Integral of $\frac{x^2+y^2}{\sqrt{1-x^2-y^2}}$

Question

In the solution to problem 12.10 1 part b at this link, they deduce that $$\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \frac{x^2+y^2}{\sqrt{1-x^2-y^2}} dy dx = \pm \frac{4 \pi}3. $$ I understood everything up to this part of the solution, but I am very confused as to how they evaluated this integral. I tried plugging these values into https://www.integral-calculator.com but I'm not getting a nice solution.

Answer 1

Note that $$\frac{x^2+y^2}{\sqrt{1-x^2-y^2}} = \frac{r^2}{\sqrt{1-r^2}}.$$ Also note that the integral is taken inside the unit circle. So we convert to polar coordinates to get $$\int_{0}^{2\pi} \int_{0}^{1} \frac{r^2}{\sqrt{1-r^2}} r \mathrm d r \mathrm d \theta,$$ from which straightforward integration techniques like substituting $r = \sin u$ easily finish the problem.