# Surface integral over ellipsoid

I've problem with this surface integral: $$\iint\limits_S {\sqrt{ \left(\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}\right)}}{dS}$$, where $$S = \{(x,y,z)\in\mathbb{R}^3: \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}= 1\}$$

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– Mhenni Benghorbal Jun 28 '13 at 18:34

Note that the outward unit normal ${\bf n}$ to the ellipsoid is given by ${\displaystyle {({x \over a^2}, {y \over b^2}, {z \over c^2}) \over \sqrt{ \left(\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}\right)}}}$.

So if ${\bf F} = ({x \over a^2}, {y \over b^2}, {z \over c^2})$, your integral is $\int_S {\bf F} \cdot {\bf n}\,dS$. By the divergence theorem, this is equal to $\int_E div\,\mathbf{F}$, where $E$ is the ellipsoid's interior. But $div \,\mathbf{F}$ is the constant ${1 \over a^2} + {1 \over b^2} + {1 \over c^2}$ and the ellipsoid has volume ${4\pi \over 3}abc$, so the integral will evaluate to $${4\pi \over 3}abc \times \bigg({1 \over a^2} + {1 \over b^2} + {1 \over c^2}\bigg)$$ $$= {4\pi \over 3} \bigg({bc \over a} + {ac \over b} + {ab \over c}\bigg)$$

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I had hoped there would be something like this. But you have pulled ${\bf n}$ out of the blue. – Christian Blatter Jun 29 '13 at 10:14
Not really... when I see a surface integral, one thing I sometimes try is to write it in the form $\int {\bf F} \cdot {\bf n}\,dS$ so that I may use the divergence theorem. So here I compute ${\bf n}$ and then try to see if there's a natural choice for ${\bf F}$. In this case it's pretty clear what ${\bf F}$ should be, once you compute ${\bf n}$. – Zarrax Jun 29 '13 at 14:21

What an interesting integral. I had to resort to referring to the first differential form of the spherical parametrization, but doing that, I am amazed at how this turns out.

We parametrize in the usual way:

$$x=a\sin{u} \cos{v}$$ $$y=b\sin{u} \sin{v}$$ $$z=c \cos{u}$$

where $u \in [0,\pi)$ and $v \in [0,2 \pi)$. The coefficients of the first differential form are

$$E=(a^2 \sin^2{v}+b^2 \cos^2{v}) \sin^2{u}$$ $$F=(b^2-a^2) \sin{u} \cos{u} \sin{v} \cos{v}$$ $$G=(a^2 \cos^2{v}+b^2 \sin^2{v}) \cos^2{u}+c^2 \sin^2{u}$$

The stated integral is equal to

$$\int_0^{\pi} du \: \int_0^{2 \pi} dv \: \sqrt{E G-F^2} \sqrt{\frac{\sin^2{u} \cos^2{v}}{a^2} + \frac{\sin^2{u} \sin^2{v}}{b^2} + \frac{\cos^2{u}}{c^2}}$$

There is an enormous amount of algebra involved in simplifying the integrand. Miraculously, it simplifies a lot, and the integral is equal to

$$\frac{1}{a b c} \int_0^{\pi} du \: \sin{u} \int_0^{2 \pi} dv\: (a^2 b^2 \cos^2{u} + b^2 c^2 \sin^2{u} \cos^2{v} + a^2 c^2 \sin^2{u} \sin^2{v})$$

I really couldn't believe this myself at first, but it does check out for the case $a=b=c$. In any case, these integrals are much easier than one would expect from first seeing this problem, and the reader should have no trouble evaluating them by hand. The result is

$$\frac{4 \pi}{3} \left ( \frac{a\, b}{c} + \frac{a\, c}{b} + \frac{b\, c}{a} \right)$$

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Let the ellipsoid $S$ be given by $${\bf x}(\theta,\phi)=(a\cos\theta\cos\phi,b\cos\theta\sin\phi,c\sin\theta)\ .$$ Then for all points $(x,y,z)\in S$ one has $$Q^2:={x^2\over a^4}+{y^2\over b^4}+{z^2\over c^4}={1\over a^2b^2c^2}\left(\cos^2\theta(b^2c^2\cos^2\phi+a^2c^2\sin^2\phi)+a^2b^2\sin^2\theta\right)\ .$$ On the other hand
$${\rm d}S=|{\bf x}_\theta\times{\rm x}_\phi|\>{\rm d}(\theta,\phi)\ ,$$ and one computes \eqalign{|{\bf x}_\theta\times{\rm x}_\phi|^2&=\cos^4\theta(b^2c^2\cos^2\phi+a^2c^2\sin^2\phi)+a^2b^2\cos^2\theta\sin^2\theta\cr &=\cos^2\theta\ (a^2b^2c^2\ \ Q^2)\ \cr} It follows that your integral ($=:J$) is given by \eqalign{J&=\int\nolimits_{\hat S} Q\ |{\bf x}_\theta\times{\rm x}_\phi|\>{\rm d}(\theta,\phi)=\int\nolimits_{\hat S}abc\ Q^2\ \cos\theta\ {\rm d}(\theta,\phi) \cr &={1\over abc}\int\nolimits_{\hat S}\cos\theta\left(\cos^2\theta(b^2c^2\cos^2\phi+a^2c^2\sin^2\phi)+a^2b^2\sin^2\theta\right)\ {\rm d}(\theta,\phi)\ ,\cr} where $\hat S=[-{\pi\over2},{\pi\over2}]\times[0,2\pi]$. Using $$\int_{-\pi/2}^{\pi/2}\cos^3\theta\ d\theta={4\over3},\quad \int_{-\pi/2}^{\pi/2}\cos\theta\sin^2\theta\ d\theta={2\over3},\quad \int_0^{2\pi}\cos^2\phi\ d\phi=\int_0^{2\pi}\sin^2\phi\ d\phi=\pi$$ we finally obtain $$J={4\pi\over3}\left({ab\over c}+{bc\over a}+{ca\over b}\right)\ .$$

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