TITLE: integral of $\cos (x+y+z)$ over a sphere
QUESTION [7 upvotes]: Let $r>0$ and $S_r=\{(x,y,z)\in\mathbb{R}^3|x^2+y^2+z^2=r^2\}$.
Show that
$$\int_{S_r}\cos x\cos y\cos z\ dS=\int_{S_r}\cos (x+y+z)\ dS$$
and find the value of the integrals.
I know that we can parametrize $S_r$ and try to solve the integrals by hand or try to expand the integrands using trigonometric identities, but I had no success in doing so.
I am asking for a hint on how to do this. Moreover, is there some way to handle with integrals over spheres that involve "cosines and sines" (before parametrization). What I mean by this is that when the integrand does not involve "cosines and sines", parametrization often helps but in a situation like the above I don't know what to do.

REPLY [4 votes]: Note that
$$\eqalign{&\cos x\cos y\cos z\cr &\qquad={1\over4}\bigl(\cos(x+y+z)+\cos(x+y-z)+\cos(y+z-x)+\cos(z+x-y)\bigr)\ .\cr}$$
Both statements then follow from the observation that
$$\int_{S_r}\cos\bigl({\bf u}\cdot{\bf x})\>dS=\int_{-r}^r\cos\bigl(|{\bf u}|\>z)\>2\pi r\>dz={4\pi r\over\sqrt{3}}\>\sin\bigl(\sqrt{3}\, r\bigr)\ .$$
Here we have used that the area of a sphere between two  parallel planes is equal to the area cut out by these planes from the  cylinder tangent to the sphere along the corresponding equator.