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!

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,
$\displaystyle \int_0^{2\pi} \int_0^{\pi/2} \int_0^{\cos\phi} \rho^3 \sin\phi \ d\rho \ d\phi \ d\theta = \frac{\pi}{10}$
