Question on surface integral of curl

[Simfish]

Homework Statement

Let F be F = ( x^2 z^2 ) i + (sin xyz) j + (e^x z) k.Find \int\int \nabla \times F \cdot n dS

where the region E is above the cone z^2 = x^2 + y^2 and inside the sphere centered at (0,0,1) and with radius 1. (so it is x^2 + y^2 + (z-1)^2 = 1).. I know that they intersect at z = 1, z = 0. So the boundary is at z = 1.

Homework Equations

So I know that I can apply Stokes Theorem here or evaluate the surface integral directly.

If I try Stokes Theorem, I apply it to the boundary x^2 + y^2 + (z-1)^2 = 1, which can be parametrized by r(\theta) = cos(\theta) i + sin(\theta) j + k, r'(\theta) = -sin(\theta) i + cos(\theta) j + 0 k

But when I try F(r(\theta)) \cdot r'(\theta), I get \int_0^{2 \pi} -sin(\theta) cos(\theta)^2 + cos(\theta) sin( cos(\theta) sin(\theta)).. Which is pretty much impossible to evaluate, since I never heard of integrating a sine function within a sine function

as for trying to evaluate the surface integral directly... I get the cross product which is...

\nabla \times F= (-x y cos(xyz))i + (2x^2 z - e^x z)j + (y z cos(xyz) k).<br /> <br /> (it should read \nabla \times F but my edit doesn&#039;t change the tex...)<br /> <br /> So I try to evaluate the surface integral through the northern hemisphere. Of x^2 + y^2 + (z-1)^2 = 1), I get...<br /> <br /> \frac{\partial g}{\partial x} = 2x, \frac{\partial g}{\partial y} = 2y<br /> <br /> So...<br /> <br /> P \frac{\partial g}{\partial x}- Q \frac{\partial g}{\partial y} + R<br /> = 2x xy cos(xyz) - 2y 2x^2 z + 2y e^y z + yz cos(xyz)... which is like totally impossible to evaluate with a surface integral.<br /> <br /> So what am I doing wrong?

 

[Simfish]

Another question incidental to this:

div (curl) F = 0. But yet, Stokes Theorem applies to a vector function. Is there any fundamental difference between \int\int \nabla \times F \cdot dS and \int\int F \cdot dS when both are vector functions? Yet, the divergence theorem applies to the second one, when \int\int F \cdot dS = \int\int\int div F dV. If there is no fundamental difference between \nabla \times F \cdot dS and F \cdot dS, then why does the divergence theorem work for the latter but not the former? If it works for the former, then \int\int \nabla \times F \cdot dS = \int\int\int div \nabla \times F = 0 (but this is clearly not the case, otherwise Stokes' Theorem would be useless. Can anyone please clarify the issue to me? Thanks

 

[Simfish]

bump...