I have the following vector field:
$\bar v= \langle y^2+z^2, x^2+z^2, x^2+y^2\rangle $ and $\bar x=x \hat i+ y\hat j+z\hat k $
Part $a$ asks to prove that $(\nabla\times \bar v)\cdot \bar x=0$.
Part $b$ states, let $\Gamma$ be any curve from $(2,0,0)$ to ($-2,0,0)$ on the sphere, $x^2+y^2+z^2=4$, use results from $a$ to calculate, $$\int_{\Gamma}^{} \bar v \cdot d\bar r$$
Since $(\nabla\times \bar v)\cdot \bar x=0$, this results in the Stokes' Theorem giving a value of $0$, since, according to the theorem, $$\int_{\Gamma}^{} \bar v \cdot d\bar r = \int{\int_{S}^{} (\nabla\times \bar v)\bar x\cdot dS}$$ Can someone elaborate on what this means, mathematically or physically? Or have I have made a conceptual mistake?
You can only apply Stokes's Theorem if you have a closed curve $\Gamma$ bounding a surface $S$. Now it is true that at the point $\vec x$ on the given sphere, the unit normal is a scalar multiple of $\vec x$, so if we take the flux of $\nabla\times\vec v$ across any portion of the sphere, we will get $0$.
Here'a suggestion: Make a closed curve by taking your curve $\Gamma$ from $(2,0,0)$ to $(-2,0,0)$ and following it with one of the semicircles $x^2+y^2=4$, $z=0$ going from $(-2,0,0)$ to $(2,0,0)$ — say $\Gamma'$. Now you have a closed curve $\bar\Gamma$ and what I said above applies. So all you have to do is do the line integral $$\int_{\Gamma'}\vec v\cdot d\vec r = \int_{\Gamma'} y^2\,dx + x^2\,dy$$ explicitly and see that its value is $0$. Oh dear, but it isn't. So the homework problem is flawed.