Cylindrical Coordinates
Cylindrical coordinates are related to rectangular coordinates as follows.
p
x2 + y 2
y
tan θ =
x
z=z
ρ=
x = ρ cos θ
y = ρ sin θ
z=z
The cylindrical coordinate vectors are defined as
1
∇ρ
|∇ρ|
1
eθ :=
∇θ
|∇θ|
1
ez :=
∇z
|∇z|
eρ :=
Thus,
eρ = p
x
i+ p
y
j = cos θ i + sin θ j
x2 + y 2
x2 + y 2
y
x
eθ = − p
i+ p
j = − sin θ i + cos θ j
2
2
2
x +y
x + y2
ez = k
The inverse relationship is as follows.
i = cos θ eρ − sin θ eθ
j = sin θ eρ + cos θ eθ
k = ez
It is worth noting that the above computations also imply the following.
∂ρ
∂x
∂ρ
∂y
∂θ
∂x
∂θ
∂y
= cos θ
= sin θ
1
= − sin θ
ρ
1
=
cos θ
ρ
1
The position vector R = xi + yj + zk is written
R = ρ eρ + z ez . (cylindrical coordinates)
If R = R(t) is a parameterized curve, then
that dedtρ = dθ
e . Thus,
dt θ
dR
dt
=
dρ
e
dt ρ
+ ρ dedtρ +
dz
e .
dt z
Since eρ = cos θ i + sin θ j, it follows
dρ
dθ
dz
dR
=
eρ + ρ eθ +
ez
dt
dt
dt
dt
Hence, dR = dρ eρ + ρ dθ eθ + dz ez and it follows that the element of volume in cylindrical coordinates is
given by
dV = ρ dρ dθ dz
If f = f (x, y, z) is a scalar field (that is, a real-valued function of three variables), then
∇f =
∂f
∂f
∂f
i+
j+
k.
∂x
∂y
∂z
If we view x and y as functions of ρ and θ and apply the chain rule, we obtain
∂f ∂ρ ∂f ∂θ
∂f ∂ρ ∂f ∂θ
∂f
∇f =
+
i+
+
j+
k.
∂ρ ∂x ∂θ ∂x
∂ρ ∂y ∂θ ∂y
∂z
Writing this in terms of ρ, θ, and the cylindrical coordinate vectors yields
∂f
1
∂f
∂f
1
∂f
∂f
∇f = cos θ
− sin θ
(cos θ eρ − sin θ eθ ) + sin θ
+ cos θ
(sin θ eρ + cos θ eθ ) +
ez .
∂ρ ρ
∂θ
∂ρ ρ
∂θ
∂z
Simplifying, we obtain the result
∇f =
∂f
1 ∂f
∂f
eρ +
eθ +
ez .
∂ρ
ρ ∂θ
∂z
If F = F(x, y, z) is a vector field (that is, a vector-valued function of three variables), then we can write
F = F1 i + F2 j + F3 k
= (cos θF1 + sin θF2 )eρ + (− sin θF1 + cos θF2 )eθ + F3 ez
Thus, F = Fρ eρ + Fθ eθ + Fz ez , where
F1 = cos θ Fρ − sin θ Fθ
F2 = sin θ Fρ + cos θ Fθ
F3 = Fz
Fρ = cos θ F1 + sin θ F2
Fθ = − sin θ F1 + cos θ F2
Fz = F3
Now we can transform ∇ · F and ∇ × F into cylindrical coordinates. To transform ∇ · F, we compute as
follows.
2
∂F1 ∂F2 ∂F3
+
+
∂x
∂y
∂z
∂F2 ∂ρ ∂F2 ∂θ
∂F1 ∂ρ ∂F1 ∂θ
∂F3
=
+
+
+
+
∂ρ ∂x
∂θ ∂x
∂ρ ∂y
∂θ ∂y
∂z
1
∂
∂
= cos θ (cos θ Fρ − sin θ Fθ ) − sin θ (cos θ Fρ − sin θ Fθ )
∂ρ
ρ
∂θ
1
∂
∂Fz
∂
+ sin θ (sin θ Fρ + cos θ Fθ ) + cos θ (sin θ Fρ + cos θ Fθ ) +
∂ρ
ρ
∂θ
∂z
∇·F=
After simplifying, we obtain
∇·F=
1 ∂
1 ∂Fθ ∂Fz
(ρFρ ) +
+
ρ ∂ρ
ρ ∂θ
∂z
∇ × F is handled similarly.
i
∂ ∂j k
∂ ∇ × F = ∂x ∂y ∂z F1 F2 F3 ∂F3 ∂F2
∂F1 ∂F3
∂F2 ∂F1
=
−
i+
−
j+
−
k
∂y
∂z
∂z
∂x
∂x
∂y
Writing the partial derivatives of F1 , and F2 in terms of Fρ , Fθ , and their partial derivatives, we obtain
∂F3 ∂F2
∂Fz ∂Fρ
1 ∂Fz ∂Fθ
−
= sin θ
−
+ cos θ
−
∂y
∂z
∂ρ
∂z
ρ ∂θ
∂z
∂Fρ ∂Fz
1 ∂Fz ∂Fθ
∂F1 ∂F3
−
= cos θ
−
+ sin θ
−
∂z
∂x
∂z
∂ρ
ρ ∂θ
∂z
∂F1 ∂F3
1
∂Fθ 1 ∂Fρ
−
= Fθ +
−
∂z
∂x
ρ
∂ρ
ρ ∂θ
Writing i, j, and k in terms of eρ , eθ , and ez and simplifying, we obtain
1
∇×F=
ρ
1 = ρ
∂Fz
∂Fθ
∂Fρ ∂Fz
∂
∂Fρ
−ρ
eρ +
−
ρeθ +
(ρFθ ) −
ez
∂θ
∂z
∂z
∂ρ
∂ρ
∂θ
eρ ρeθ ez ∂
∂ ∂
∂ρ
∂θ
∂z Fρ ρFθ Fz 3