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As unit normal on the surface is $$\hat{n} = \frac{ax\hat{i}+by\hat{j}+cz\hat{k}}{\sqrt{a^2x^2+b^2y^2+c^2z^2}}$$ I took $$\overrightarrow{F} = \frac{\hat{i}}{ax}$$

so that $$\overrightarrow{F}\cdot \hat{n} = \frac{1}{\sqrt{a^2x^2+b^2y^2+c^2z^2}}$$

Hence the given integral is equal to

$$ \int \overrightarrow{F}\cdot dS = \int div \overrightarrow{F} dV = \int -\frac{1}{ax^2} dV $$ over ellipsoid.

I can't think of any method to compute this last integral. Please give some ideas.

Ng Chung Tak
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Prince Kumar
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    The problem with your calculation is that $\vec{F}$ that you picked is not defined at all points inside the ellipsoid (i.e., it is not defined on the plane $x=0$). Therefore, you cannot use the Divergence Theorem. One possible choice is $\vec{F}:=\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$, so that $\vec{F}\cdot \hat{n}=\dfrac{1}{\sqrt{a^2x^2+b^2y^2+c^2z^2}}$ on the ellipsoid's surface. As $\vec{\nabla}\cdot\vec{r}=3$, the integral is just three times the volume of the ellipsoid, which is $\dfrac{4\pi}{3}\left(\dfrac1{\sqrt{abc}}\right)$. Multiply that by $3$ to get the correct answer. – Batominovski Aug 17 '18 at 20:50

1 Answers1

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You should choose suitable $\mathbf{F}$,

\begin{align} \mathbf{F} &= \frac{(x,y,z)}{ax^2+by^2+cz^2} \\ \nabla \cdot \mathbf{F} &= \frac{1}{ax^2+by^2+cz^2} \\ \end{align}

Using reduced coordinates:

\begin{align} (x,y,z) &= \left( \frac{x'}{\sqrt{a}}, \frac{y'}{\sqrt{b}}, \frac{z'}{\sqrt{c}} \right) \\ dV &= \frac{dx' \, dy' \, dz'}{\sqrt{abc}} \\ \iiint \nabla \cdot \mathbf{F} \, dV &= \iiint \color{red}{\frac{1}{x'^2+y'^2+z'^2}} \times \frac{dx' \, dy' \, dz'}{\sqrt{abc}} \\ &= \frac{1}{\sqrt{abc}} \int_0^{2\pi} d\phi \int_0^{\pi} \sin \theta \, d\theta \int_0^1 \color{red}{\frac{1}{r^2}} \times r^2 \, dr \\ &= \frac{4\pi}{\sqrt{abc}} \end{align}

Compare the results of another answer of mine here.

Ng Chung Tak
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