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!
A: Your mistake is ignoring that $\rho_v$ depends on $r$ in the first integration step.
You integrated $r^2$ to $\frac 13 r^3$ as if $\rho_v$ was constant – but it's not!
$\rho_v = 2r$ makes the function under the integral equal $2r^3\sin\theta$, and only the sine and $2$ are terms constant (independent) in respect to r. So the first step of integration is:
$$\int_0^{2\pi} \int_0^{\pi} \int_0^1 \rho_v r^2 \sin(\theta) \,dr\,d\theta\,d\phi =\\
\int_0^{2\pi} \int_0^{\pi} \int_0^1 2 r^3 \sin(\theta) \,dr\,d\theta\,d\phi =\\
\int_0^{2\pi} \int_0^{\pi} \left(2 \sin(\theta) \int_0^1 r^3 \,dr\right)\,d\theta\,d\phi =\\
\int_0^{2\pi} \int_0^{\pi} \left(2 \sin(\theta) \left[\frac 14 r^4\right]_0^1\right)\,d\theta\,d\phi =\\
\int_0^{2\pi} \int_0^{\pi} \left(2 \sin(\theta) \cdot \frac 14 (1^4-0^4)\right)\,d\theta\,d\phi =\\
\int_0^{2\pi} \int_0^{\pi} \frac 12 \sin(\theta) \,d\theta\,d\phi =\\
\frac 12 \int_0^{2\pi} \int_0^{\pi} \sin(\theta) \,d\theta\,d\phi $$
Now, the whole internal integral expression is independent on $\phi$, hence it is a constant factor in the external integral and we can exclude it (similary to $\frac 12$ above):
$$= \frac 12 \left(\int_0^\pi \sin(\theta) \,d\theta\right) \cdot \left(\int_0^{2\pi} \,d\phi\right) = \\
\frac 12 \left(\int_0^\pi \sin(\theta) \,d\theta\right) \cdot \big(\,\phi\,\big)_0^{2\pi} = \\
\frac 12 \left(\int_0^\pi \sin(\theta) \,d\theta\right) \cdot \left(2\pi -
 0\right) = \\
\pi\,\int_0^\pi \sin(\theta) \,d\theta =\\
\pi\, \Big[-\cos(\theta)\Big]_0^\pi =\\
\pi\, \Big((-\cos(\pi))-(-\cos(0))\Big) =\\
\pi\, \Big((-(-1))-(-1)\Big) =\\
\pi\, (1+1) =\\
2\pi$$
