Math Subject GRE 1268 Problem 64 Flux of Vector Field 
What is the value of the flux of the vector field F, defined on $R^3$ by $F(x,y,z) = (x,y,z)$ through the surface $z=\sqrt{1-x^2-y^2}$ oriented with upward-pointing normal vector field?
  $$\begin{matrix}
  (\text{A})\; 0              \quad&
  (\text{B})\; \frac{2\pi}{3} \quad&
  (\text{C})\; \pi            \quad&
  (\text{D})\; \frac{4\pi}{3} \quad&
  (\text{E})\; 2\pi 
\end{matrix}$$

Given that this is a GRE question, I'd like to know how it might be done as quickly as possible.   
 A: (E).  Over the surface, the vector field itself is unit and normal, so the flux should just be the area of the surface, $2\pi$.
(ETA: To elaborate, the surface is the upper unit hemisphere.  The value of the vector field at each point is the unit vector from the origin to that point, which is coincident with the normal at that point, so the dot product is unity.  Thus, the flux should be the surface area, which is half the surface area of the unit sphere.)
A: This is a cute application of the divergence theorem:
$$
\iint_{\partial V} F\cdot dA \;\; =\;\; \iiint_V \nabla\cdot F dV.
$$
Note that the surface is a hemisphere and $\nabla \cdot F = 3$.  We can take the volume to include the unit disc at $z=0$, but since the infinitesimal flux $F\cdot dA = (x,y,z)\cdot (-\hat{k}) = -z = 0$, this part of the surface contributes nothing to the flux.  Since $\nabla\cdot F$ is constant in this volume, and the volume is $\frac{1}{2} \frac{4\pi}{3} = \frac{2\pi}{3}$ you find that the flux is given as $3\cdot \frac{2\pi}{3} = 2\pi$.  The answer is E.
