Calculate $\int\int_S\nabla\times\textbf F \cdot d\textbf S$ where $S$ is defined by $z=e^{1-x^2-y^2}, z\ge1$ and $\textbf F = (-y,x,xyz)$. Orient $S$ so the normal is pointing downwards.
My attempt: $\nabla\times\textbf F = (xz,-yz,2)$
I parameterize my surface: $\textbf R(x,y)=(x,y,e^{1-x^2-y^2}))$.
I know $d\textbf S=\textbf R_x\times \textbf R_y dxdy$. $\textbf R_x\times \textbf R_y = (2xe^{1-x^2-y^2}, 2ye^{1-x^2-y^2},1)$. I just take the negative of this vector so the normal faces downwards.
I then switched to cylindrical coordinates where $x=1cos\theta, y=1sin\theta, z=z$. Evaluating the integrand results in: $$\int_0^{2\pi}\int_1^e(2cos^2\theta - 2sin^2\theta+2)dzd\theta$$ I got e as an upper bound by looking at the surface. I'm not sure if this is right however. Setting $r=1$ does not seem right to me. I will provide a graphic of the surface. I think I could have just used the divergence theorem since it is a closed surface, but I want to be able to do the surface integral.
$z=e^{1-x^2-y^2}$">