Trouble setting up the integrals for this problem.
Find the volume of the solid bounded by $x^2 + y^2 = 1, z = 0$, $z = 6$, $y\geq 1/2$. Use integration with Spherical coordinates. (Hint: Use two triple integrals and tangent inverse)
Using Mathematica, the answer is this and the solid looks like this.
The integral that I have setup, which doesn't evaluate to the correct answer, is 
When $z=6$ and $y=\frac{1}{2}$, we get $\rho\cos\phi=6, \rho\sin\phi\sin\theta=\frac{1}{2}$; so $\tan\phi\sin\theta=\frac{1}{12}$ gives
$\tan\phi=\frac{1}{12}\csc\theta$ and $\phi=\tan^{-1}\left(\frac{1}{12}\csc\theta\right)$.
When $z=6$ and $r=1$, we get $\rho\cos\phi=6, \rho\sin\phi=1$ so $\tan\phi=\frac{1}{6}$ and $\phi=\tan^{-1}\frac{1}{6}$.
When $z=6, \rho=6\sec\phi$; and when $y=\frac{1}{2}, \rho\sin\phi\sin\theta=\frac{1}{2}\implies \rho=\frac{1}{2}\csc\phi\csc\theta$.
This gives $\displaystyle V=\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\int_{\tan^{-1}\frac{1}{12}\csc\theta}^{\tan^{-1}\frac{1}{6}}\int_{\frac{1}{2}\csc\phi\csc\theta}^{6\sec\phi}\rho^2\sin\phi \;d\rho d\phi d\theta\;+$
$\displaystyle\hspace{1.1 in}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\int_{\tan^{-1}\frac{1}{6}}^{\frac{\pi}{2}}\int_{\frac{1}{2}\csc\phi\csc\theta}^{\csc\phi}\rho^2\sin\phi \;d\rho d\phi d\theta$.
(I believe the first integral gives $\frac{2\pi}{3}-\sqrt{3}+\frac{1}{8}\ln(7+4\sqrt{3})$ and
the second integral gives $\hspace{.4 in}\frac{4\pi}{3}-\frac{\sqrt{3}}{2}-\frac{1}{8}\ln(7+4\sqrt{3})$, for a sum of $2\pi-\frac{3\sqrt{3}}{2}$.)