Evaluate $$I=\iiint_V \sqrt{x^2+y^2+z^2}\, dV\,,$$ where $V: x^2 + y^2 + z^2 \leq 2z$.
I tried using the cylindrical coordinates to arrive at:
$$I = \int\int_R\int_{z=1-\sqrt{1-r^2}}^{1+\sqrt{1-r^2}}r\cdot\sqrt{r^2+z^2}\,dz\,dr\,d\theta$$ where $R = r \leq 1$
But this is very tough to evaluate. Is there a better way to do this?
In spherical coordinates we have that
$$x^2+y^2+z^2 \leq 2z \implies \rho \leq 2\cos\phi$$
so let's set up in the integral in this coordinate system. As an added bonus, let's do the angular integral first:
$$I = \int_0^{2\pi} \int_0^2 \int_0^{\cos^{-1}\left(\frac{\rho}{2}\right)} \rho^3\sin\phi \:d\phi \:d\rho \:d\theta = 2\pi\int_0^2\rho^3-\frac{1}{2}\rho^4\:d\rho = \frac{8\pi}{5}$$