The math optional, made finite. Daily Practice

← 2013 Paper 1

UPSC Maths 2013 Paper 1 Q8d — Solution

15 marks · Section B

Question

Use Stokes’ theorem to evaluate the line integral ∫C(−y3 dx+x3 dy−z3 dz)\displaystyle\int_C(-y^{3}\,dx+x^{3}\,dy-z^{3}\,dz), where CC is the intersection of the cylinder x2+y2=1x^{2}+y^{2}=1 and the plane x+y+z=1x+y+z=1.

Technique

Stokes’ theorem with the plane as the surface; project to the xyxy-disk for the area integral; polar coordinates.

Solution

Strategy. Stokes: ∫CF⃗⋅dr⃗=∬S(∇×F⃗)⋅n^ dS\displaystyle\int_{C}\vec F\cdot d\vec r=\iint_{S}(\nabla\times\vec F)\cdot\hat n\,dS for any oriented surface SS with boundary CC. Pick SS as the portion of the plane x+y+z=1x+y+z=1 inside the cylinder.

Step 1 — Vector field and curl

F⃗=(−y3, x3, −z3).\vec F=(-y^{3},\,x^{3},\,-z^{3}). ∇×F⃗=(∂y(−z3)−∂z(x3), ∂z(−y3)−∂x(−z3), ∂x(x3)−∂y(−y3))=(0, 0, 3x2+3y2).\nabla\times\vec F=\Bigl(\partial_y(-z^{3})-\partial_z(x^{3}),\,\partial_z(-y^{3})-\partial_x(-z^{3}),\,\partial_x(x^{3})-\partial_y(-y^{3})\Bigr)=\bigl(0,\,0,\,3x^{2}+3y^{2}\bigr).

So ∇×F⃗=(0,0,3(x2+y2))\nabla\times\vec F=(0,0,3(x^{2}+y^{2})).

Step 2 — Choose the surface SS

SS = disk-like region in the plane x+y+z=1x+y+z=1 bounded by CC. Project onto the xyxy-plane: the projection is the unit disk D={x2+y2≤1}D=\{x^{2}+y^{2}\le 1\}.

Outward unit normal to the plane: n^=(1,1,1)3\hat n=\dfrac{(1,1,1)}{\sqrt 3}.

Step 3 — Surface element

The plane is z=1−x−yz=1-x-y, so zx=zy=−1z_x=z_y=-1. Then

dS=1+zx2+zy2 dx dy=1+1+1 dx dy=3 dx dy.dS=\sqrt{1+z_x^{2}+z_y^{2}}\,dx\,dy=\sqrt{1+1+1}\,dx\,dy=\sqrt 3\,dx\,dy.

Step 4 — Compute the flux

(∇×F⃗)⋅n^=(0,0,3(x2+y2))⋅(1,1,1)3=3(x2+y2)3=3 (x2+y2).(\nabla\times\vec F)\cdot\hat n=\bigl(0,0,3(x^{2}+y^{2})\bigr)\cdot\frac{(1,1,1)}{\sqrt 3}=\frac{3(x^{2}+y^{2})}{\sqrt 3}=\sqrt 3\,(x^{2}+y^{2}). ∬S(∇×F⃗)⋅n^ dS=∬D3(x2+y2)⋅3 dx dy=3∬D(x2+y2) dx dy.\iint_{S}(\nabla\times\vec F)\cdot\hat n\,dS=\iint_{D}\sqrt 3(x^{2}+y^{2})\cdot\sqrt 3\,dx\,dy=3\iint_{D}(x^{2}+y^{2})\,dx\,dy.

Step 5 — Polar integral

∬D(x2+y2) dx dy=∫02π∫01r2⋅r dr dθ=2π⋅r44∣01=π2.\iint_{D}(x^{2}+y^{2})\,dx\,dy=\int_{0}^{2\pi}\int_{0}^{1}r^{2}\cdot r\,dr\,d\theta=2\pi\cdot\frac{r^{4}}{4}\Big|_{0}^{1}=\frac{\pi}{2}.

Step 6 — Combine

∫CF⃗⋅dr⃗=3⋅π2=3π2.\int_{C}\vec F\cdot d\vec r=3\cdot\frac{\pi}{2}=\frac{3\pi}{2}.

Answer

  ∫C(−y3 dx+x3 dy−z3 dz)=3π2.  \boxed{\;\int_{C}(-y^{3}\,dx+x^{3}\,dy-z^{3}\,dz)=\frac{3\pi}{2}.\;}

We've mapped all 13 years of this exam. Get new solutions, tools, and guides as we release them — free.