Let $f$ be defined on $Q = [ -1,1 ] \times [ -1, 1 ]$ as follows: $f(x,y) = x^2+y^2$ if $x^2 + y^2 \le 1,$ and otherwise, $f(x,y) = 0.$ Compute the volume of this ordinate set.
I tried expressing this volume as $$\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} x^2+y^2 dy dx,$$
but after computing the inner integral, the resulting outer integral seems a bit complicated. It should still be solvable, but is there a nicer way to do this?
Yes! Absolutely.
You can use polar coordinates to express the same integral. You are basically integrating over the unit circle, since anything outside of $x^2 + y^2 \leq 1$ is $0$.
One thing to note is that $\,dx\,dy = r\,dr\,d\theta$, when making that conversion.
So to go over the whole unit circle we would have $0 \leq r \leq 1$, and $0 \leq \theta \leq 2\pi$.
And of course, your integrand is $x^2+y^2$, which is equal to $r^2$.
So the new integral is $\displaystyle \int_0^{2\pi}\int_0^1 r^3 \,dr\,d\theta$, which is super easy comparatively.