Double Integrals in Polar Coordinates — Question 1

Question 1 – Solution

Strategy. Radial symmetry turns the integrand into $r^2$ and the area element into $r\,dr\,d\theta$.

Step 1: Conversion The disk is $0\le r\le 2$, $0\le\theta\le 2\pi$, and $x^2+y^2=r^2$. Hence $$\iint_D(x^2+y^2)dA=\int_0^{2\pi}\int_0^2r^2(r\,dr\,d\theta).$$

Step 2: Evaluate $$\int_0^{2\pi}\left[\frac{r^4}{4}\right]_0^2d\theta =\int_0^{2\pi}4\,d\theta=\boxed{8\pi}.$$

Verification A thin polar cell has radial side $dr$ and arc-length side approximately $r\,d\theta$, so its area is $r\,dr\,d\theta$. Omitting $r$ would give the wrong units and value.