Divergence and Stokes Could anyone help with this problem?
Evaluate $\iint_S \textbf{F}$ $\cdot$ $\textbf{n}$ $d \alpha$ using Stoke's Theorem or the 
Divergence Theorem.
c) S is the truncated cone $y=2\sqrt{x^2+z^2}$, $2 \le y \le 4$, $ \textbf{n}$ is the outward-pointing normal, and $\textbf{F}(x,y,z)=(x,-2y,z)$
From the comments I am guessing the answer depends on whether we include the end caps or not. I am not sure if we have to include them. In class we used the divergence theorem to solve this problem and we got 0, but the back of the book has $28\pi$. 
*It's not homework. 
 A: Only one of the theorems (Stokes or Divergence) can apply.  Which one applies depends on your situation.


*

*Stokes' Theorem applies if your surface has a boundary curve.  This would mean that your truncated cone (frustum) does not include the end caps.

*The Divergence Theorem applies if your surface is closed. This would mean that your frustum includes the end caps.
The way your problem is stated, it would seem as if the end caps are not included, meaning that the Divergence Theorem does not apply.  However, while Stokes' Theorem does apply, it's difficult to see how it could be used practically here.  My advice would then be to compute the integral directly, probably via a sort of rotated cylindrical coordinate system.
If the end caps are included, then you can use the Divergence Theorem and get an answer of $0$.  This is because if $\mathbf{F}(x,y,z) = (x, -2y, z)$, then $\text{div }\mathbf{F} = 1 - 2 + 1 = 0$, so $\iint \mathbf{F}\cdot \mathbf{n}\, dS = \iiint \text{div }\mathbf{F}\,dV = 0$.

As an interesting aside, note that if both of the theorems applied to a surface $S$, then we would have
$$\int_C \mathbf{F}\cdot d\mathbf{r} = \iint_S \text{curl }\mathbf{F}\cdot \mathbf{n}\,dS = \iiint_E \text{div}(\text{curl }\mathbf{F})\,dV = 0,$$
and the integral of any vector field over the boundary curve $C$ would be zero.  This is a fancy way of saying that "the boundary of a boundary is zero."  In other words, if a volume has a boundary surface, then that surface cannot have a boundary curve.
