Problem: Evaluate the integral
$$\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \frac{2}{(1+x^2+y^2)^2}dydx$$
I am having a hard times understanding this concept and I couldn't solve this question. I was stuck at changing the cartesian to polar. Thank you for helping.
Recall polar coordinates are defined as $(x, y) \mapsto (r \cos(\phi), r \sin(\phi))$. It is an easy exercise to show that the determinant of the Jacobian reads $r(\cos(\phi)^2+\sin(\phi)^2)$.
Let us denote the integral by $I$. Substitute according to the rule given above and obtain $I = \int_{r=0}^{1} \int_{\phi=0}^{2\pi} \frac{2}{\left(1+r^2\right)^2} r d\phi dr$.
I'll leave it as a tiny exercise for your to derive the boundaries of the intergral. In case you're not sure how to obtain them, a sketch might help!
Now we see, that we can directly carry out the integral over $\phi$ and obtain $I = \int_{r=0}^{1} \frac{4\pi}{\left(1+r^2\right)^2} r dr$.
It remains to calculate the integral over $r$. A simple substitution $t:=1+r^2$ yields $I = \int_{t=1}^{2} \frac{2\pi}{t^2} dt = \pi$, where I leave the intermediate steps (like correctly determining the boundaries I gave) as an exercise.