Surface Integral

If we consider the function $f(x,y,z) = \tfrac{x^2}{4} + \tfrac{y^2}{\pi^2} + \tfrac{z^2}{e^2}$ , so that the surface $\Omega$ is the ellipsoid $f(x,y,z) = 1$ , then since $\mathbf{x} \cdot \nabla\,f \; = \; 2f$ we see that $\left(\tfrac{x^2}{16} + \tfrac{y^2}{\pi^4} + \tfrac{z^2}{e^4}\right)^{-\frac12}\,dS \; = \; \mathbf{x} \cdot d\mathbf{S}$ on the surface $\Omega$ and hence $I \; = \; \iint_{\Omega} \frac{\mathbf{x} \cdot d\mathbf{S}}{|\mathbf{x}|^3} \; = \; -\iint_\Omega \nabla \phi\,\cdot d\mathbf{S}$ where $\phi(\mathbf{x}) = \tfrac{1}{|\mathbf{x}|}$ . Since $\phi$ is harmonic except at the origin, we deduce from Green's Theorem that $I \; = \; -\iint_{C_u} \nabla \phi \cdot d\mathbf{S}$ where $C_u$ is a sphere centred at the origin with radius $u > 0$ . But this means that $I \; = \; \iint_{C_u} u^{-3} \times u \times u^2d\Omega \; = \; \iint_{C_u}\,d\Omega \; = \; 4\pi$ where $d\Omega$ represents integration over solid angle. Thus the integral is equal to $\boxed{4\pi}$

$\begin{aligned} I&=\iint_\Omega\left(x^2+y^2+z^2\right)^{-3/2}\left(\frac{x^2}{16}+\frac{y^2}{\pi^4}+\frac{z^2}{e^4}\right)^{-1/2}dS\\ &=2\iint\limits_{\frac{x^2}{4}+\frac{y^2}{\pi^2}\le 1}\left[\left(x^2+y^2+z^2\right)^{-3/2}\left(\frac{x^2}{16}+\frac{y^2}{\pi^4}+\frac{z^2}{e^4}\right)^{-1/2}·\frac{e\left(\frac{x^2}{16}+\frac{y^2}{\pi^4}+\frac{z^2}{e^4}\right)^{1/2}}{\sqrt{1-\frac{x^2}{4}-\frac{y^2}{\pi^2}}}\right]_{z=e\sqrt{1-\frac{x^2}{4}-\frac{y^2}{\pi^2}}}dxdy\\ &=2e\iint\limits_{\frac{x^2}{4}+\frac{y^2}{\pi^2}\le 1}\left[x^2+y^2+e^2\left(1-\frac{x^2}{4}+\frac{y^2}{\pi^2}\right)\right]^{-3/2}\left(1-\frac{x^2}{4}+\frac{y^2}{\pi^2}\right)^{-1/2}dxdy&\color{#3D99F6}\text{Transform variables to }x=2r\cos\theta,y=\pi r\sin\theta\\ &=16\pi e\int_0^{\pi/2}d\theta\int_0^1[e^2+r^2(4\cos^2\theta+\pi^2\sin^2\theta-e^2)]^{-3/2}(1-r^2)^{-1/2}rdr&\color{#3D99F6}\text{Transform variables to }s=r^2,\ ds=2rdr\\ &=8\pi e\int_0^{\pi/2}d\theta\int_0^1[e^2+s(4\cos^2\theta+\pi^2\sin^2\theta-e^2)]^{-3/2}(1-s)^{1/2}ds\\ &=\frac{8\pi}{e^2}\int_0^{\pi/2}d\theta\int_0^1\left[1+s\frac{4\cos^2\theta+\pi^2\sin^2\theta-e^2}{e^2}\right]^{-3/2}(1-s)^{-1/2}ds \end{aligned}$ Set $\Large p=\frac{4\cos^2\theta+\pi^2\sin^2\theta-e^2}{e^2}$ , we have $\begin{aligned} J(p)&=\int_0^1(1+ps)^{-3/2}(1-s)^{-1/2}ds&\color{#3D99F6}\text{Transform variables to }t=(1-s)^{1/2},\ dt=-\frac12(1-s)^{-1/2}ds\\ &=2\int_0^1(1+pt^2)^{-3/2}dt, \end{aligned}$ where $p>-1$ .

When $p=0$ , $J(0)=2$ .

When $p>0$ , $\begin{aligned} J(p)&=\frac{2}{(1+p)^{3/2}}\int_0^1\left(1-\frac{p}{1+p}t^2\right)^{-3/2}dt&\color{#3D99F6}\text{Transform variables to }\sqrt\frac{p}{1+p}t=\sin\phi,\ \sqrt\frac{p}{1+p}dt=\cos\phi d\phi\\ &=\frac{2}{(1+p)^{3/2}}\int_0^{arcsin\sqrt\frac{p}{1+p}}\sqrt\frac{1+p}{p}\sec^2\phi d\phi\\ &=\left[\frac{2}{(1+p)\sqrt p}\tan\phi\right]_0^{arcsin\sqrt\frac{p}{1+p}}\\ &=\frac{2}{1+p}. \end{aligned}$ When $p<0$ ,

$\begin{aligned} J(p)&=\frac{2}{(1+p)^{3/2}}\int_0^1\left(1+\frac{-p}{1+p}t^2\right)^{-3/2}dt&\color{#3D99F6}\text{Transform variables to }\sqrt\frac{-p}{1+p}t=\tan\phi,\ \sqrt\frac{-p}{1+p}dt=\sec^2\phi d\phi\\ &=\frac{2}{(1+p)^{3/2}}\int_0^{arctan\sqrt\frac{-p}{1+p}}\sqrt\frac{1+p}{-p}\cos\phi d\phi\\ &=\left[\frac{2}{(1+p)\sqrt {-p}}\sin\phi\right]_0^{arctan\sqrt\frac{-p}{1+p}}\\ &=\frac{2}{1+p}. \end{aligned}$ In a word, $J(p)=\frac{2}{1+p}$ , $p>-1$ .

Hence $\begin{aligned} I&=\frac{8\pi}{e^2}\int_0^{\pi/2}J\left(\frac{4\cos^2\theta+\pi^2\sin^2\theta-e^2}{e^2}\right)d\theta\\ &=\frac{8\pi}{e^2}\int_0^{\pi/2}\frac{2e^2}{4\cos^2\theta+\pi^2\sin^2\theta}d\theta\\ &=8\pi\int_0^{\pi/2}\frac{d\theta}{4\cos^2\theta+\pi^2\sin^2\theta}\\ &=8\pi\int_0^{\pi/2}\frac{d\tan\theta}{4+\pi^2\tan^2\theta}\\ &=8\int_0^{\pi/2}\frac{d\left(\frac{\pi}{2}\tan\theta\right)}{1+\left(\frac{\pi}{2}\tan\theta\right)^2}\\ &=\left[8\arctan\left(\frac{\pi}{2}\tan\theta\right)\right]_0^{\frac{\pi}{2}-}\\ &=4\pi\approx\boxed{12.57}. \end{aligned}$ Remark: The surface integral

$I=\displaystyle\iint_\Omega\left(x^2+y^2+z^2\right)^{-3/2}\left(\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}\right)^{-1/2}dS \text{, where }\Omega\text{ denotes ellipsoid }\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$

has the same answer.