← 2018 Paper 1

UPSC 2018 Maths Optional Paper 1 Q6d — Step-by-Step Solution

12 marks · Section B

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

Question

If SS is the surface of the sphere x2+y2+z2=a2x^2+y^2+z^2=a^2, then evaluate

∬S[(x+z) dy dz+(y+z) dz dx+(x+y) dx dy]\iint_S\big[(x+z)\,dy\,dz+(y+z)\,dz\,dx+(x+y)\,dx\,dy\big]

using Gauss’ divergence theorem.

Technique

Recognise the flux form; ∇⋅F⃗=2\nabla\cdot\vec F=2 (constant); flux =2×=2\times volume of ball =2⋅43πa3=2\cdot\frac43\pi a^3.

Solution

Step 1 — Identify the vector field

The integral is the flux ∬SF⃗⋅n^ dS\displaystyle\iint_S\vec F\cdot\hat n\,dS of

F⃗=(x+z) i^+(y+z) j^+(x+y) k^,\vec F=(x+z)\,\hat i+(y+z)\,\hat j+(x+y)\,\hat k,

since dy dz, dz dx, dx dydy\,dz,\ dz\,dx,\ dx\,dy are the components of n^ dS\hat n\,dS.

Step 2 — Compute the divergence

∇⋅F⃗=∂∂x(x+z)+∂∂y(y+z)+∂∂z(x+y)=1+1+0=2.\nabla\cdot\vec F=\frac{\partial}{\partial x}(x+z)+\frac{\partial}{\partial y}(y+z)+\frac{\partial}{\partial z}(x+y)=1+1+0=2.

Note the third component x+yx+y has no zz-dependence, so its zz-derivative is 00 — the divergence is 22, not 33.

Step 3 — Apply Gauss’ theorem

By the divergence theorem, with VV the solid ball of radius aa (volume 43πa3\tfrac43\pi a^3):

∬SF⃗⋅n^ dS=∭V(∇⋅F⃗) dV=∭V2 dV=2⋅43πa3.\iint_S\vec F\cdot\hat n\,dS=\iiint_V(\nabla\cdot\vec F)\,dV=\iiint_V 2\,dV=2\cdot\frac43\pi a^3.

Answer

  ∬S[(x+z) dy dz+(y+z) dz dx+(x+y) dx dy]=83πa3.  \boxed{\;\iint_S\big[(x+z)\,dy\,dz+(y+z)\,dz\,dx+(x+y)\,dx\,dy\big]=\frac{8}{3}\pi a^3.\;}
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