Mathematical Tools for Physics

9—Vector Calculus 1 260

steps still more. Do this for the other sides, add, and you get the result. It all looks very simple when you do it
this way, but what if you need to do it in cylindrical coordinates?

∆r

∆z

r∆θ

When everything is small, the volume is close to a rectangular box, so its volume isV = (∆r)(∆z)(r∆θ).
Go through the simple version for the calculation of the surface integral. The top and bottom present nothing
significantly different from the rectangular case.

[

vz(r 0 ,θ 0 ,z 0 + ∆z)−vz(r 0 ,θ 0 ,z 0 )

]

(∆r)(r 0 ∆θ)÷r 0 ∆r 0 ∆θ∆z−→

∂vz
∂z
The curved faces of constantrare a bit different, because the areas of the two opposing faces aren’t the
same.
[
vr(r 0 + ∆r,θ 0 ,z 0 )(r 0 + ∆r)∆θ∆z−vr(r 0 ,θ 0 ,z 0 )r 0 ∆θ∆z

]

÷r 0 ∆r∆θ∆z−→

1

r

∂(rvr)

∂r

A bit more complex than the rectangular case, but not too bad.
Now for the constantθsides. Here the areas of the two faces are the same, so even though they are not
precisely parallel to each other this doesn’t cause any difficulties.

[

vθ(r 0 ,θ 0 + ∆θ,z 0 )−vz(r 0 ,θ 0 ,z 0 )

]

(∆r)(∆z)÷r 0 ∆r∆θ∆z−→

1

r

∂vθ

∂θ

The sum of all these terms is the divergence expressed in cylindrical coordinates.

div~v=

1

r

∂(rvr)

∂r

+

1

r

∂vθ

∂θ

+

∂vz

∂z

(15)

The corresponding expression in spherical coordinates is found in exactly the same way, problem 4.

div~v=

1

r^2

∂(r^2 vr)

∂r

+

1

rsinθ

∂(sinθvθ)

∂θ

+

1

rsinθ

∂vφ

∂φ

(16)