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?
 A: Firstly, note that 
$$\int_0^{2 \pi} \int_0^4 r(\sqrt{25-r^2} -3)dr d \theta = 2 \pi \int_0^4 r(\sqrt{25-r^2} -3)dr$$
Now, call $5 \sin u = r$. So
$$\int_0^4 r(\sqrt{25-r^2} -3)dr = \int_0^{\arcsin \frac{4}{5}} 5 \sin u(5 \sqrt{1- \sin^2 u} -3) 5 \cos u \ du =$$
$$= 25 \int_0^{\arcsin \frac{4}{5}} \sin u \cos u (5 \cos u - 3) du = $$
Call $\cos u = t$ (so that $\sin u \ du = -dt$) and get
$$=25 \int_0^{\frac{3}{5}} t (3 - 5t) \ dt = \frac{9}{2}$$
So the final answer is
$$\int_0^{2 \pi} \int_0^4 r(\sqrt{25-r^2} -3)dr d \theta = 9 \pi$$
A: 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
$$\ \int_{5}^3 -\frac{1}{2}(\sqrt t-3)dt$$
A: Observe that if $\;f\;$ is a differentiable function, then
$$\int\sqrt x\;dx=\frac23x\sqrt x+C\implies \int f'(x)\sqrt{f(x)}\;dx=\frac23 f(x)\sqrt{f(x)}+C$$
Here, we have 
$$f(r):=25-r^2\;,\;\;\;f'(r)=-2r\implies$$
$$\int r\sqrt{25-r^2}\;dr=-\frac12\int(25-r^2)'\sqrt{25-r%2}\;dr=-\frac13(25-r^2)\sqrt{25-r^2}+C$$
No need of direct substitution in these cases.
