Discussion
The flux of the vector field \(\vec{u} = x\hat{i} + y\hat{j} + z\hat{k}\) flowing out through the surface of the ellipsoid
\(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1,\) \(a \gt b \gt c \gt 0,\)
is
Answer: D
The outward flux is \(4\pi abc\).
By the divergence theorem, the flux through the closed ellipsoid equals the integral of \(\nabla\cdot\vec{u}\) over the solid it bounds. Here
\[ \begin{aligned} \nabla\cdot\vec{u} &= 1 + 1 + 1 = 3 \\ \text{flux} &= 3\cdot\tfrac{4}{3}\pi abc = 4\pi abc \end{aligned} \]
using the volume \(\frac{4}{3}\pi abc\) of the ellipsoid. The condition \(a \gt b \gt c\) plays no part. \(\pi abc\), \(2\pi abc\) and \(3\pi abc\) come from using the wrong volume or dropping the factor 3 from the divergence; the volume alone would be \(\frac{4}{3}\pi abc\), which is not offered.
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