[QUOTE=HeheZz;510745]Verify Stokes Theorem ∬(∇xF).N dA where surface S is the paraboloid z = 0.5(x^2 + y^2) bound by the plane z=2, C is its boundary, and the vector field F = 3yi - xzj + yzk
I had found (∇xF) = (z+x)i + (-z-3)k
r = [u, v, 0.5(u^2 + v^2)]
Therefore N= ru X rv = -ui -uj +k[/tex]
First this is wrong because this N is not a unit vector.
It seems to me simpler to use , from the start: so that and .Therefore (∇xF).N = [(z+x), 0, (-z-3)].[-x, -y, 1]
After that I substitute x = rcos(θ), y = rsin(θ), z = 0.5r^2
Thus ∫(0-2)∫(0-2pi) (∇xF).N r dθdr
But I cant seems to get the answer. Can anyone help? Help would greatly appreciated
There is no need to calculate "N" and "dS" separately since both would involve calculating the length of that vector which would then cancel. Instead .
The parts involving only and are easy- the integral with respect to is 0.
That leaves only .
By the way, notice the negative sign in front of that. Doing a vector integration over a surface involves a choice of orientation. I chose to orient with "z component upward". Choosing z component downard will reverse the sign. Of course, in using Stoke's theorem, you also have to choose an orientation for the path integral around the boundary- which direction you will integrate around the boundary. An orientation for one implies a corresponding orientation for the other.