Finding the mass using sphere coordinates I need to find the mass of $E$, where $E$ is outside the sphere $x^2 + y^2 + z^2 = 2z$, but inside the hemisphere $x^2 + y^2 + z^2 = 4$, $z\geq 0$ and its density is $p(x,y,z)=z\sqrt{x^2 +y^2 +z^2}$.
 A: Okay! For further study you can also refer to 
http://mathworld.wolfram.com/SphericalCoordinates.html
but for how we wanna solve the question in spherical coordinates. First we need to convert any thing to spherical equivalent for example converting $x^2+y^2+z^2=2z$ in spherical coordinates considering $x^2+y^2+z^2=R^2$ and $z=R\cos\theta$ is:
$$R^2=2R\cos\theta\to R=2\cos\theta$$
also the hemisphere is $R=2$ where $z>0$ results in $\theta<\frac{\pi}{2}$ and the density function is $p(R,\theta,\phi)=R^2\cos\theta$. Now let's get to corresponding integral due to the problem. Before going to write it note that $dxdydz=R^2\sin\theta d\theta d \phi dR$:
$$I=\iiint{p(R,\theta,\phi)}R^2\sin\theta d\theta d \phi dR=\int_{0}^{2\pi}\int_{0}^{{\pi\over 2}}\int_{2\cos\theta}^{2}R^4\sin\theta\cos\theta dRd\theta d\phi$$
Now the integral is simple to follow:
$$I=2\pi\int_{0}^{{\pi\over 2}}\int_{2\cos\theta}^{2}R^4\sin\theta\cos\theta dRd\theta=2\pi\int_{0}^{{\pi\over 2}}\frac{32}{5}(1-\cos^5\theta)\sin\theta\cos\theta d\theta$$$$=\frac{64\pi}{5}\int_{0}^{1}u-u^6du=\frac{32\pi}{7}$$
