# 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?

• Your Cartesian coordinate representation of $D$ is also incorrect (You're describing a rectangular region). Commented Dec 16, 2017 at 11:37

No, it is not correct. The polar representation you give yields a quarter circle of radius $\sqrt{3\pi}$ on the first quadrant, which is obviously not the region described by the question.
The region $D$ you are concerned about I highlighted in yellow. It is easy to see that $x^2+y^2=2\pi$ and $x^2+y^2=3\pi$ are just circles of radius $\sqrt{2\pi}$ and $\sqrt{3\pi}$ respectively. Similarly, it can be seen that $y=\sqrt{3}\cdot x$ and $x=0$ correspond to $\theta=60^{\circ}=\pi/3$ (Since $\arctan(\sqrt{3})=\pi/3$) and $\theta=90^{\circ}=\pi/2$ respectively. Hence, the region can be described in polar coordinates as: $$D=\{(r,\theta)\in \mathbb{R}^2\mid \sqrt{2\pi}\leq r\leq \sqrt{3\pi}, \pi/3\leq \theta\leq \pi/2\}$$ Hence, converting the double integral to polar coordinates gives (Don't forget the Jacobian): $$\iint_D \sin(x^2+y^2)~dA=\int_{\pi/3}^{\pi/2} \int_{\sqrt{2\pi}}^{\sqrt{3\pi}} \sin(r^2)\cdot r~dr~d\theta$$ This should be easy to evaluate.
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 ${\pi/6}$