<s>[INST] Transform $\iint_D \sin(x^2+y^2)~dA$ to polar coordinates and evaluate the polar integral. 
Transform the given integral in Cartesian coordinates to polar coordinates and evaluate the polar integral:
(b) $\iint_D \sin(x^2+y^2)~dA$, where $D$ is the region in the first quadrant bounded by the lines $x=0$ and $y=\sqrt{3}\cdot x$ and the circles $x^2+y^2=2\pi$ and $x^2+y^2=3\pi$.

In Cartesian coordinates I know $0\le y\le \sqrt{3 \pi}, \space0\le x \le \frac{\sqrt{3\pi}}{4}$
I also know $x=r\cos\theta$, $y=r\sin\theta$
How should I proceed from here?
$r$ remains the same $r= \sqrt{3\pi}$?
So $0\le r\le\sqrt{3\pi}$, $0\le\theta\le \frac{\pi}{2}$. Is this correct?
 [/INST] Given the above integral with the given bounds converted to polar coordinates, $$\iint_D \sin(x^2+y^2)~dA=\int_{\pi/3}^{\pi/2} \int_{\sqrt{2\pi}}^{\sqrt{3\pi}} \sin(r^2)\cdot r~dr~d\theta$$ evaluated gives ${\pi/6}$
</s>