11.2 Evaluation of double integrals in polar form

Double Integrals in Polar Coordinates

Conversion from Cartesian to Polar Coordinates

A double integral in Cartesian coordinates $\iint f(x,y)\, dA$ can be rewritten in polar coordinates using the substitutions $x = r\cos\theta$ and $y = r\sin\theta$. The function $f(x,y)$ becomes $f(r\cos\theta,\, r\sin\theta)$, and the area element transforms as well.

The key piece of the conversion is the Jacobian determinant, which equals $r$. This factor accounts for the fact that the area element in polar coordinates is not a simple rectangle. Instead, it's a small "wedge" whose size depends on how far you are from the origin. The Cartesian area element $dA = dx\,dy$ becomes:

$dA = r\,dr\,d\theta$

So the full converted integral is:

$\iint f(r\cos\theta,\, r\sin\theta)\; r\,dr\,d\theta$

That extra $r$ is easy to forget, and leaving it out is one of the most common mistakes on exams.

Benefits and Applications of Polar Coordinates

Polar coordinates simplify double integrals when the region of integration has circular, annular, or rotational symmetry. If you're integrating over a disk, a ring, or a sector, polar coordinates will almost always produce cleaner limits than Cartesian.

Certain functions are also much easier to handle in polar form. For example, $e^{-x^2 - y^2}$ looks difficult in Cartesian coordinates, but in polar form it becomes $e^{-r^2}$, which is straightforward to integrate with respect to $r$. Functions involving $x^2 + y^2$ generally benefit from the substitution $x^2 + y^2 = r^2$.

These techniques appear frequently in physics and engineering, particularly when modeling electromagnetic fields, gravitational potentials, fluid flow, and other systems with radial symmetry.

Evaluating Polar Double Integrals

Setting Up the Integration Limits

Before computing anything, sketch the region of integration in the $r\theta$-plane. This step prevents most limit-setting errors.

1. Identify the angular limits. Determine the range of $\theta$ that sweeps across the entire region. For a full disk, $0 \leq \theta \leq 2\pi$. For a quarter-disk in the first quadrant, $0 \leq \theta \leq \pi/2$.
2. Identify the radial limits. For each angle $\theta$, determine where $r$ starts (the inner boundary) and where it ends (the outer boundary). These limits can be constants (e.g., $1 \leq r \leq 3$ for an annular region) or functions of $\theta$ (e.g., $0 \leq r \leq 2\cos\theta$ for a circle offset from the origin).
3. Write the integral. The standard setup with $r$ as the inner integral is:

$\int_{\theta_1}^{\theta_2}\int_{r_1(\theta)}^{r_2(\theta)} f(r\cos\theta,\, r\sin\theta)\; r\,dr\,d\theta$

The radial limits can depend on $\theta$, but the angular limits should be constants.

Evaluating the Radial and Angular Integrals

Work from the inside out. Here's a concrete example: evaluate $\int_0^{\pi/4}\int_0^2 r^2\,dr\,d\theta$.

Step 1: Evaluate the inner (radial) integral. Treat $\theta$ as a constant and integrate with respect to $r$:

$\int_0^2 r^2\,dr = \frac{1}{3}r^3\Big|_0^2 = \frac{8}{3}$

Step 2: Evaluate the outer (angular) integral. Substitute the result from Step 1 and integrate with respect to $\theta$:

$\int_0^{\pi/4} \frac{8}{3}\,d\theta = \frac{8}{3}\,\theta\Big|_0^{\pi/4} = \frac{8}{3}\cdot\frac{\pi}{4} = \frac{2\pi}{3}$

When the radial limits depend on $\theta$, the result of the inner integral will be a function of $\theta$ rather than a constant, so the outer integral requires more work.

Interpreting the Area Element

The polar area element $dA = r\,dr\,d\theta$ represents a small curved patch, not a true rectangle. Its radial width is $dr$, and its arc length is $r\,d\theta$. The product of these gives the area of the patch.

Notice that as $r$ increases, the patch gets larger even if $dr$ and $d\theta$ stay the same. This is why the factor of $r$ appears: regions farther from the origin contribute more area per unit of $dr\,d\theta$. Near the origin (small $r$), the patches are tiny; far from the origin (large $r$), they're much bigger.

To find the total area of a polar region (without any additional function), you integrate just the area element:

$A = \iint r\,dr\,d\theta$

This is equivalent to setting $f = 1$ in the double integral.