So I am presented with a double integral which is can be evaluated easily enough as is.
$$\int _0^{\frac{1}{2}}\int _{\sqrt{3}y}^{\sqrt{1-y^2}}xy^2 dxdy.$$
What I'm curious about is converting this into polar form then integrating it, and this is where I'm having trouble with.
I try to establish the bounds for $r$ but I end up cancelling $r$ out for the bounds, and when I convert $xy^2$ into polar I get a weird mess I can't really simplify. What I try to integrate doesn't come out to be an "area".
Is there something simple I'm missing out on?
Best Answer
Did you draw a picture of the region? That is pretty much essential. The region of integration is a sector of the unit circle, with $0\le\theta\le \pi/6$. (Note that $x=\sqrt3y$ is the line $y=\frac1{\sqrt3}x$, which indeed makes an angle of $\pi/6$ with the $x$-axis. The equation $x=\sqrt{1-y^2}$ gives a portion of the unit circle.) Thus, you end up with the integral $$\int_0^{\pi/6}\int_0^1 r^3\cos\theta\sin^2\theta\cdot r\,dr\,d\theta.$$