<s>[INST] Solution Verification: Find the mass of $V$. (Triple integral) 
We define $V=\{(x,y,z)\in\mathbb{R}^3: x^2+y^2+z^2\le z$ } 
Mass Density is given by $g(x,y,z)=\sqrt{x^2+y^2+z^2}$ , 
Find the Mass of $V$.

My Work:
So I need to find $I=\iiint_{V}\sqrt{x^2+y^2+z^2}dxdydz$, and in order to do that, I will move to spherical coordinations: 
$x=rsin\phi cos\theta$ 
$y=rsin\phi sin\theta$ 
$z=rcos\phi$.
And so $x^2+y^2+z^2=r^2\le z = rcos\theta \Longrightarrow r\le cos\theta \Longrightarrow -\frac{\pi}{2} \le\theta \le \frac{\pi}{2}$ ($cos\theta \ge 0)$. 
and so there's no restrictions on $0 \le \phi \le \pi$. and $0 \le r\le cos\theta$. 
So: 
$$I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{\pi}\int^{cos\theta}_{0}r.r^2sin\phi drd\phi d\theta=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}[\int_0^{\pi}\frac{cos^4\theta}{4}sin\phi ]d\phi d\theta=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}[\int^{\pi}_0\frac{1+cos^2(2\theta)}{2}sin\phi]d\phi d\theta = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}[\int^{\pi}_0\frac{3+cos(4\theta)}{4}sin\phi]d\phi d\theta = \frac{1}{4} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}[\int^{\pi}_0(3sin\phi+cos(4\theta)sin\phi)]d\phi d\theta = \frac{1}{4} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}[(\pi3sin\phi+sin(4\pi)sin\phi)]d\phi d\theta = \frac{1}{4} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}[(\pi3sin\phi)]d\phi=\frac{3}{4}\pi (-cos(-\pi) - cos(\pi))=\frac{6\pi}{4}$$I would really appreciate any feedback about my solution, even the smallest mistakes, and would love to get an approval of my answer, Thanks in advance!
 [/INST] Your approach is correct but you have a mistake in your work.
$V=\{(x,y,z)\in\mathbb{R}^3: x^2+y^2+z^2\le z\}$
As you defined $x = \rho \cos\theta \sin\phi, y = \rho \sin\theta \sin\phi, z = \rho \cos\phi$
$x^2+y^2+z^2 \leq z \implies \rho \leq \cos\phi, 0 \leq \phi \leq \dfrac{\pi}{2}$ and $0 \leq \theta \leq 2\pi$
So the integral should be,
$\displaystyle \int_0^{2\pi} \int_0^{\pi/2} \int_0^{\cos\phi} \rho^3 \sin\phi \ d\rho \ d\phi \ d\theta = \frac{\pi}{10}$
</s>