9—Vector Calculus 1 264cylindrical: ∇=ˆr∂
∂r+ˆθ1
r∂
∂θ+ˆz∂
∂z(24)
spherical: ∇=ˆr∂
∂r+ˆθ1
r∂
∂θ+φˆ1
rsinθ∂
∂φIn all nine of these components, the denominator (e.g.rsinθ dφ) is the element of displacement along the
direction indicated.
9.6 Shorter Cut for div and curl
There is another way to compute the divergence and curl in cylindrical and rectangular coordinates. A direct
application of Eqs. ( 13 ), ( 22 ), and ( 24 ) gets the the result quickly. The only caution is that you have to be
careful that the unit vectors areinsidethe derivative, so you have to differentiate them too.
∇.~vis the divergence of~v, and in cylindrical coordinates
∇.~v=(
ˆr∂
∂r+ˆθ1
r∂
∂θ+zˆ∂
∂z)
.
(
ˆrvr+ˆθvθ+ˆzvz)
The unit vectorsˆr,θˆ, andˆzdon’t change as you alterrorz. They do change as you alterθ. (except forˆz).
∂ˆr
∂r=
∂θˆ
∂r=
∂ˆz
∂r=
∂ˆr
∂z=
∂ˆθ
∂z=
∂ˆz
∂z=
∂ˆz
∂θ= 0
Next come∂ˆr/∂θand∂θ/∂θˆ. This is problem8.20. You can do this by first showing that
ˆr=ˆxcosθ+yˆsinθ and ˆθ=−ˆxsinθ+ˆycosθand differentiating with respect toθ. This gives
∂ˆr/∂θ=θ,ˆ and ∂θ/∂θˆ =−ˆrPut these together and you have
∇.~v=∂vr
∂r+ˆθ.1
r∂
∂θ(
ˆrvr+ˆθvθ)
+
∂vz
∂z
=∂vr
∂r+ˆθ.1
r(
vrdˆr
dθ+ˆθ∂vθ
∂θ)
+
∂vz
∂z=∂vr
∂r+
1
rvr+1
r∂vθ
∂θ+
∂vz
∂z