Mathematical Tools for Physics

9—Vector Calculus 1 261

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.orthogonal coordinates

9.4 Integral Representation of Curl
The calculation of the divergence was facilitated by the fact the the equation ( 5 ) could be manipulated into the
form of an integral, Eq. ( 9 ). Is there a similar expression for the curl? Yes.

curl~v= lim

V→ 0

1

V

∮

dA~×~v (17)

For the divergence there was a logical and orderly development to derive Eq. ( 9 ) from ( 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 (18)

To evaluate the integral use a sphere of radiusRcentered at the origin, makingˆn=rˆ. You also need the identity
A~×(B~×C~) =B~(A~.C~)−C~(A~.B).

dA~×(~ω×~r) =~ω(dA~.~r)−~r(~ω.dA~) (19)