$\iint (\bigtriangledown \times V) \cdot n d\sigma $ over the part of the surface $z = 9 - x^2 - 9y^2 $ above the xy-plane, if $V = 2xy\ i + (x^2-2x)\ j -x^2z^2\ k$
My attempt:
The surface is an ellipse in the xy plane, $x^2 + 9y^2 = 9$
Using Stokes's Theorem : $\iint (\bigtriangledown \times V) \cdot n d\sigma = \oint V \cdot dr$
I tried to take the line integral of $V\cdot dr$ and get my solution but I end up with a ridiculous integrand down the line.
So doing the other integral (surface integral) would be the way to go for this problem.
I calculated the curl and did its dot product with the gradient of the given surface which is n to get my integrand, $36xyz^2-2$ dS which I don't know how to solve because its a double integral with 3 variables in the integrand.
At the foot of the capping surface, $x^2+9 y^2=9$ at $z=0$, let $x=3 \cos{t}$, $y=\sin{t}$, $dx = -3 \sin{t}$, $dy = \cos{t}$.
$$V = (6 \cos{t} \sin{t}, 9 \cos^2{t}-6 \cos{t})$$
$$V \cdot dr = -18 \cos{t} \sin^2{t} + 9 \cos^3{t} - 6 \cos^2{t} = 27 \cos^3{t} - 6 \cos^2{t}-18 \cos{t}$$
By Stokes' theorem, the integral is equal to
$$\oint_{x^2+9 y^2=9} V \cdot dr = \int_0^{2 \pi} dt \: (27 \cos^3{t} - 6 \cos^2{t}-18 \cos{t})$$
Only the $\cos^2{t}$ term is nonzero (its integral is $\pi$) and the result is $- 6 \pi$.