Why am I evaluating this polar integral wrong? I have: $$ \int_{0}^{6} \int_{0}^{y}xdxdy.$$
I drew a picture already, which is just a triangle in the first quadrant. I then changed the cartesian coordinates into polar coordinates, which came out to be:
$$ \int_{pi/4}^{pi/2} \int_{0}^{6}(rcos(\theta))rdrd\theta.$$ 
But when I go to solve this, I keep on getting $36(2\sqrt{2})$ instead of my book's answer, which is just $36$. 
Help is greatly appreciated!
 A: That region is a triangle, which isn't trivially easy to describe in polar coordinates, but it's far from impossible.
Your bounds of $\theta$ are correct; it will range from $\pi/4$ to $\pi/2$. As for $r$, the bounds you have denote a circular region (try plotting the equations $r=0$ and $r=6$ to see what I mean there). Your lower bound of $r$ is fine, since that just means our region includes the origin. As for the upper bound, however, you'll want to describe the line $y=6$ in polar coordinates, because that's the true limit to how far we can get from the origin while staying inside the triangle. That's not too hard, though, given that $y=r\sin{\theta}$. $$r\sin{\theta}=6\implies r=6\csc{\theta}$$
Therefore, the integral will look like this in polar coordinates: $$\int_{\pi/4}^{\pi/2} \int_0^{6\csc{\theta}}r^2cos{\theta}\,drd\theta$$
Evaluating, we find that $$\int_{\pi/4}^{\pi/2} \int_0^{6\csc{\theta}}r^2cos{\theta}\,drd\theta=\int_{\pi/4}^{\pi/2}\Big{[}\frac{r^3}{3}\cos{\theta}\Big{]}_0^{6\csc{\theta}}\,d\theta=72\int_{\pi/4}^{\pi/2}\frac{\cos{\theta}}{\sin^3{\theta}}\,d\theta$$ $$u=\sin{\theta}, \,du=\cos{\theta}d\theta; \, \theta=\pi/4\implies u=\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}, \,\theta=\pi/2\implies u=\sin{\frac{\pi}{2}}=1$$ $$72\int_{\pi/4}^{\pi/2}\frac{\cos{\theta}}{\sin^3{\theta}}\,d\theta=72\int_{\sqrt{2}/2}^1u^{-3}\,du=72\Big{[}-\frac{1}{2}u^{-2}\Big{]}_{\frac{\sqrt{2}}{2}}^1=-36\Big{(}1-\frac{4}{2}\Big{)}=36$$
