Triple Integral Property: \bigtriangledown \times F

[Treadstone 71]

"Suppose that a smooth vector field F(x,y,z) given on a region D has the property that on the bounding surface S it is perpenticular to the surface. Show that
\int\int\int_D \bigtriangledown \times F dV = 0
in the sense that each component of \bigtriangledown \times F has integral 0 over D."
I'm really stumped on this one.

 

[Physics Monkey]

You can proceed by examing each component of the curl of F in turn. Take the x component for instance, it looks something like
(\vec{\nabla}\times\vec{F})_x = \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}. Since integration is linear you can consider each piece in turn. Can you find a way to represent
<br /> \frac{\partial F_z}{\partial y}<br />
as a divergence. Hint: to make progress, apply Gauss' Theorem to the result.

 

[Treadstone 71]

I suppose you can express it as \bigtriangledown(0,F_3,0) so by Gauss's theorem,

\int\int\int_D\bigtriangledown(0,F_3,0)dV=\int\int_{\partial D}(0,F_3,0)d\vec{S}.

If the field were tangent to the surface, the dot product between the normal vector and it would be zero, but since it's already perpenticular...

 

[Physics Monkey]

You aren't quite there yet, but you're close. In particular, you can't just conclude that
\int\int _{\partial D}(0,F_3,0)d\vec{S}
equals zero. However, when you combine this term with the other term in the curl, then maybe you can say something ...

 

[Treadstone 71]

It boils down to

\int\int \bigtriangledown(0,F_3,0)-\bigtriangledown(0,0,F_2)\vec{N}dS

The integrand becomes

\bigtriangledown(0,F_3,F_2)\vec{N}

Since the normal is parallel to the field, it becomes (F3F2-F2F3)|N|, and this is 0.

Did I get this right? If so thanks for the help!

 

[Physics Monkey]

I don't know why you have the gradient operator still there, but yes, you've pretty much got it.