← 2022 Paper 1

UPSC 2022 Maths Optional Paper 1 Q7a — Step-by-Step Solution

20 marks · Section B

Stokes' theorem · Vector Analysis · asked 10× in 14 yrs · Read the full method →

Question

Verify Stokes’ theorem for $\vec F=x\hat\imath+z^2\hat\jmath+y^2\hat k$ over the plane surface $x+y+z=1$ lying in the first octant.

Technique

Direct computation of both sides. The integrand $(2y-2z)$ has nice cancellation when restricted to the plane $z=1-x-y$.

Solution

Setup. $S$: triangle on plane $x+y+z=1$ in first octant, vertices $(1,0,0),(0,1,0),(0,0,1)$.

Stokes’ theorem: $\displaystyle\oint_C\vec F\cdot d\vec r=\iint_S(\nabla\times\vec F)\cdot\hat n\,dS$.

Step 1 — Compute $\nabla\times\vec F$

$\vec F=(x,z^2,y^2)$.

$\nabla\times\vec F=\det\begin{pmatrix}\hat\imath & \hat\jmath & \hat k\\ \partial_x & \partial_y & \partial_z\\ x & z^2 & y^2\end{pmatrix}=(\partial_y(y^2)-\partial_z(z^2),\;\partial_z(x)-\partial_x(y^2),\;\partial_x(z^2)-\partial_y(x))$.

$=(2y-2z,\;0-0,\;0-0)=(2y-2z,0,0)$.

Step 2 — Surface integral

$\hat n$ on plane $x+y+z=1$: $\hat n=(1,1,1)/\sqrt 3$ (outward, pointing away from origin).

$(\nabla\times\vec F)\cdot\hat n=(2y-2z)\cdot 1/\sqrt 3+0+0=(2y-2z)/\sqrt 3$.

Surface area element on the plane: parametrise by $(x,y)$ with $z=1-x-y$. The Jacobian of the map from $(x,y)$ to surface is $\sqrt{1+(\partial z/\partial x)^2+(\partial z/\partial y)^2}=\sqrt{1+1+1}=\sqrt 3$.

So $dS=\sqrt 3\,dx\,dy$.

Surface integral:

where $T_{xy}$ is the projection of $S$ onto $xy$-plane: triangle with vertices $(0,0),(1,0),(0,1)$.

Compute (using vertical strips):

Inner: $\int_0^{1-x}(2x+4y-2)\,dy=[(2x-2)y+2y^2]_0^{1-x}=(2x-2)(1-x)+2(1-x)^2$.

$=(2x-2)(1-x)+2(1-x)^2=(1-x)[(2x-2)+2(1-x)]=(1-x)[2x-2+2-2x]=(1-x)\cdot 0=0$.

So inner integral $=0$ for all $x$!

Outer: $\int_0^1 0\,dx=0$.

Surface integral $=0$.

Step 3 — Line integral $\oint_C\vec F\cdot d\vec r$

$C$ is the boundary of $S$: triangle $(1,0,0)\to(0,1,0)\to(0,0,1)\to(1,0,0)$ (orientation chosen consistent with $\hat n$ pointing outward from origin — by right-hand rule, this is CCW when viewed from the $\hat n$ direction).

Edge 1: $(1,0,0)\to(0,1,0)$. Parametrise: $\vec r(t)=(1-t,t,0)$, $t:0\to 1$. $d\vec r=(-1,1,0)\,dt$.

$\vec F=(x,z^2,y^2)=(1-t,0,t^2)$.

$\vec F\cdot d\vec r=(1-t)(-1)+0+0=-(1-t)=t-1$.

$\int_0^1(t-1)\,dt=1/2-1=-1/2$.

Edge 2: $(0,1,0)\to(0,0,1)$. $\vec r(t)=(0,1-t,t)$. $d\vec r=(0,-1,1)\,dt$.

$\vec F=(0,t^2,(1-t)^2)$.

$\vec F\cdot d\vec r=0+t^2(-1)+(1-t)^2(1)=-t^2+(1-t)^2$.

$(1-t)^2-t^2=1-2t+t^2-t^2=1-2t$.

$\int_0^1(1-2t)\,dt=1-1=0$.

Edge 3: $(0,0,1)\to(1,0,0)$. $\vec r(t)=(t,0,1-t)$. $d\vec r=(1,0,-1)\,dt$.

$\vec F=(t,(1-t)^2,0)$.

$\vec F\cdot d\vec r=t(1)+(1-t)^2(0)+0(-1)=t$.

$\int_0^1 t\,dt=1/2$.

Total: $-1/2+0+1/2=0$.

Step 4 — Compare

Both line integral and surface integral equal $0$. Stokes’ theorem verified.

Answer