So I'm very rusty on Stokes' theorem, and I have a test about line integrals, surface integrals and such. While practicing, I came across the question, Evaluate the following Line integral:
$$\int_{C} \vec{F} \cdot d\vec{r}$$ $$ \vec{F}(x,y,z) = (3z - 2y)\hat{i} + (4x - 3y)\hat{j} + (x+2y)\hat{k}, \hspace{.5cm} C: x^2 + y^2 = 1,\hspace{0.5cm} z = 3$$ now I want to use Stokes' theorem to evaluate this line integral.
$$\int_{C} \vec{F} \cdot d\vec{r} = \iint_\limits{S}( \nabla \times \vec{F}) \cdot\hat{n}dS$$
So I calculate $\nabla \times \vec{F} = 2\hat{i} - 2\hat{j} + 6\hat{k}$. When I looked up how to find a unit normal vector to a surface $S$, I found that $\hat{n} = \frac{\nabla G}{||\nabla G||}$, for some $G$ whose level surface is $S$. However, I can't figure out what the equation to the surface is. Can the surface be any smooth surface whose boundary is $C$? In that case can I use $S: z = 3$ as my surface? In which case $G(x,y,z) = 3 - z$. However, doing so results in $\hat{n} = -\hat{k}$, and $dS = dA$. This results in the integral $$\iint\limits_{R} -6dA$$. This integral is equal to the area of the Region $R$ times $-6$, which is equal to $-6\pi$. However, the answer key claims that the answer is $6\pi$. Am I missing something?