I need to calculate $$ \iint_S \textbf{J}\cdot\hat{\textbf{n}}\; dS $$ Where S is the circular surface centered at the origin, area $A$, in the $yz$-plane, with unit normal having a negative $x$-component.
$$\textbf{J} = -\sigma\nabla\phi(\textbf{r}), \;\; \phi(\textbf{r}) = V(1-x/L)$$
What I'm struggling with here is getting a parametric representation of this surface for this integral in terms of u and v. I can very easily parametrise the curve of the boundary of this surface but obviously for this integral I need a representation of the surface. I tried:
$$ S:\; (u,v) \rightarrow \; \left(v, \sqrt\frac{A}{\pi}\cos(u), \sqrt\frac{A}{\pi}\sin(u)\right) \;\; 0\leq u\leq 2\pi, \; v = 0 $$
But then calculating the normal, $\textbf{v}_{u} \times \textbf{v}_{u}$, I get a normal to the cylinder with this circle as it's end. I know I need to get $(-1, 0, 0)$ as my unit normal vector.
Any help would be much appreciated!
A parametrization of the disk S is $$S:\; (u,v) \rightarrow \; \left(0, v\cos(u), v\sin(u)\right) \;\;\text{with}\;\; 0\leq u\leq 2\pi, \; 0\leq v \leq \sqrt\frac{A}{\pi}.$$ Then $$\textbf{S}_{u} \times \textbf{S}_{v}=\left(0, -v\sin(u), v\cos(u)\right) \times \left(0, \cos(u), \sin(u)\right)=(-v,0,0)\implies \textbf{n}=(-1,0,0).$$
Actually, you don't need it, because $$\textbf{J} = -\sigma\nabla(V(1-x/L))=-\sigma(-V/L,0,0)$$ is constant and therefore $$\iint_S \textbf{J}\cdot\hat{\textbf{n}}\; dS=-\sigma(-V/L,0,0)\cdot(-1,0,0)A=\frac{\sigma VA}{L}.$$