Question regarding Stokes' theorem

Question

I have a very simple question. The Calculus book I am using provides this question (along with a solution) as an exercise in Stokes' theorem.

First of all, I have no idea why they have a picture of a triangular domain. I'll let that one slide as a typo. But why is it that the force vector $\vec F$ is simply $\langle 0, 0, 1 \rangle$? Is that another mistake, or is there just something I'm not seeing? If so, would somebody mind explaining?

Here is the question/solution given by the book:

Answer 1

(Edited)

I think something went wrong here in your book; maybe two different exercices were mixed up.

There is no need to introduce the surface $S_1$ at all.

We are given a flow field ${\bf v}$ that possesses a vector potential ${\bf A}$: $${\bf v}={\rm curl}({\bf A})\ .$$

By Stokes' theorem the flow of ${\bf v}$ through $S$ can then be related to ${\bf A}$ as follows:

$$\int_S {\bf v}\cdot{\bf n}\ {\rm d}\omega =\int_S{\rm curl}({\bf A})\cdot{\bf n}\ {\rm d}\omega=\int_{\partial S}{\bf A}\cdot d{\bf x}=25\ .$$

Answer 2

I believe this problem is an application of Stoke's theorem. In vector calculus terms it can be translated to $$\int_{S}\nabla\times X\cdot dS=\oint_{\partial S}X\cdot dr$$where $X$ is a vector field defined on the the surface $S$ with boundary $\partial S$.

If I am not being confused, your book proposed that assuming $F=\nabla\times A$ and the integral of $A$ on the boundary circle is $25$. If this is true, then the left hand side implies the surface integral of $F$ on $S$ should be equal to $25$ as well, and no computation is needed.

The book seem to use $D$ as a dissection of the surface $S$ by cutting along $xz$ plane. But this makes little sense. $F$ and $A$ are only defined on the surface, we do not know their value in the inside at all. The integration of $F$ over the circle is the standard one, but as you pointed out it is suspicious because we have no reason to believe $F$ must have value $(0,0,1)$ as a constant vector field - we do not know $A$. Since the "solution" does not use the number $25$ at all, I suggest the solution should be discarded and you may move on.