Discussion
Consider a cube of unit edge length and sides parallel to co-ordinate axes, with its centroid at the point (1, 2, 3). The surface integral \(\int_{A} \vec{F} \cdot d\vec{A}\) of a vector field \(\vec{F} = 3x\hat{i} + 5y\hat{j} + 6z\hat{k}\) over the entire surface \(A\) of the cube is ______.
Answer: A
The surface integral is 14.
By the divergence (Gauss) theorem, the outward flux through a closed surface equals the volume integral of the divergence inside it:
\[ \begin{aligned} \nabla \cdot \vec{F} &= \frac{\partial (3x)}{\partial x} + \frac{\partial (5y)}{\partial y} + \frac{\partial (6z)}{\partial z} \\ &= 3 + 5 + 6 = 14 \end{aligned} \]
The divergence is a constant, so the flux is 14 times the volume of the cube, which is 1. The position of the centroid at (1, 2, 3) does not matter, because the divergence is the same everywhere.
27, 28 and 31 cannot arise: with a constant divergence of 14 and a volume of 1, the flux is 14 wherever the cube sits; 28, for instance, is double the true flux.
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