Mathematical Tools for Physics

9—Vector Calculus 1 262

Choose a spherical coordinate system with thez-axis along~ω.

dA~=ˆndA=r dA,ˆ and ~ω.dA~=ω dAcosθ

∮

dA~×~v=

∮

~ωRdA−~r ω dAcosθ

=~ωR 4 πR^2 −ω

∫π

0

R^2 sinθ dθ

∫ 2 π

0

dφˆz Rcosθcosθ

=~ωR 4 πR^2 −ωˆz 2 πR^3

∫π

0

sinθdθcos^2 θ

=~ωR 4 πR^2 −ωˆz 2 πR^3

∫ 1

− 1

cos^2 θ dcosθ=~ωR 4 πR^2 −ωˆz 2 πR^3.

2

3

=~ω

8

3

πR^3

Divide by the volume of the sphere and you have 2 ~ω as promised. In the first term on the first line of the
calculation,~ωRis a constant over the surface so you can pull it out of the integral. In the second term,~rhas
components in theˆx,ˆy, andzˆdirections; the first two of these integrate to zero because for every vector with a
positiveˆx-component there is a negative one. Same foryˆ. All that is left of~risˆz Rcosθ.

The Curl in Components
With the integral representation, Eq. ( 17 ), available for the curl, the process is much like that for computing
the divergence. Start with rectangular of course. Use the same equation, Eq. ( 10 ) and the same picture that
accompanied that equation. With the experience gained from computing the divergence however, you don’t have
to go through all the complications of the first calculation. Use the simpler form that followed.
In Eq. ( 14 ) you have~v.∆A=vx∆y∆zon the right face and on the left face. This time replace the dot
with a cross (in the right order).
On the right,
∆A~×~v= ∆y∆zxˆ×~v(x 0 + ∆x,y 0 + ∆y/ 2 ,z 0 + ∆z/2) (20)

On the left it is
∆A~×~v= ∆y∆zˆx×~v(x 0 ,y 0 + ∆y/ 2 ,z 0 + ∆z/2) (21)