Recipe

Convert and evaluate a double integral in polar coordinates

Use when: R is a disk, sector, or annulus, or the integrand or boundary involves x^2+y^2 as a single block

Goal: the double integral's value, computed in r,\theta with no leftover x's or y's anywhere

Taught in Double integrals in polar coordinates

1. 1

   Check that R's boundary is naturally a circle, sector, or ring, or that the integrand simplifies once x^2+y^2 is replaced by r^2.

   Example

   Find \displaystyle\iint_R \sqrt{x^2+y^2}\,dA where R is the quarter disk x^2+y^2\le9 in the first quadrant. Both the region and the integrand are built from x^2+y^2, the polar signature.

   Why: converting costs an extra factor of r in the integrand, so it's only worth paying when the boundary or the integrand actually simplifies in return.

2. 2

   Rewrite R's boundary using x^2+y^2=r^2, giving a range for r (often constants) and a range for \theta.

   Example

   x^2+y^2\le9 becomes 0\le r\le3. The first quadrant is 0\le\theta\le\pi/2.

   Why: a boundary built from circles and rays becomes a plain rectangle in r,\theta, exactly the same simplification a Type I or Type II description gives for a curved region in x,y.

3. 3

   Replace every x^2+y^2 with r^2, and any lone x or y with r\cos\theta or r\sin\theta.

   Example

   \sqrt{x^2+y^2} becomes \sqrt{r^2}=r.

   Why: the substitution x=r\cos\theta,\ y=r\sin\theta is what defines polar coordinates in the first place, so every occurrence of x and y has to be rewritten, not just the ones that look convenient.

4. 4

   Write dA=r\,dr\,d\theta, multiplying the converted integrand by r before integrating; never leave it as dr\,d\theta.

   Example

   The integral becomes \displaystyle\int_0^{\pi/2}\!\int_0^3 r\cdot r\,dr\,d\theta = \int_0^{\pi/2}\!\int_0^3 r^2\,dr\,d\theta.

   Why: a polar "rectangle" of size \Delta r\times\Delta\theta is really a thin wedge whose true area grows with distance from the pole, and that extra r is precisely the correction the ordinary product \Delta r\,\Delta\theta misses.

5. 5

   Integrate with respect to r first (bounds usually constants), then with respect to \theta.

   Example

   Inner: \displaystyle\int_0^3 r^2\,dr = \left[\frac{r^3}{3}\right]_0^3 = 9. Outer: \displaystyle\int_0^{\pi/2} 9\,d\theta = \boxed{\frac{9\pi}{2}}.

   Why: this is the same two-step iterated evaluation as any other double integral, just carried out in the coordinates that make both bounds constant.