I would like to solve the following problem:
Let $B_1$ be the unit ball in $R^3$ and
$A := \delta B_1 \cap(\{x>0, z=0\}\cup\{x=0, z>0\})$.
Let $F(x,y,z) := (-y+e^{x+z}, 0, e^{x+z})$. Calculate
$\oint_A F\cdot v dS(x)$ for a choice of $v$ as a tangent on $A$.
I am not sure if the statement "...for a choice of v as a tangent on $A$." is meant is it seems to be: The creator of the task wants me to choose a tangent and then calculate the surface integral.
I would rather expect the creator to mean $v$ being the unit normal field of $A$ so that this task would be solvable by using the gaussian theorem.
So I have two questions:
What interpretation do you think is the right one?
If it is the first one: How to calculate the integral?
If it is the second (the gaussian) one: I am struggling to find a proper set $B$ in $R^3$ so that $A$ would be the boundary of B and we would be able to calculate the integral of $div F$ over $B$.
Well, as far as I understand, this is the section of the sphere referred to (in red):
From the description that you gave, it appears that the question explicitly states a choice of a tangent (unit tangent, I imagine). This depends on the parametrization of the surface. In this case, the surface was parametrized (for a general radius, $R$) as: $$\left\{\begin{array}\\ x=R\sin\theta\cos\phi \\ y=R\sin\theta\sin\phi \\ z=R\cos\theta\end{array}\right.,\text{ For }\theta\in\left[0,\frac{\pi}{2}\right]\text{ and }\phi\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$ According to this parametrization, we allow a notation of $Z^i=(Z^1,Z^2,Z^3)=(x,y,z)$ to represent the embedding (or, ambient) coordinates and $S^\alpha=(S^1,S^2)=(\theta,\phi)$ to be the surface coordinates. In this, an upper index and lower index represent summation (Einstein Summation Convention) ie. $A^iB_i=A^1B_1+A^2B_2+A^3B_3+...+A^iB_i$. According to this, the tangent vectors (one for each coordinate direction, $\theta$ & $\phi$) on the surface are given by: $$\textbf{S}_\alpha=\frac{\partial Z^i}{\partial S^\alpha}\vec{\textbf{Z}}_i=\frac{\partial Z^1}{\partial S^\alpha}\vec{\textbf{Z}}_1+\frac{\partial Z^2}{\partial S^\alpha}\vec{\textbf{Z}}_2+\frac{\partial Z^3}{\partial S^\alpha}\vec{\textbf{Z}}_3$$ Where $\vec{\textbf{Z}}_i=(\vec{\textbf{Z}}_1,\vec{\textbf{Z}}_2,\vec{\textbf{Z}}_3)=(\hat{x},\hat{y},\hat{z})$. Therefore, the tangent vectors are: $$\textbf{S}_\theta=\left[\begin{array}\\R\cos\theta\cos\phi \\R\cos\theta\sin\phi\\-R\cos\theta\end{array}\right]\text{ and }\textbf{S}_\phi=\left[\begin{array}\\-R\sin\theta\sin\phi \\ R\sin\theta\cos\phi \\ 0\end{array}\right]$$ Using these, you can find out any tangent vector you wish, normalize it by dividing it by its length, and use it in the integral above to execute the surface integral.