I have evaluated $\int_{S}\frac{z}{(x^2+y^2+z^2)^{3/2}}dV$, where $S$ is the region bounded below by the sphere $x^2+y^2+z^2=2z$ and above by the sphere $x^2+y^2+z^2=1$.
Using spherical coordinates I get:
\begin{align*} \int_{S}\frac{z}{(x^2+y^2+z^2)^{3/2}}dV&=\int_{\theta=0}^{\theta=2\pi}\int_{\phi=0}^{\phi=\pi/3}\int_{\rho=1/2\sec(\phi)}^{\rho=1}\frac{\rho\cos(\phi)}{\rho^3}\rho^2\sin(\phi)d\rho d\phi d\theta\\&+\int_{\theta=0}^{\theta=2\pi}\int_{\phi=\pi/3}^{\phi=\pi/2}\int_{\rho=0}^{\rho=2\cos(\phi)}\frac{\rho\cos(\phi)}{\rho^3}\rho^2\sin(\phi)d\rho d\phi d\theta\\ &=2\pi\int_{\phi=0}^{\phi=\pi/3}\cos(\phi)\sin(\phi)[1-\frac{1}{2}\sec(\phi)]d\phi\\ &+2\pi\int_{\phi=\pi/3}^{\phi=\pi/2}\cos(\phi)\sin(\phi)\cdot 2\cos(\phi)d\phi d\theta\\&=2\pi\cdot\frac{1}{8}+2\pi\cdot\frac{1}{12}=\frac{\pi}{4}+\frac{\pi}{6}=\frac{5}{12}\pi \end{align*}
but the book I took this exercise from says that the solution should instead be $\frac{11}{12}\pi$; even so, I cannot find my mistake so I would be grateful if someone would point out where I have gone wrong. Thanks.
Your solution is missing one integral,
$$\int_0^{2\pi} \int_0^{\tfrac\pi3} \int_0^{\tfrac12\sec(\phi)} \cos(\phi)\sin(\phi) \, d\rho\, d\phi \, d\theta = \frac\pi2$$
which corresponds to the space above the cone defined by $\phi=\dfrac\pi3$ and below the plane $\rho=\dfrac12 \sec(\phi)$.