# Question on surface integral of curl

1. Oct 1, 2007

### Simfish

1. The problem statement, all variables and given/known data

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.

2. Relevant 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...

$$\nabla \times [tex]F$$$$= (-x y cos(xyz))i + (2x^2 z - e^x z)j + (y z cos(xyz) k).$$

(it should read \nabla \times F but my edit doesn't change the tex...)

So I try to evaluate the surface integral through the northern hemisphere. Of $$x^2 + y^2 + (z-1)^2 = 1)$$, I get...

$$\frac{\partial g}{\partial x} = 2x, \frac{\partial g}{\partial y} = 2y$$

So...

$$P \frac{\partial g}{\partial x}- Q \frac{\partial g}{\partial y} + R$$
$$= 2x xy cos(xyz) - 2y 2x^2 z + 2y e^y z + yz cos(xyz)$$... which is like totally impossible to evaluate with a surface integral.

So what am I doing wrong?

Last edited: Oct 1, 2007
2. Oct 1, 2007

### 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

3. Oct 3, 2007

bump...