Double Integrals in Polar Coordinates
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 5.3: “Double Integrals in Polar Coordinates” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/5-3-double-integrals-in-polar-coordinates |
| Textbook used in class | Stewart, Calculus, Section 15.4: “Double Integrals in Polar Coordinates” (Examples 1, 2, 3) |
Quick Reference
Substitution: $x = r\cos\theta$, $y = r\sin\theta$, $x^2+y^2 = r^2$.
Key formula: $$\iint_D f(x,y)\,dA = \iint_{D'} f(r\cos\theta, r\sin\theta)\,r\,dr\,d\theta.$$
Disk $x^2+y^2 \leq a^2$: $0 \leq r \leq a$, $0 \leq \theta \leq 2\pi$.
Do not forget the extra $r$ in $dA = r\,dr\,d\theta$.
Motivation
When the region is a disk or a sector and the integrand involves $x^2+y^2$, the Cartesian iterated integral is difficult or impossible. Polar coordinates replace these expressions with simple powers of $r$, transforming the integral into something computable.
The factor $r$ in $dA = r\,dr\,d\theta$ comes from the Jacobian of the transformation from $(r,\theta)$ to $(x,y)$: a thin polar “rectangle” of width $dr$ and angular span $d\theta$ at radius $r$ has area $r\,dr\,d\theta$, not $dr\,d\theta$.
Key Concept
When to use polar: the integrand contains $x^2+y^2$, or the region is a full disk, annulus, or angular sector. Attempting to use Cartesian coordinates for such problems typically produces integrands like $\sqrt{a^2-x^2}$ that are hard to integrate.
Worked Examples
Example 1. Evaluate $\iint_D e^{x^2+y^2}\,dA$ where $D$ is the disk $x^2+y^2 \leq 4$. (Stewart 15.4, Example 1.)
In polar: $e^{x^2+y^2} = e^{r^2}$. Limits: $0 \leq r \leq 2$, $0 \leq \theta \leq 2\pi$.
$$\int_0^{2\pi}\int_0^2 e^{r^2}\,r\,dr\,d\theta = \int_0^{2\pi}\left[\frac{e^{r^2}}{2}\right]_0^2 d\theta = \int_0^{2\pi}\frac{e^4-1}{2}\,d\theta = \pi(e^4-1).$$
Example 2. Find the volume under $z = \sqrt{4-x^2-y^2}$ and above the disk $x^2+y^2 \leq 4$ (a hemisphere of radius 2). (Stewart 15.4.)
$$V = \int_0^{2\pi}\int_0^2 \sqrt{4-r^2}\,r\,dr\,d\theta.$$
Let $u = 4-r^2$, $du = -2r\,dr$: $$\int_0^2\sqrt{4-r^2}\,r\,dr = \left[-\frac{(4-r^2)^{3/2}}{3}\right]_0^2 = 0 + \frac{8}{3} = \frac{8}{3}.$$
$$V = 2\pi\cdot\frac{8}{3} = \frac{16\pi}{3}.$$
(Matches the formula $\frac{2}{3}\pi r^3$ for a hemisphere of radius 2: $\frac{2}{3}\pi(8) = \frac{16\pi}{3}$.)
$dA = dr\,d\theta$ in polar coordinates. The area element in polar coordinates is $dA = r\,dr\,d\theta$, NOT $dr\,d\theta$. The extra factor of $r$ is the Jacobian of the polar transformation. Omitting it scales the answer by the wrong amount. Memory check: a polar “rectangle” at distance $r$ from the origin has width $dr$ radially and arc length $r\,d\theta$ tangentially, giving area $r\,dr\,d\theta$.
Leveled Practice
Problem 1. Evaluate $\iint_D (x^2+y^2)\,dA$ where $D$ is the annular region $1 \leq x^2+y^2 \leq 4$.
Show answer
In polar: $1 \leq r \leq 2$, $0 \leq \theta \leq 2\pi$.
$$\int_0^{2\pi}\int_1^2 r^2 \cdot r\,dr\,d\theta = 2\pi\int_1^2 r^3\,dr = 2\pi\left[\frac{r^4}{4}\right]_1^2 = 2\pi\cdot\frac{16-1}{4} = \frac{15\pi}{2}.$$