# evaluating the given integral by changing to polar coordinates (why is my answer wrong?)

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.

• i just saw that derivative of cosine is -sine >.< – 2316354654 Oct 31 '17 at 7:26
• but that just changes my answer to positive 10/3, and the answer is 1250/3... what am I doing wrong??? – 2316354654 Oct 31 '17 at 7:29
• math.meta.stackexchange.com/q/5020/306553 mathjax reference – Siong Thye Goh Oct 31 '17 at 7:41

You have forgotten to integrate with respect to $r$ and I have no idea where does your leading $5$ comes from.
$$\int_0^5r^4 \,dr =\frac{5^5}{5}=5^4$$