Double Integrals in Polar Coordinates

Some regions and integrands have circular symmetry that makes Cartesian coordinates cumbersome. Polar coordinates $(r, \theta)$ — where $x = r\cos\theta$ and $y = r\sin\theta$ — naturally describe disks, sectors, and annuli. This note shows how to set up and evaluate double integrals in polar coordinates.

1. The polar rectangle

Recall from Polar Coordinates that a point in the plane is described by $(r, \theta)$ with $r \ge 0$ and $\theta \in [0, 2\pi)$. Just as Cartesian rectangles $[a,b]\times[c,d]$ are the building block for Cartesian integrals, polar rectangles are the natural building block for polar integrals.

A polar rectangle is the set

$R = \{\,(r,\theta) : a \le r \le b,\;\; \alpha \le \theta \le \beta\,\},$

where $0 \le a < b$ and $0 \le \alpha < \beta \le 2\pi$.

In Cartesian coordinates, $R$ is the region between two arcs of circles (radii $a$ and $b$) and two radial lines (angles $\alpha$ and $\beta$) — sometimes called an annular sector.

Use the sliders below to explore different polar rectangles.

2. Why we cannot use $dr\,d\theta$ naively

A natural first guess for the polar area element might be $dr\,d\theta$ (by analogy with $dx\,dy$). Let us check this against a known area.

Take the full disk of radius $R$: $0 \le r \le R$, $0 \le \theta \le 2\pi$. The area of a disk is $\pi R^2$.

Naive attempt:

$\int_0^{2\pi}\!\int_0^R dr\,d\theta = \int_0^{2\pi} R\,d\theta = 2\pi R.$

This gives $2\pi R$, not $\pi R^2$. The naive element $dr\,d\theta$ is wrong.

The interactive graph above displays the "true area" versus the "naive area" computed without the $r$ factor — you can see they differ whenever $a \ne 0$ or the sector is not degenerate.

3. The polar area element: $dA = r\,dr\,d\theta$

The correct area element in polar coordinates is $r\,dr\,d\theta$. Here is where it comes from.

Geometry of a thin polar rectangle. Consider the polar rectangle with $r \in [r_0, r_0 + \Delta r]$ and $\theta \in [\theta_0, \theta_0 + \Delta\theta]$.

Its area is the area of the outer sector minus the area of the inner sector:

$\Delta A = \frac{1}{2}(r_0+\Delta r)^2\,\Delta\theta - \frac{1}{2}r_0^2\,\Delta\theta = \frac{\Delta\theta}{2}\bigl[(r_0+\Delta r)^2 - r_0^2\bigr].$

Expanding:

$\Delta A = \frac{\Delta\theta}{2}\bigl[2r_0\,\Delta r + (\Delta r)^2\bigr] \approx r_0\,\Delta r\,\Delta\theta \quad \text{as } \Delta r \to 0.$

In the limit, $\Delta A \to r\,dr\,d\theta$. The factor of $r$ accounts for the fact that arcs at large $r$ are much longer than arcs near the origin.

$\boxed{dA = r\,dr\,d\theta.}$

Verification. For the disk of radius $R$:

$\int_0^{2\pi}\!\int_0^R r\,dr\,d\theta = \int_0^{2\pi} \frac{R^2}{2}\,d\theta = \pi R^2. \checkmark$

4. The change-of-variables formula for polar coordinates

Let $f(x,y)$ be continuous on the polar rectangle $R = \{a \le r \le b,\, \alpha \le \theta \le \beta\}$. Then

$\boxed{\iint_R f(x,y)\,dA = \int_\alpha^\beta\!\int_a^b f(r\cos\theta,\, r\sin\theta)\; r\,dr\,d\theta.}$

How to use this:

1. Replace $x \to r\cos\theta$ and $y \to r\sin\theta$ in the integrand.
2. Replace $dA \to r\,dr\,d\theta$.
3. Set the limits: $r$ from $a$ to $b$, $\theta$ from $\alpha$ to $\beta$.

Example. Evaluate $\displaystyle\iint_D (x^2+y^2)\,dA$ where $D$ is the disk $x^2+y^2 \le 4$.

In polar: $x^2+y^2 = r^2$, $D$ corresponds to $0 \le r \le 2$, $0 \le \theta \le 2\pi$.

$\int_0^{2\pi}\!\int_0^2 r^2 \cdot r\,dr\,d\theta = \int_0^{2\pi}\!\int_0^2 r^3\,dr\,d\theta = \int_0^{2\pi} \frac{r^4}{4}\Bigr|_0^2\,d\theta = \int_0^{2\pi} 4\,d\theta = 8\pi.$

5. General polar domains

We are not limited to polar rectangles. A general polar domain is described by

$D = \{\,(r,\theta) : \alpha \le \theta \le \beta,\;\; h_1(\theta) \le r \le h_2(\theta)\,\},$

where $h_1$ and $h_2$ are continuous and $0 \le h_1(\theta) \le h_2(\theta)$.

This is the polar analogue of a Type I region (sweep $\theta$ from $\alpha$ to $\beta$; for each $\theta$, $r$ runs from $h_1(\theta)$ to $h_2(\theta)$).

$\boxed{\iint_D f(x,y)\,dA = \int_\alpha^\beta\!\int_{h_1(\theta)}^{h_2(\theta)} f(r\cos\theta, r\sin\theta)\; r\,dr\,d\theta.}$

Example. Compute the area of one petal of the rose $r = \cos(2\theta)$.

The first petal lies in $-\pi/4 \le \theta \le \pi/4$ where $\cos(2\theta) \ge 0$. So $h_1 = 0$, $h_2 = \cos(2\theta)$:

$\text{Area} = \int_{-\pi/4}^{\pi/4}\!\int_0^{\cos(2\theta)} r\,dr\,d\theta = \int_{-\pi/4}^{\pi/4} \frac{\cos^2(2\theta)}{2}\,d\theta = \frac{1}{2}\int_{-\pi/4}^{\pi/4}\frac{1+\cos(4\theta)}{2}\,d\theta = \frac{\pi}{8}.$

6. When to use polar coordinates

Polar coordinates simplify a double integral when:

• The domain is a disk, annulus, sector, or described by an equation like $x^2+y^2 = R^2$.
• The integrand contains $x^2+y^2$, $\sqrt{x^2+y^2}$, $\arctan(y/x)$, or similar combinations.
• The domain has a circular boundary.

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

7. Exercises