Suppose, $F$ be a smooth vector field. Now, we want to evaluate $\iint(\nabla \times F)\cdot dS$ where $S=\{(x,y,z)|x^2+y^2+z^2=1,z\le0\}$ i.e. $S$ is lower half part of unit sphere.
Now suppose we add the lower part on the $xy$-plane (which is $\{(x,y,z)|x^2+y^2\le1,z=0\}$$=S''$(say)) to S and call it $S'$. Now, $S'$ forms a closed and bounded region in $\Bbb{R}^3$.
So, $\iint(\nabla \times F)\cdot dS'=\iiint\nabla\cdot (\nabla \times F)dV=0$ (by Gauss Divergence Theorem and the fact $\nabla\cdot (\nabla \times F)=0$)
Now, since $S\cup S''=S'$, we can write $\iint(\nabla \times F)\cdot dS=\iint(\nabla \times F)\cdot dS'-\iint(\nabla \times F)\cdot dS''=-\iint(\nabla \times F)\cdot dS''$ $\cdots$ (1)
Now recall $S''=\{(x,y,z)|x^2+y^2\le1,z=0\}$, from (1) and by Stokes' Theorem, we have $\iint(\nabla \times F)\cdot dS=-\oint_{c} F.dr$ where $c(t)=(\cos t,\sin t), t\in[0,2\pi]$ (since outward normal for $S''$ is $\hat k$ and by right hand rule orientation of the boundary of $''$ will be counter clockwise).
Thus, we finally get $\iint(\nabla \times F)\cdot dS=-\oint_{c} F.dr$ $\cdots$ (2)
Again, if we apply Stokes' Theorem directly in the problem we will get
$\iint(\nabla \times F)\cdot dS=\oint_{c} F.dr$ where $c$ is the boundary parametrized by $c(t)=(\cos t,\sin t), t\in[0,2\pi]$.
Comparing this and (2), we have $\oint_{c} F.dr=-\oint_{c} F.dr$!!!
Can anybody clear this confussion? Where is the falacy?
Best Answer
Your closed surface $S'$, as you’ve correctly noticed, is made up of two separate oriented components $S$ and $S''$ which share the same border, but induce on it two opposite orientations: the border of $S''$ is $\partial S''$, the unit circle traversed in the counterclockwise sense, the border of $S$ is $\partial S$, the same as $\partial S''$ but traversed in the clockwise sense. As the integrand $\nabla \times F$ is the same on both surface components, Stokes’ theorem ensures that the two contributions cancel out: $$\iint_{S'} (\nabla \times F)\cdot da = \oint_{\partial S} F \cdot d\ell + \oint_{\partial S''} F \cdot d\ell = \oint_{\partial S} F \cdot d\ell - \oint_{\partial S} F \cdot d\ell = 0. $$ Just like computing the volume integral of the divergence had indicated.