Mathematical Tools for Physics

13—Vector Calculus 2 403

Use exactly the same reasoning that leads from the definition of the divergence to Eqs. ( 12 ) and ( 13 ) (see
problem 6 ), and this leads to the analog of Gauss’s theorem, but with cross products.
∮

S

dA~×~v=

∫

V

curl~v dV (15)

This isn’t yet in a form that is all that convenient, and a special case is both easier to interpret and more useful
in applications. First apply it to a particular volume, one that is very thin and small. Take a tiny disk of height
∆h, with top and bottom area∆A 1. Letˆn 1 be the unit normal vector out of the top area. For small enough
values of these dimensions, the right side of Eq. ( 14 ) is simply the value of the vectorcurl~vinside the volume
times the volume∆A 1 ∆hitself.

∮

S

dA~×~v=

∫

V

curl~v dV = curl~v∆A 1 ∆h

ˆn 1

Take the dot product of both sides withnˆ 1 , and the parts of the surface integral from the top and the bottom
faces disappear. That’s just the statement that on the top and the bottom,dA~is in the direction of±ˆn 1 , so the
cross product makesdA~×~vperpendicular toˆn 1.
I’m using the subscript 1 for the top surface and I’ll use 2 for the surface around the edge. Otherwise it’s
too easy to get the notation mixed up.
Now look atdA~×~varound the thin edge. The element of area has height∆hand length∆`along the
arc. Callˆn 2 the unit normal out of the edge.

∆A~ 2 = ∆h∆`ˆn 2

ˆn 2

d~`

The productˆn 1 .∆A~ 2 ×~v=ˆn 1 .ˆn 2 ×~v∆h∆=ˆn 1 ׈n 2 .~v∆h∆, using the property of the triple scalar product.
The productˆn 1 ׈n 2 is in the direction along the arc of the edge, so

nˆ 1 ׈n 2 ∆`= ∆~` (16)