I need to calculate the following integral $$\iint_D \left(\sqrt{a^2 - x^2 -y^2} - \sqrt{x^2 + y^2}\right)\:\mathrm{d}x\:\mathrm{d}y$$
where $D$ is the disk $x^2 + y^2 \leq a^2$
Using the transformation to polar coordinates $x=r\cos \theta$ and $y = r \sin \theta$, we have the new integration region $$ 0 < r < a$$ $$ 0 < \theta < 2 \pi$$
and the integral will be
$$ \int_0^{2\pi} \int_0^a \left(\sqrt{a^2 - r^2} - r\right)r \:\mathrm{d}r\:\mathrm{d}\theta$$
But I am getting $0$ as answer, which is not the given answer.
What am I doing wrong? Or maybe is the answer wrong? Thanks in advance!
Using polar co-ordinates as you have, we have:
$$\iint_{D}\left(\sqrt{a^{2}-x^{2}-y^{2}}-\sqrt{x^{2}+y^{2}}\right)\:\mathrm{d}A=\int_{0}^{2\pi}\int_{0}^{a}r\left(\sqrt{a^{2}-r^{2}}-r\right)\:\mathrm{d}r\:\mathrm{d}\theta$$
Expanding the integrand we have:
$$2\pi\int_{0}^{a}(r\sqrt{a^{2}-r^{2}}-r^{2})\:\mathrm{d}r=2\pi\left[-\frac{1}{3}(a^{2}-r^{2})^{\frac{3}{2}}-\frac{r^{3}}{3}\right]_{0}^{a}$$
Expanding this out we get:
$$2\pi\left(-\frac{a^{3}}{3}+\frac{a^{3}}{3}\right)=2\pi\left(0\right)=0$$
Which is what you have calculated.