← 2023 Paper 1

UPSC 2023 Maths Optional Paper 1 Q6c — Step-by-Step Solution

20 marks · Section B

Gauss divergence theorem · Vector Analysis · asked 9× in 13 yrs · Read the full method →

Question

Evaluate the integral

where $S$ is the upper part of the surface $4x^2+4y^2+4z^2=1$ above the plane $z=0$ and bounded by the $xy$-plane. Hence, verify Gauss-Divergence theorem.

Technique

Use divergence theorem to convert the surface integral on $S$ (a hemisphere) to a volume integral on the half-ball, plus a flat-disk contribution that vanishes (since the integrand has $z^2$ factor on $\hat k$-component).

Solution

Note on $S$. The equation $4x^2+4y^2+4z^2=1$ is the sphere $x^2+y^2+z^2=1/4$, i.e., radius $1/2$. $S$ is the upper hemisphere ($z\ge 0$).

Strategy. Compute the volume integral via the divergence theorem (much easier), then deduce the surface integral over $S$ alone.

Step 1 — Apply the divergence theorem to the closed half-ball

Let $V$ be the upper half-ball ($x^2+y^2+z^2\le 1/4,\,z\ge 0$). Its boundary is $S$ (hemispherical part) ∪ $D$ (the disk $z=0,\,x^2+y^2\le 1/4$).

Divergence of $\vec F=(3y^2z^2,\,4z^2x^2,\,z^2y^2)$:

Step 2 — Compute the volume integral

In spherical coordinates $x=r\sin\psi\cos\phi,\,y=r\sin\psi\sin\phi,\,z=r\cos\psi$, with $r\in[0,1/2],\,\psi\in[0,\pi/2],\,\phi\in[0,2\pi]$. $dV=r^2\sin\psi\,dr\,d\psi\,d\phi$.

$2zy^2=2(r\cos\psi)(r\sin\psi\sin\phi)^2=2r^3\cos\psi\sin^2\psi\sin^2\phi.$

Compute each:

• $\int_0^{1/2}r^5\,dr=[r^6/6]_0^{1/2}=1/(6\cdot 64)=1/384$.
• $\int_0^{\pi/2}\cos\psi\sin^3\psi\,d\psi=[\sin^4\psi/4]_0^{\pi/2}=1/4$.
• $\int_0^{2\pi}\sin^2\phi\,d\phi=\pi$.

Total: $2\cdot\frac{1}{384}\cdot\frac{1}{4}\cdot\pi=\frac{2\pi}{384\cdot 4}=\frac{\pi}{768}$.

Step 3 — Compute the integral over the disk $D$

On $D$: $z=0$, outward normal $\hat n=-\hat k$ (pointing out of $V$, downward).

$\vec F\cdot(-\hat k)=-z^2y^2|_{z=0}=0.$

$\iint_D\vec F\cdot\hat n\,dS=0.$

Step 4 — Conclude

Answer