Interaction of an atom with another atom that is distance $r$ away from it is given by $$ V=-\frac{C}{r^6}. $$ If you first use polar coordinates for the total interaction with a an infinite plane that is placed distance $h$ away from it you get for unit density $\frac{C\pi}{2h^4}$. Integrating all the slabs from $h$ to infinity, the interaction of an atom that is place above a semi-infinite crystal surface at the height $h$, you easily get $$ V=\frac{C\pi}{6h^3} $$ with unit density.
Now the question in my textbook that I use for self-study is: Reproduce the same result using an appropriate volume integration in spherical coordinates. This is my (wrong?) solution: $$ V=\int^{2\pi}_0 d\phi \int_0^{\frac{\pi}{2}}d\theta\int^{\infty}-\frac{C}{r^4}\sin\theta dr $$ where I let the last integral start at $r=\frac{h}{\sin\theta}$ (I cannot get it to display inside the block above somehow), because there is no interaction in the space between the atom and the surface of the crystal.
To me this seems all crystal clear, until I try to integrate the above and get: \begin{align} & =-2C\pi\int^{\frac{\pi}{2}}_0 d\theta*(\frac{r^{-3}}{3}|_{\frac{h}{\sin\theta}}^\infty) \\ & =-2C\pi\int^{\frac{\pi}{2}}_0 d\theta \frac{h^{-3}}{3}\sin^4\theta dr \end{align} where you see with the $\int^{\frac{\pi}{2}}_0 d\theta \sin^4\theta=\frac{3\pi}{8}$ it goes wrong.
I have been repeating the same steps many times, also on the previous way with polar coordinates.
Can someone show me where I get it wrong?
So I had my picture wrong, it should be $\frac{h}{\cos\theta}$, then it all works out