Determine the volume of $A:=\{(x,y,z)\in \mathbb R^3 : \sqrt{x^2+y^2}\leq f(z)\}$

Question

Let $f\in L^2(\mathbb R)$ and $f\geq0$. Determine $A:=\{(x,y,z)\in \mathbb R^3 : \sqrt{x^2+y^2}\leq f(z)\}$.

"Normal" substitution $(x=rcos(\phi),y=rsin(\phi))$ did not help a lot, since I dont have any information about f(z)

Answer 1

You were on the right track. Using polar coordinates, we find that the volume of $A$ is $$\int_0^{2\pi} \int_{-\infty}^\infty \int_0^{f(z)} r\, dr\, dz\, d\theta = \pi \int_{-\infty}^{\infty} f(z)^2\, dz.$$