UPSC 2020 Maths Optional Paper 1 Q8b — Step-by-Step Solution

15 marks · Section B

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

Question

Evaluate the surface integral $\iint_S \nabla\times\vec F\cdot\hat n\,dS$ for $\vec F=y\hat i+(x-2xz)\hat j-xy\hat k$ and $S$ is the surface of the sphere $x^2+y^2+z^2=a^2$ above the $xy$ -plane.

Technique

Stokes’ theorem replaces the hemisphere flux by the circulation around the bounding circle $z=0$ ; the integrand collapses to the exact differential $d(xy)$ , hence $0$ . Cross-checked by direct spherical integration and the cap-vs-disk (solenoidal-curl) argument.

Solution

$S$ is the upper hemisphere; its boundary $\partial S$ is the circle $x^2+y^2=a^2$ , $z=0$ . By Stokes’ theorem the flux of the curl through $S$ equals the circulation of $\vec F$ around $\partial S$ :

\iint_S(\nabla\times\vec F)\cdot\hat n\,dS=\oint_{\partial S}\vec F\cdot d\vec r .

Step 1 — (For reference) the curl

\nabla\times\vec F=\Big(\partial_y(-xy)-\partial_z(x-2xz)\Big)\hat i+\Big(\partial_z(y)-\partial_x(-xy)\Big)\hat j+\Big(\partial_x(x-2xz)-\partial_y(y)\Big)\hat k

=(-x-(-2x))\hat i+(0-(-y))\hat j+((1-2z)-1)\hat k=x\,\hat i+y\,\hat j-2z\,\hat k .

Step 2 — Reduce to the boundary circle

On $\partial S$ we have $z=0$ , so there $\vec F=y\,\hat i+x\,\hat j-xy\,\hat k$ , and along the curve $dz=0$ . Hence

\vec F\cdot d\vec r=y\,dx+x\,dy=d(xy).

Since $xy$ is single-valued and the boundary is a closed curve,

\oint_{\partial S}\vec F\cdot d\vec r=\oint_{\partial S}d(xy)=0 .

(Explicitly, with $x=a\cos t,\ y=a\sin t,\ t:0\to2\pi$ for the upward-normal orientation: $\int_0^{2\pi}\big[a\sin t(-a\sin t)+a\cos t(a\cos t)\big]dt=\int_0^{2\pi}a^2\cos 2t\,dt=0$ .)

Answer

\boxed{\,\iint_S(\nabla\times\vec F)\cdot\hat n\,dS=0\,}