Mathematical Tools for Physics - Department of Physics - University

9—Vector Calculus 1 222

The corresponding expression in spherical coordinates is found in exactly the same way, prob-
lem9.4.

div~v=

1

r^2

∂(r^2 vr)

∂r

+

1

rsinθ

∂(sinθvθ)

∂θ

+

1

rsinθ

∂vφ

∂φ

(9.16)

These are the three commonly occurring coordinates system, though the same simplified method
will work in any other orthogonal coordinate system. The coordinate system is orthogonal if the surfaces
made by setting the value of the respective coordinates to a constant intersect at right angles. In the

spherical example this means that a surface of constantris a sphere. A surface of constantθis a

half-plane starting from thez-axis. These intersect perpendicular to each other. If you set the third

coordinate,φ, to a constant you have a cone that intersects the other two at right angles. Look back

to section8.8.

9.4 Integral Representation of Curl
The calculation of the divergence was facilitated by the fact that the equation (9.5) could be manipulated
into the form of an integral, Eq. (9.9). Is there a similar expression for the curl? Yes.

curl~v= lim

V→ 0

1

V

∮

dA~×~v (9.17)

For the divergence there was a logical and orderly development to derive Eq. (9.9) from (9.5). Is there
a similar intuitively clear path here? I don’t know of one. The best that I can do is to show that it
gives the right answer.

And what’s that surface integral doing with a×instead of a.? No mistake. Just replace the
dot product by a cross product in the definition of the integral. This time however you have to watch
the order of the factors.

~ω

θ

ˆndA

To verify that this does give the correct answer, use a vector field that represents pure rigid body

rotation. You’re going to take the limit as∆V → 0 , so it may as well be uniform. The velocity field

for this is the same as from problem7.5.

~v=~ω×~r (9.18)

To evaluate the integral use a sphere of radiusRcentered at the origin, makingnˆ=ˆr. You also need

the identityA~×(B~×C~) =B~(A~.C~)−C~(A~.B).

dA~×(~ω×~r) =~ω(dA~.~r)−~r(~ω.dA~) (9.19)

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

dA~=ˆndA=rdA,ˆ and ~ω.dA~=ωdAcosθ