[QUOTE=HeheZz;510745]Verify Stokes Theorem ∬(∇xF).NdA 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)]

ThereforeN= ru X rv = -ui-uj+k[/tex]

First this is wrong because thisNis not aunitvector.

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).Nr dθdr

But I cant seems to get the answer. Can anyone help? Help would greatly appreciated

Now, .

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 .

so .

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 oforientation. 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.