Solve this triple integral I am trying to solve this triple integral:
$$\int_0^{ 2\pi} \int_0^4 \int_3^\sqrt{25-r^2} r \: dz \: dr \: d \theta$$.
I get stuck at 
$$\int_0^{ 2\pi} \int_0^4 \ r(\sqrt{25-r^2}-3)  \: dr \: d \theta$$
Should I use u-sub to find the rest? And if so, how?

Given
$$\ \int_{0}^4 r(\sqrt{25-r^2}-3)dr$$
make the substitution
$$\ 25-r^2=t, -2rdr=dt; \implies rdr=-\frac{1}{2}dt$$
so you get
$$\ \int_{5}^3 -\frac{1}{2}(\sqrt t-3)dt$$
