Surface Integral of a dot product in spherical coordinates let $k$ and $l$ be two distinct radial unit vectors.  let $k$ be fixed. Allow that $l$ may point in each and any direction on a sphere with equal probability.  Am i correct that the expected value of $k \cdot l$, $\left<k \cdot l\right> $, is 
$\left<k \cdot l\right> = \frac{\int_0^{2\,\pi}\int_0^{\pi}\, k \cdot l \, \sin{(\phi)} \, d\phi \,   d\theta}{4\pi}$  ?
How can I evaluate this integral?
 A: Note that as $k$ is fixed, we can without loss of generality specify a Cartesian coordinate system with $k$ corresponding to $(0,0,1)$, i.e the usual unit vector $k$. In spherical coordinates $k$ corresponds to the unit vector with $r=1$ and $\phi=0$ (note $\theta$ is arbitrary). 
Your expression for the expected value is correct. Note that integrating over the probability space gives $$\int_0^{2\pi}\int_0^\pi \sin\phi\, d\phi\, d\theta = 4\pi,$$ so the uniform probability density function for picking a vector $l$ with coordinates $\phi,\theta$ is $p(\phi,\theta)=\frac{1}{4\pi}$.
Note that in Cartesian coordinates $l=(cos\theta\sin\phi,\sin\theta\sin\phi,\cos\phi)$ and so $k\cdot l=\cos\phi$. See if you can evaluate the integral now using the substitution rule! You shouldn't be surprised by the answer given the symmetry of the sphere and the fact that $k\cdot l$ is negative if the angle between the vectors is greater than $\frac{\pi}{2}$.
You could also think about the expected value when $l$ is restricted to the half sphere around $k$, i.e $\phi\in[0,\frac{\pi}{2}]$ (remember to adjust the probability density function!)        
