Discussion
The value of the surface integral
∯S z dx dy
where S is the external surface of the sphere \(x^{2} + y^{2} + z^{2} = R^{2}\) is
Answer: C
The surface integral equals \(\dfrac{4\pi}{3}R^{3}\).
The integral \(\oint z\,dx\,dy\) over a closed surface is the flux of the vector field \(\mathbf{F} = (0, 0, z)\). By the divergence theorem it equals the volume integral of \(\nabla \cdot \mathbf{F} = \partial z/\partial z = 1\):
\[ \begin{aligned} \oint_{S} z\,dx\,dy &= \iiint_{V} 1\,dV \\ &= \frac{4}{3}\pi R^{3} \end{aligned} \]
which is the volume of the sphere. Zero would be the answer only if the divergence vanished; \(4\pi R^{3}\) is three times the volume (the flux of the full position vector \(x\,\hat{i} + y\,\hat{j} + z\,\hat{k}\)), and \(\pi R^{3}\) matches no step.
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