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UPSC 2021 Maths Optional Paper 1 Q6c — Step-by-Step Solution

15 marks · Section B

Line integrals · Vector Analysis · asked 8× in 14 yrs · Read the full method →

Question

Evaluate $\int_C\vec F\cdot d\vec r$ where $C$ is an arbitrary closed curve in the $xy$-plane and $\vec F=\dfrac{-y\hat\imath+x\hat\jmath}{x^2+y^2}$.

Technique

Check curl-free ($\partial Q/\partial x=\partial P/\partial y$); split into cases based on whether origin (the singularity) is enclosed; for enclosed case, deform contour to a small circle.

Solution

Setup. $P=-y/(x^2+y^2)$, $Q=x/(x^2+y^2)$. Check $\partial Q/\partial x=\partial P/\partial y$ (would mean $\vec F$ is “locally” conservative, but…).

$\partial Q/\partial x=\dfrac{(x^2+y^2)-x\cdot 2x}{(x^2+y^2)^2}=\dfrac{y^2-x^2}{(x^2+y^2)^2}$.

$\partial P/\partial y=\dfrac{-(x^2+y^2)+y\cdot 2y}{(x^2+y^2)^2}=\dfrac{y^2-x^2}{(x^2+y^2)^2}$.

So $\partial Q/\partial x=\partial P/\partial y$ everywhere except at the origin (where the field is undefined).

Step 1 — Cases

The integral $\oint_C\vec F\cdot d\vec r$ depends on whether $C$ encloses the origin:

Case A: $C$ does NOT enclose the origin. $\vec F$ is smooth and “closed” (curl-free in the plane) on the region inside $C$. By Green’s theorem, $\oint_C\vec F\cdot d\vec r=\iint_{R}(\partial Q/\partial x-\partial P/\partial y)\,dA=0$.

Case B: $C$ encloses the origin. Cannot apply Green’s directly (singularity inside). Replace $C$ with a small circle $C_\epsilon$ of radius $\epsilon$ around origin. By Stokes-like argument, $\oint_C=\oint_{C_\epsilon}$ (since the region between them has $\partial Q/\partial x-\partial P/\partial y=0$).

Compute $\oint_{C_\epsilon}$: on the circle, $x=\epsilon\cos\theta$, $y=\epsilon\sin\theta$, $dx=-\epsilon\sin\theta\,d\theta$, $dy=\epsilon\cos\theta\,d\theta$.

$P\,dx+Q\,dy=\dfrac{-y\,dx+x\,dy}{x^2+y^2}=\dfrac{\epsilon^2\sin^2\theta+\epsilon^2\cos^2\theta}{\epsilon^2}\,d\theta=d\theta$.

$\oint_{C_\epsilon}=\int_0^{2\pi}d\theta=2\pi$ (for counter-clockwise orientation).

Step 2 — Result

Answer