# Evaluate the surface integral $\int \int z \cos \gamma \, dS$ over an unit sphere.

Evaluate $\iint z \cos \gamma \, dS$, over the outside of the $unit$ $sphere$ centred at origin, where γ is the inclination of the normal surface at any point of the $unit$ $sphere$ with the z-axis.

How do I deal with this $\gamma$ term? I know for a point $(x,y,z)$ on the unit sphere $\cos \gamma = z$. So can I replace the term $z \cos \gamma$ by $z^2$? I am confused about this because the $z$ in the integrand is a scalar function and not a point on the surface.

• Each "surface element" $dS$ sits at a certain point $(x,y,z)$ in space. Take the $z$ of this point when dealing with $z^2\>dS$. – Christian Blatter Dec 23 '14 at 12:43
The quick way is without parametrization: View $S^2$ as union of infinitesimal lampshades having radius $\rho=\sin\theta$ and width $d\theta$. The area element is then given by $${\rm d}S=2\pi\rho\>d\theta=2\pi\sin\theta\>d\theta\ ;$$ furthermore we have $$z\cos\theta=\cos^2\theta\ .$$ The integral in question then becomes $$J:=\int_{S^2}z\>\cos\theta\>{\rm d}S=2\pi\int_0^\pi\cos^2\theta\sin\theta\>d\theta=-{2\pi\over3}\cos^3\theta\biggr|_0^\pi={4\pi\over3}\ .$$ But you can also parametrize $S^2$ as follows: (\phi,\theta)\mapsto\left\{\eqalign{x&=\sin\theta\cos\phi \cr y&=\sin\theta\sin\phi\cr z&=\cos\theta\cr}\right.\qquad(0\leq\phi\leq2\pi, \ 0\leq\theta\leq \pi)\ . The resulting ${\rm d}S$ (you have to compute it!) will then be $${\rm d}S=\sin\theta\>{\rm d}(\phi,\theta)\ ,$$ so that we obtain $$J=\int_0^{2\pi}\int_0^\pi \cos^2\theta\>\sin\theta\>d\theta\>d\phi\ ,$$ leading to the same end value.