Evaluate the given integral by changing to polar coordinates.
$\int\int_D x^2y \, dA$, where $D$ is the top half of the disk with center the origin and radius $5$.
when i plugged everything in, I got the double integral $$\int_0^\pi \int_0^5 r^4\cos^2\theta \sin \theta \,dr \,d\theta.$$ then I used $u$ substitution and got $5\int u^2 \,du =\left. \frac{5u^3}{3}\right\vert_{\theta=0}^{\theta=\pi} =-5/3 - 5/3 = -10/3$ but that's definitely wrong. I looked at multiple examples, what am I doing wrong here? I'm pretty sure i have the right method but something's going wrong.
$\endgroup$ 31 Answer
$\begingroup$You have forgotten to integrate with respect to $r$ and I have no idea where does your leading $5$ comes from.
$$\int_0^5r^4 \,dr =\frac{5^5}{5}=5^4$$
\begin{align} \int_0^\pi \cos^2 \theta \sin \theta \, d\theta &= \left. -\frac{\cos^3\theta}{3} \right\vert_0^\pi \end{align}
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