So I have tried to solve this problem, but I'm running into a problem, because the top circle (intersection of the function with z=1) when you project it onto the xy plane is smaller than the circle that it creates with the plane z=0. so when you try to compute the integral you only get the cylinder part :/ can someone please help me with this I've been stuck on this question for over a day now....
Here is my attempt :
Maybe I have to change the integrations, like change it from dzdydx to dydxdz or dxdydz etc.. But I'm not sure how to do that
In cylinder coordinates with $r^2=x^2+y^2$ and Jacobian $r$: $$ V=\int_0^1 dz \int_{x^2+y^2\le 1/(z+1)^2} dx dy = \int_0^1 dz \int_{r=0}^{1/(z+1)} r dr \int_0^{2\pi} d\phi $$ $$ = 2\pi \int_0^1 dz \int_{r=0}^{1/(z+1)} r dr = \pi \int_0^1 dz r^2\mid _{0}^{1/(z+1)} $$ $$ = \pi \int_0^1 dz \frac{1}{(z+1)^2} = \pi \int_1^2 dz \frac{1}{z^2} = \pi [-\frac{1}{z}]\mid_1^2=\frac{\pi}{2}. $$