UPSC 2017 Maths Optional Paper 1 Q8c-i — Step-by-Step Solution

9 marks · Section B

Gauss divergence theorem · Vector Analysis · asked 10× in 14 yrs · Read the full method →

Question

Evaluate the integral $\displaystyle\iint_S\vec F\cdot\hat n\,ds$ where $\vec F=3xy^2\hat i+(yx^2-y^3)\hat j+3zx^2\hat k$ and $S$ is a surface of the cylinder $y^2+z^2\le 4,\ -3\le x\le 3$ , using divergence theorem.

Technique

Divergence theorem; $\nabla\cdot\vec F=4x^2$ ; integrate over cylinder (disc area $4\pi$ ) $\times\int_{-3}^3x^2dx$ .

Solution

The solid is the cylinder of radius $2$ with axis along the $x$ -axis: $y^2+z^2\le4,\ -3\le x\le3$ . By Gauss’s divergence theorem,

\iint_S\vec F\cdot\hat n\,dS=\iiint_V(\nabla\cdot\vec F)\,dV.

Step 1 — Divergence

\nabla\cdot\vec F=\frac{\partial}{\partial x}(3xy^2)+\frac{\partial}{\partial y}(yx^2-y^3)+\frac{\partial}{\partial z}(3zx^2).

=3y^2+(x^2-3y^2)+3x^2=4x^2.

(The $3y^2$ and $-3y^2$ cancel cleanly, leaving $4x^2$ .)

Step 2 — Integrate over the cylinder

Use coordinates with the disc $y^2+z^2\le4$ (radius $2$ , area $\pi\cdot2^2=4\pi$ ) in the cross-section and $x\in[-3,3]$ . Since $4x^2$ is independent of $y,z$ ,

\iiint_V 4x^2\,dV=\int_{-3}^{3}4x^2\Big(\iint_{y^2+z^2\le4}dy\,dz\Big)dx=\int_{-3}^{3}4x^2\,(4\pi)\,dx.

=16\pi\int_{-3}^{3}x^2\,dx=16\pi\cdot\Big[\frac{x^3}{3}\Big]_{-3}^{3}=16\pi\cdot\frac{54}{3}=16\pi\cdot18=288\pi.

Answer

\boxed{\;\iint_S\vec F\cdot\hat n\,dS=288\pi.\;}