Calculating triple integral in spherical coordinates

Question

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?

Answer 1

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!

Answer 2

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): $$= \frac 12 \left(\int_0^\pi \sin(\theta) \,d\theta\right) \cdot \left(\int_0^{2\pi} \,d\phi\right) = \\ \frac 12 \left(\int_0^\pi \sin(\theta) \,d\theta\right) \cdot \big(\,\phi\,\big)_0^{2\pi} = \\ \frac 12 \left(\int_0^\pi \sin(\theta) \,d\theta\right) \cdot \left(2\pi - 0\right) = \\ \pi\,\int_0^\pi \sin(\theta) \,d\theta =\\ \pi\, \Big[-\cos(\theta)\Big]_0^\pi =\\ \pi\, \Big((-\cos(\pi))-(-\cos(0))\Big) =\\ \pi\, \Big((-(-1))-(-1)\Big) =\\ \pi\, (1+1) =\\ 2\pi$$