Let N be the upwards unit normal on the surface $M:z=e^{-x^{2}-y^{2}},x^{2}+y^{2}\leq1$. Consider the vector field u=$(x^{4}-y^{3},cos(x),sin(z))$
Im asked to compute the flux integral $\int \int_{M}curl(u)\cdot NdS$.
This problem have made me lose allmost all my confidence about the subject. By direct calculation I don't get anywhere. Then I use Stokes theorem and calculate the line integral around the circle $(cos(t),sin(t), e^{-1})$. I get as well complicated terms to integrate like $cos(cos(t))$.
Then i tried with (because the surface of integration doesn't matter as long the boundary is the same.., RIGHT?) calculating the integral over the disk with radius 1 on $z=e^{-1}$, because they share the boundary. But again the intregals get overcomplicated. I would be happy seeing how you would solve this problem. The answer is $\frac{3\pi}{4}$. Thanks
For your field $\vec{F}=\left(x^{4}-y^{3},\cos x,\sin z\right)$
$$\vec{\nabla}\times\vec{F}=\left(0,0,-\sin x+3y^{2}\right)$$
As you said, you can calculate the integral over the disk $D=\big\{\left(x,y,e^{-1}\right):x^{2}+y^{2}\leq 1\big\}$ so
$$\iint_{M}\left(\vec{\nabla}\times\vec{F}\right)\cdot\vec{n}{\rm d}S=\iint_{D}\left(\vec{\nabla}\times\vec{F}\right)\cdot\vec{n}{\rm d}S=$$
$$=\iint_{D}\left(-\sin x+3y^{2}\right){\rm d}S=3\iint_{D}y^{2}{\rm d}S=$$
since $\sin x$ is odd and the region is symmetric. Now you can covert to polar coordinates $x=r\cos\theta$ and $y=r\sin\theta$ to get
$$=3\int_{0}^{2\pi}\int_{0}^{1}r^{2}\sin^{2}\theta r{\rm d}r{\rm d}\theta=\frac{3\pi}{4}$$
as wanted.