1. ## Applying Stokes theorem

Let $\displaystyle S_G$ be the surface of the cap (see figure). Let $\displaystyle C$ be the differentiable curve of the cap in the $\displaystyle xy$ plane. Let $\displaystyle S_A$ be the portion of the plane xy that C contains. And let $\displaystyle A$ be the value of the area of $\displaystyle S_A$. Consider $\displaystyle S_G$ oriented outward. Determine with arrows the sense of the curve $\displaystyle C$ and the value of $\displaystyle \iint _{S_G} curl F dS$, where $\displaystyle F=((\ln (z+1))^2xy, (\ln (z+1))^2 \frac{x^2}{2}+x(z+1), \cos z )$.
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My attempt :
The arrow is easy to draw : counter clockwise.
I've calculated the curl of $\displaystyle F$, but now I realize it is likely useless. Anyway here it comes : $\displaystyle curl F =\left ( \frac{x^2 \ln (z+1)}{2z+2} \right ) \vec i + \left( \frac{xy \ln (z+1)}{z+1} \right) \vec j +(z+1) \vec k$.
I'm sure I have to calculate $\displaystyle \int _C FdS$ (which is worth $\displaystyle \iint _{S_G} curl F dS$ by Stokes' theorem) but how to proceed? I have almost no information about $\displaystyle C$, I mean I don't know its parametrization. The only way I think I might solve the problem is to realize something obvious that imply that $\displaystyle \int _C FdS=0$, but I don't see anything.

2. Originally Posted by arbolis
Let $\displaystyle S_G$ be the surface of the cap (see figure). Let $\displaystyle C$ be the differentiable curve of the cap in the $\displaystyle xy$ plane. Let $\displaystyle S_A$ be the portion of the plane xy that C contains. And let $\displaystyle A$ be the value of the area of $\displaystyle S_A$. Consider $\displaystyle S_G$ oriented outward. Determine with arrows the sense of the curve $\displaystyle C$ and the value of $\displaystyle \iint _{S_G} curl F dS$, where $\displaystyle F=((\ln (z+1))^2xy, (\ln (z+1))^2 \frac{x^2}{2}+x(z+1), \cos z )$.
this is a nice problem! in order to find the integral, you'll need to use Stoke's theorem first and then Green's theorem. note that in $\displaystyle xy$ plane we have $\displaystyle z = 0$ and thus $\displaystyle \ln(z + 1) = dz = 0.$

so by Stoke's theorem: $\displaystyle \iint_{S_G} curl (F) \cdot \bold{n} \ dS = \int_C F \cdot d \bold{r}=\int_C xdy.$ on the other hand, by Green's theorem: $\displaystyle \int_C xdy = \int_C 0 \cdot dx + xdy = \iint_{S_A} dxdy =A.$ thus $\displaystyle \iint_{S_G} curl (F) \cdot \bold{n} \ dS =A.$

3. Thanks for the answer NCA, I wouldn't have find it by myself within the next 6 remaining days my exam. The problem is indeed nice because it is different from most.
Just a little clarification of my first post : I mean C has the clockwise sense and not anticlockwise as I said. I sketched it clockwise but I don't know why I said anticlockwise.