Vector Calculus; Divergence and Stokes Question I've been working on these two questions for sometime and I'm a little stumped/unsure if my answer is correct.
The questions are a little long and I'm terrible with formatting anyways so I uploaded a screenshot: http://i.imgur.com/FeSodSt.png


*So I'm pretty sure the two vectors are supposed to be dr. Integrating it, I got the original r. Rewriting as dS, I got $r_1 = <rsint,-rcost,0>$ and  $r_2 = <rsint,rcost,0>$ and the respective normal vectors as <0,0,r> and <0,0,-r>


This would suggest curl F dot n would be some positive number for equation one and negative for equation two, suggesting that the statement is true.


*I think you can Divergence Theorem for this, however you'd have to take some circle around the original on z=0 with some radius that infinitely small. Because Div F is zero, the flux is just the surface integral of that infinitely small circle. However, because of the orientation of the two pieces, the integral would be the flipped signs of each other, making the statement false.


I've been sitting on these for a while and I'm not quite sure if what I've thought is right or not. I'd appreciate any feedback and if I explained something unclearly, I'd be happy to clarify. Thanks.
 A: The given equations in the problem you link represent line-integrals. The vectors next to $\vec{F}$ are the velocity vectors of a curve which parametrizes the unit-circle. The formula $\langle \cos \theta, \sin \theta, 0 \rangle$ gives us the circle $x^2+y^2=1$ at $z=0$ oriented counter-clockwise (CCW) as viewed from positive $z$-axis. On the other hand,replacing $\theta$ with $-\theta$ makes the curve go clockwise(CW) and gives us the formula $\langle \cos \theta, -\sin \theta, 0 \rangle$. In either case, you can apply Stokes Theorem:
$$ \iint_S \left(\nabla \times \vec{F} \right) \cdot d\vec{S} = \int_{\partial S} \vec{F} \cdot d\vec{r}$$
to the unit disk $S = \{ (x,y,0) \ | \ x^2+y^2 \leq 1 \}$. To use Stokes Theorem you must consistently orient $S$ and its boundary $\partial S$. For the CW curve the normal to the disk points down. For the CCW oriented circle, the disk must be oriented upwards. Then, the reason for the inequality should be easy enough to see.
Added: for question 6, I think you can just use an arbitrary sphere. The surface can be split into a N and S hemi-spherical shells. Of course the union of these shells gives the surface of the sphere. But, as you anticipate, use divergence theorem to see the net-flux through both hemispheres is the integral of the divergence through the bulk-volume of the sphere (hence zero). That is: $0=\iiint_B \nabla \cdot \vec{F} dV = \iint_{\partial B} \vec{F} \cdot \vec{dS}$ However, using $S_S \cup S_N = \partial B$, 
$$ \iint_{\partial B} \vec{F} \cdot \vec{dS} =  \iint_{S_S} \vec{F} \cdot \vec{dS}+\iint_{S_N} \vec{F} \cdot \vec{dS} = 0.$$
Now, the $S_S$ and $S_N$ are given out-ward orientations as to be consistent with the divergence theorem's use here. But, you can flip the normal on one which gives a minus and hence the claim. I think you were going in this direction, I don't see the need for an infinitesimal argument since you're given the divergence is zero everywhere (I think I read that). Hope this helps.
