Discussion
The value of the integral
\[ \oint\!\!\oint_{S} \vec{r} \cdot \vec{n}\, dS \]
over the closed surface S bounding a volume V, where \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\) is the position vector and \(\vec{n}\) is the normal to the surface S, is
Answer: C
The integral equals 3V.
By the divergence (Gauss) theorem, the flux of a vector field through a closed surface equals the volume integral of its divergence:
\[ \begin{aligned} \oint\!\!\oint_{S} \vec{r} \cdot \vec{n}\, dS &= \iiint_{V} \nabla \cdot \vec{r}\, dV \\ \nabla \cdot \vec{r} &= 1 + 1 + 1 = 3 \\ \text{flux} &= 3V \end{aligned} \]
Check with a sphere of radius a: \(\vec{r} \cdot \vec{n} = a\) everywhere on the surface, so the flux is \(a \times 4\pi a^{2} = 3 \times \frac{4}{3}\pi a^{3}\). V and 2V count only one or two of the three coordinate directions, and 4V has no basis.
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