I need to find: $$\int_0^2 \int_0^{\sqrt{3}x} f(\sqrt{x^2+y^2})dydx$$ in polar coordinates.
Since $x=r\cos(\theta)$ and $y=r\sin(\theta)$, I got:
$y=\sqrt{3}x \iff r\sin(\theta)=\sqrt{3}r\cos(\theta)\iff \tan(\theta)=\sqrt{3}\iff \theta=\arctan(\sqrt{3})$
From this I conclude that $0 \leq \theta \leq \arctan(\sqrt{3})$.
My problem is I'm not sure where the radius is, I thought about from $0$ to $2\sqrt{3}$ (from graphing $y=\sqrt{3}$) but I think it is wrong.
Is there something I'm missing in order to get the $r$ interval?
Since $x\in[0,\,2]$ while $y\in[0,\,x\sqrt{3}]$, $r\in[0,\,2x]\subseteq[0,\,4]$. The integration region is $0\le r\le4,\,0\le\theta\le\pi/3$ (you'd already worked out the second part), so the integral is $\int_0^{\pi/3}d\theta\int_0^4rf(r)dr=\tfrac{\pi}{3}\int_0^4rf(r)dr$.