← 2014 Paper 1

UPSC 2014 Maths Optional Paper 1 Q6c — Step-by-Step Solution

20 marks · Section B

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

Question

Evaluate by Stokes’ theorem

∫Γ(y dx+z dy+x dz),\int_\Gamma(y\,dx+z\,dy+x\,dz),

where Γ\Gamma is the curve given by x2+y2+z2−2ax−2ay=0,  x+y=2ax^{2}+y^{2}+z^{2}-2ax-2ay=0,\;x+y=2a, starting from (2a,0,0)(2a,0,0) and then going below the zz-plane.

Technique

Stokes’ theorem with a great-circle boundary; careful right-hand-rule orientation matching.

Solution

Strategy. Identify Γ\Gamma as the intersection of a sphere and a plane (a great circle); apply Stokes’ theorem with S=S= the disk in the plane bounded by Γ\Gamma; carefully match orientations.

Step 1 — Identify Γ\Gamma

Sphere: (x−a)2+(y−a)2+z2=2a2(x-a)^{2}+(y-a)^{2}+z^{2}=2a^{2}. Centre (a,a,0)(a,a,0), radius 2 a\sqrt 2\,a.

Plane: x+y=2ax+y=2a passes through the centre (check: a+a=2aa+a=2a ✓), so Γ\Gamma is a great circle of the sphere.

Step 2 — Curl of F⃗=(y,z,x)\vec F=(y,z,x)

∇×F⃗=(∂yx−∂zz, ∂zy−∂xx, ∂xz−∂yy)=(0−1, 0−1, 0−1)=(−1,−1,−1).\nabla\times\vec F=\bigl(\partial_y x-\partial_z z,\,\partial_z y-\partial_x x,\,\partial_x z-\partial_y y\bigr)=(0-1,\,0-1,\,0-1)=(-1,-1,-1).

Step 3 — Choose surface and normal

Let SS = the disk in the plane x+y=2ax+y=2a bounded by Γ\Gamma. The disk has radius 2 a\sqrt 2\,a, so area π(2 a)2=2πa2\pi(\sqrt 2\,a)^{2}=2\pi a^{2}.

Unit normal to plane x+y=2ax+y=2a: n^=(1,1,0)/2\hat n=(1,1,0)/\sqrt 2.

Step 4 — Match orientation

Parametrise Γ\Gamma: with u⃗=(1,−1,0)/2\vec u=(1,-1,0)/\sqrt 2 (from centre (a,a,0)(a,a,0) toward (2a,0,0)(2a,0,0)) and v⃗=(0,0,1)\vec v=(0,0,1) (perpendicular to both n^\hat n and u⃗\vec u),

r⃗(ϕ)=(a,a,0)+2 a[u⃗cos⁡ϕ+v⃗sin⁡ϕ]=(a+acos⁡ϕ, a−acos⁡ϕ, 2 asin⁡ϕ).\vec r(\phi)=(a,a,0)+\sqrt 2\,a\bigl[\vec u\cos\phi+\vec v\sin\phi\bigr]=(a+a\cos\phi,\,a-a\cos\phi,\,\sqrt 2\,a\sin\phi).

At ϕ=0\phi=0: r⃗=(2a,0,0)\vec r=(2a,0,0) ✓. Tangent at ϕ=0\phi=0: r⃗′(0)=(0,0,2 a)\vec r'(0)=(0,0,\sqrt 2\,a), pointing in +z^+\hat z direction.

The question’s orientation goes “below the zz-plane” at the start, i.e., tangent in −z^-\hat z direction. This corresponds to decreasing ϕ\phi (or equivalently, ϕ\phi from 00 to −2π-2\pi).

By the right-hand rule, this orientation of Γ\Gamma corresponds to n^=+(1,1,0)/2\hat n=+(1,1,0)/\sqrt 2 (so that n^\hat n, u⃗\vec u, r⃗′(ϕ=0)∣question’s direction=−z^\vec r'(\phi=0)|_{\text{question's direction}}=-\hat z form a right-handed triad: n^×u⃗=(1,1,0)/2×(1,−1,0)/2=(0,0,−1)\hat n\times\vec u=(1,1,0)/\sqrt 2\times(1,-1,0)/\sqrt 2=(0,0,-1) ✓).

Step 5 — Apply Stokes’ theorem

∫ΓF⃗⋅dr⃗=∬S(∇×F⃗)⋅n^ dS.\int_\Gamma\vec F\cdot d\vec r=\iint_S(\nabla\times\vec F)\cdot\hat n\,dS.

(∇×F⃗)⋅n^=(−1,−1,−1)⋅(1,1,0)/2=−2/2=−2(\nabla\times\vec F)\cdot\hat n=(-1,-1,-1)\cdot(1,1,0)/\sqrt 2=-2/\sqrt 2=-\sqrt 2.

∫ΓF⃗⋅dr⃗=−2⋅2πa2=−22 πa2.\int_\Gamma\vec F\cdot d\vec r=-\sqrt 2\cdot 2\pi a^{2}=-2\sqrt 2\,\pi a^{2}.

Answer

  ∫Γ(y dx+z dy+x dz)=−22 πa2.  \boxed{\;\int_\Gamma(y\,dx+z\,dy+x\,dz)=-2\sqrt 2\,\pi a^{2}.\;}
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