UPSC Maths 2013 Paper 1 Q8c — Solution

15 marks · Section B

Question

By using Divergence Theorem of Gauss, evaluate the surface integral

\iint_{S}(a^{2}x^{2}+b^{2}y^{2}+c^{2}z^{2})^{-1/2}\,dS,

where $S$ is the surface of the ellipsoid $ax^{2}+by^{2}+cz^{2}=1$ , $a,b,c$ being all positive constants.

Technique

Spot $\vec r\cdot\hat n$ structure in the integrand (using the surface equation to simplify the numerator); divergence theorem reduces to $3\cdot\text{Vol}$ .

Solution

Strategy. Recognise the integrand as $\vec r\cdot\hat n$ for a clever choice of $\vec r$ ; then divergence theorem converts to a volume integral.

Step 1 — Outward unit normal on the ellipsoid

For the level surface $F(x,y,z)=ax^{2}+by^{2}+cz^{2}-1=0$ :

\nabla F=(2ax,\,2by,\,2cz),\quad|\nabla F|=2\sqrt{a^{2}x^{2}+b^{2}y^{2}+c^{2}z^{2}}.

Unit outward normal:

\hat n=\frac{(ax,\,by,\,cz)}{\sqrt{a^{2}x^{2}+b^{2}y^{2}+c^{2}z^{2}}}.

Step 2 — Recognise the integrand

Take $\vec r=(x,y,z)$ . Then on $S$ ,

\vec r\cdot\hat n=\frac{ax^{2}+by^{2}+cz^{2}}{\sqrt{a^{2}x^{2}+b^{2}y^{2}+c^{2}z^{2}}}=\frac{1}{\sqrt{a^{2}x^{2}+b^{2}y^{2}+c^{2}z^{2}}},

where the last step uses $ax^{2}+by^{2}+cz^{2}=1$ on $S$ .

I=\iint_{S}\vec r\cdot\hat n\,dS.

Step 3 — Apply divergence theorem

I=\iint_{S}\vec r\cdot\hat n\,dS=\iiint_{V}\nabla\cdot\vec r\,dV=\iiint_{V}3\,dV=3\,\text{Vol}(V).

Step 4 — Volume of the ellipsoid

$ax^{2}+by^{2}+cz^{2}=1$ is $\dfrac{x^{2}}{1/a}+\dfrac{y^{2}}{1/b}+\dfrac{z^{2}}{1/c}=1$ , an ellipsoid with semi-axes $1/\sqrt a,\,1/\sqrt b,\,1/\sqrt c$ . So

\text{Vol}(V)=\frac{4\pi}{3}\cdot\frac{1}{\sqrt a}\cdot\frac{1}{\sqrt b}\cdot\frac{1}{\sqrt c}=\frac{4\pi}{3\sqrt{abc}}.

Step 5 — Final value

I=3\cdot\frac{4\pi}{3\sqrt{abc}}=\frac{4\pi}{\sqrt{abc}}.

Answer

\boxed{\;I=\dfrac{4\pi}{\sqrt{abc}}.\;}