# Triple Integral

"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
Homework Helper
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
$$\frac{\partial F_z}{\partial y}$$
as a divergence. Hint: to make progress, apply Gauss' Theorem to the result.

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
Homework Helper
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 ...

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!

Last edited:
Physics Monkey