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UPSC 2021 Maths Optional Paper 1 Q8c — Step-by-Step Solution

15 marks · Section B

Stokes' theorem · Vector Analysis · asked 10× in 13 yrs · Read the full method →

Question

Using Stokes’ theorem, evaluate $\iint_S(\nabla\times\vec F)\cdot\hat n\,dS$ where $\vec F=(x^2+y-4)\hat\imath+3xy\hat\jmath+(2xy+z^2)\hat k$ and $S$ is paraboloid $z=4-(x^2+y^2)$ above $xy$-plane.

Technique

Stokes’ theorem reduces surface integral to line integral around boundary; parametrise boundary circle; integrate trigonometric polynomial.

Solution

Stokes’ theorem: $\iint_S(\nabla\times\vec F)\cdot\hat n\,dS=\oint_C\vec F\cdot d\vec r$ where $C$ is the boundary of $S$.

Boundary $C$: intersection of paraboloid with $xy$-plane $(z=0)$, i.e., $x^2+y^2=4$, the circle of radius 2.

Step 1 — Parametrise $C$

$x=2\cos\phi,\;y=2\sin\phi,\;z=0$, $\phi\in[0,2\pi]$ (counter-clockwise).

$d\vec r=(-2\sin\phi,2\cos\phi,0)\,d\phi$.

Step 2 — Evaluate $\vec F$ on $C$

$\vec F=(x^2+y-4,\;3xy,\;2xy+z^2)$ at $z=0$, $x^2+y^2=4$:

$x^2+y-4=4\cos^2\phi+2\sin\phi-4=-4\sin^2\phi+2\sin\phi=2\sin\phi(1-2\sin\phi)$.

Hmm: $4\cos^2\phi-4=-4\sin^2\phi$, so $x^2+y-4=-4\sin^2\phi+2\sin\phi$.

$3xy=12\sin\phi\cos\phi=6\sin 2\phi$.

$2xy+z^2=12\sin\phi\cos\phi+0=6\sin 2\phi$.

Step 3 — Compute $\vec F\cdot d\vec r$

$\vec F\cdot d\vec r=(x^2+y-4)(-2\sin\phi)+3xy(2\cos\phi)+0$.

$=(2\sin\phi(1-2\sin\phi))(-2\sin\phi)+6\sin 2\phi\cdot 2\cos\phi\cdot...$

Let me redo step by step.

$\vec F$ has components $F_1=-4\sin^2\phi+2\sin\phi$, $F_2=6\sin 2\phi$, $F_3=6\sin 2\phi$ (but only $F_1$ and $F_2$ matter since $dz=0$).

$\vec F\cdot d\vec r=F_1\cdot(-2\sin\phi)+F_2\cdot(2\cos\phi)$ $=(-4\sin^2\phi+2\sin\phi)(-2\sin\phi)+6\sin 2\phi\cdot 2\cos\phi$ $=8\sin^3\phi-4\sin^2\phi+12\sin 2\phi\cos\phi$.

$\sin 2\phi\cos\phi=2\sin\phi\cos^2\phi$.

$=8\sin^3\phi-4\sin^2\phi+24\sin\phi\cos^2\phi$.

$\cos^2\phi=1-\sin^2\phi$: $=8\sin^3\phi-4\sin^2\phi+24\sin\phi(1-\sin^2\phi)$ $=8\sin^3\phi-4\sin^2\phi+24\sin\phi-24\sin^3\phi$ $=-16\sin^3\phi-4\sin^2\phi+24\sin\phi$.

Step 4 — Integrate over $[0,2\pi]$

$\int_0^{2\pi}\sin\phi\,d\phi=0$. $\int_0^{2\pi}\sin^3\phi\,d\phi=0$ (odd function on symmetric interval). $\int_0^{2\pi}\sin^2\phi\,d\phi=\pi$.

So $\oint=-16\cdot 0-4\pi+24\cdot 0=-4\pi$.

Answer