I need to calculate
$$\iiint _V\sqrt{x^2+y^2+z^2} \,dx \,dy\, dz$$
where $V$ is the ball
$$x^2+y^2+z^2 \leq 4z \Leftrightarrow x^2 + y^2+(z-2)^2 \leq 4$$
The hint is to use origin centered spherical coordinates.
So, after substitution I get:
$$
r \leq 2 \cos \phi.
$$
This obviously implies that $0\leq r \leq 2\cos \phi$. But as far as I can understand, it also implies that $\phi$, which is always bounded in $\left[ 0,\pi \right] $, now satisfies $0\leq \phi \leq \frac{\pi}{2}$.
Is it true that indeed $\theta$, which has no constraints on it, will satisfy $0\leq \theta \leq 2\pi$?
Is it true that the final integral is
$$
\int_0^{2\pi}d\theta \int_0^{\frac{\pi}{2}} \int_0^{2\cos \phi} r^3\sin\phi \,dr\,d\phi ?
$$
Thanks a lot!
Best Answer
Yes; except for the fact that, as has already been pointed out in a comment, it's $r\le4\cos\phi$, not $2\cos\phi$, it all looks good.