Stokes' Theorem and the surfaces -- don't I have to evaluate more than 1 surface integral? If I have $F = (2y,3x,-z^2)$ and
If I have the hemisphere defined by $x^2 + y^2 + z^2 = 9$, for $z \geq 0$, and $C$ is the boundary that is the circle $x^2 + y^2 = 9$ at $z=0$, then by Stokes' Theorem,
$$\int_{C} F \cdot \mathrm{d}{\textbf{r}} = \iint_{S} \mathrm{curl}F \cdot \mathrm{d}{\textbf{S}}.$$
Now my question is, will we have to evaluate two surface integrals for the RHS? Namely, the "curved" surface of the hemisphere for $z\geq 0$, and the second surface being the "flat" circle at the bottom of the hemisphere. My source only evaluated the first one, without any mention of the second one, but it really feels like I'd have to.
Also, a lot of these questions really seem like it doesn't matter if I evaluate the second surface or not.
 A: The surface is not closed. It could be but it is not. For example the point right in the middle of the bottom would be $(0,0,0)$ but this point doesn't satisfy the equation. Hence the surface is only the upper hemisphere without bottom.
Also note that if it was closed and had a bottom part, then the boundary would not exist and you couldn't use Stokes. (You could try to apply Divergence Theorem in this situation since we would deal with a closed surface.)

Regarding the fact the "it doesn't matter to evaluate the integral on the bottom", it actually does. In your particular case with $F$, you indeed get $0$, but let's simply change your $F$ into $G = (2y,3x,-z^2 + \color{red}{1})$
$$
s(r,\theta) = (r\cos \theta, r\sin\theta, 0), \ \ 0 \leq r \leq 3,\ 0 \leq \theta < 2\pi \\
s_r = (\cos \theta, \sin\theta,0) \\
s_\theta (-r\sin\theta, r\cos\theta,0) \\
s_\theta \times s_r  = (0,0,-r)
$$
Then 
$$
\iint_{bottom} \vec{G} \cdot d\vec{n} = \int_0^3 dr \int_0^{2\pi} d\theta (2r\sin\theta, 3r\cos\theta,1) \cdot (0,0,-r) = -2\pi \int_0^3 r dr = -9\pi
$$
Where $\vec{n}$ is the outgoing normal (pointing downward). So we see that the integral is not necessarily $0$.
A: The idea of Stokes' Theorem is that an integral (of $\vec{F}$) around a closed path is related to the integral (of $\nabla \times \vec{F}$) over any capping surface.  We imagine the closed path to be the mouth of a bottle and the capping surface is any surface that covers the opening.  The surface can be flat, or it can bulge out or it can bulge inward.  We are free to pick any one capping surface.  Usually we pick a surface on which the $(\nabla \times \vec{F}) \cdot d\hat{S}$ is easy to evaluate.  If we want the value of the integral around a closed path we do not use two capping surfaces.
