← 2022 Paper 1

UPSC 2022 Maths Optional Paper 1 Q8c — Step-by-Step Solution

15 marks · Section B

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

Question

Using Gauss’ divergence theorem, evaluate ∬SF⃗⋅n^ dS\iint_S\vec F\cdot\hat n\,dS where F⃗=xı^−yȷ^+(z2−1)k^\vec F=x\hat\imath-y\hat\jmath+(z^2-1)\hat k and SS is the cylinder formed by z=0, z=1, x2+y2=4z=0,\,z=1,\,x^2+y^2=4.

Technique

Gauss’ divergence: convert closed-surface flux integral to volume integral of divergence. Divergence here is 2z2z, integration trivial.

Solution

Setup. SS is the closed surface of the cylinder: bottom disk (z=0z=0), top disk (z=1z=1), lateral curved surface (x2+y2=4x^2+y^2=4 between z=0z=0 and z=1z=1). Volume VV enclosed: {x2+y2≤4,  0≤z≤1}\{x^2+y^2\le 4,\;0\le z\le 1\}.

Gauss’ theorem: ∬SF⃗⋅n^ dS=∭V∇⋅F⃗ dV\iint_S\vec F\cdot\hat n\,dS=\iiint_V\nabla\cdot\vec F\,dV.

Step 1 — Divergence

∇⋅F⃗=∂x(x)+∂y(−y)+∂z(z2−1)=1+(−1)+2z=2z\nabla\cdot\vec F=\partial_x(x)+\partial_y(-y)+\partial_z(z^2-1)=1+(-1)+2z=2z.

Step 2 — Volume integral

∭V2z dV=2∫01z dz⋅∬x2+y2≤4dA=2⋅12⋅π⋅22=4π.\iiint_V 2z\,dV=2\int_0^1 z\,dz\cdot\iint_{x^2+y^2\le 4}dA=2\cdot\dfrac{1}{2}\cdot\pi\cdot 2^2=4\pi.

(Disk area =πr2=4π=\pi r^2=4\pi; ∫01z dz=1/2\int_0^1 z\,dz=1/2; factor 2 from the divergence.)

Answer

  ∬SF⃗⋅n^ dS=4π.  \boxed{\;\iint_S\vec F\cdot\hat n\,dS=4\pi.\;}
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