Delta r$ thick in the radial direction and, along the arc, $r\,\Delta\theta$ long (arc length is radius times angle). Its area is about $(\Delta r)(r\,\Delta\theta) = r\,\Delta r\,\Delta\theta$. Pieces far from the origin are bigger than pieces near it, even with the same $\Delta r$ and $\Delta\theta$, and the factor $r$ accounts for that. (Exactly: the area between radii $r_1$ and $r_2$ is $\frac{1}{2}(r_2^2 - r_1^2)\Delta\theta = \bar{r}\,\Delta r\,\Delta\theta$, with $\bar{r}$ the middle radius.)

A quarter ring. The red polar box is $\Delta r$ wide and $r\,\Delta\theta$ long: sweep the ray outward and watch the boxes grow with $r$.

When should you switch to polar? Look at both the region and the integrand. Disks, rings, sectors and regions bounded by curves like $r = 2\cos\theta$ have simple limits in polar form, and integrands built from $x^2 + y^2$ become functions of $r$ alone. The circle $x^2 + y^2 = 2x$ is $r^2 = 2r\cos\theta$, that is, $r = 2\cos\theta$ with $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$. Its area is $$\int_{-\pi/2}^{\pi/2} \int_0^{2\cos\theta} r\,dr\,d\theta = \int_{-\pi/2}^{\pi/2} 2\cos^2\theta\,d\theta = \pi,$$ as it should be for a circle of radius ${}1$.

Worked examples

Ejemplo resuelto: Over a disk

Evaluate $\iint_D (x^2 + y^2)\,dA$ over the disk $x^2 + y^2 \le 4$.

$x^2 + y^2 = r^2$, and the disk is ${}0 \le r \le 2$, ${}0 \le \theta \le 2\pi$.
$$\int_0^{2\pi}\int_0^2 r^2 \cdot r\,dr\,d\theta = \int_0^{2\pi} \left[\frac{r^4}{4}\right]_0^2 d\theta = \int_0^{2\pi} 4\,d\theta = 8\pi.$$

Respuesta: ${}8\pi$

Ejemplo resuelto: A Gaussian

Evaluate $\iint_D e^{-(x^2 + y^2)}\,dA$ over the disk of radius $R$.

$\int_0^{2\pi}\int_0^R e^{-r^2}\,r\,dr\,d\theta$: the $r$ from $dA$ makes the substitution $u = r^2$ work.
$\int_0^R e^{-r^2}r\,dr = \frac{1}{2}\left(1 - e^{-R^2}\right)$, so the integral is $\pi\left(1 - e^{-R^2}\right)$.
As $R \to \infty$ this tends to $\pi$, the famous fact behind $\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}$.

Respuesta: $\pi\left(1 - e^{-R^2}\right)$

Ejemplo resuelto: Half a ring

Evaluate $\iint_D y\,dA$ over the upper half of the ring ${}1 \le x^2 + y^2 \le 4$.

${}1 \le r \le 2$, ${}0 \le \theta \le \pi$, and $y = r\sin\theta$.
$$\int_0^{\pi}\int_1^2 r\sin\theta \cdot r\,dr\,d\theta = \left(\int_1^2 r^2\,dr\right)\left(\int_0^{\pi}\sin\theta\,d\theta\right) = \frac{7}{3} \cdot 2 = \frac{14}{3}.$$

Respuesta: $\frac{14}{3}$

Error común

Forgetting the extra factor $r$. Writing $dA = dr\,d\theta$ treats every polar box as the same size. Quick check: $\int_0^{2\pi}\int_0^R dr\,d\theta = 2\pi R$, which is a length, not the disk's area $\pi R^2$; with the factor $r$ you get $\pi R^2$. Also use the radius (${}2$), not $r^2$ (${}4$), as the upper limit for $r$.

Worked examples

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