Q:

Calculating triple integral in spherical coordinates

I am having trouble solving this triple integral:
$$\int_0^{2\pi} \int_0^{\pi} \int_0^1 \rho_v r^2 \sin(\theta) \,dr\,d\theta\,d\phi$$
I've done:
$$\int_0^{2\pi} \int_0^{\pi}  \frac{1}{3}\rho_v \sin(\theta) \,d\theta\,d\phi$$
$$\int_0^{2\pi} \frac{2}{3}\rho_v\,d\phi$$
$\frac{4\pi}{3}\rho_v$ and I am told that $\rho_v = 2r$ which gives $\frac{4\pi}{3}\times2r$. When $r = 1$ the answer is meant to be $2\pi$ but as you can see, that is not what I get. 
Can anyone see my mistake?

A:

The  $ρ_v=2r$ should be subbed in at the start, giving:
$$\int_0^{2\pi} \int_0^\pi \int_0^1  2r^3 \sin(\theta) \,dr\,d\theta\,d\phi$$
Then the answer comes nicely!

