Topic 14 / 17intermediate

Polar, Cylindrical & Spherical Coordinates

Convert between rectangular and polar/cylindrical/spherical coordinates and use the right system to simplify integrals.

Choosing the right coordinate system can turn an impossible integral into an easy one. **Polar** for circular regions in 2D, **cylindrical** for circular symmetry around the $z$ -axis, and **spherical** for ball-like or radial symmetry in 3D.

Polar coordinates (2D)

Conversions:

x = r cos θ, y = r sin θ, r = x^{2} + y^{2}, tan θ = y / x

. Differential:

d A = r d r d θ

— **the extra

r

factor is critical**.

Key takeaways

Region a disk or annulus → use polar.
Don't forget the $r$ factor in $d A$ .

Cylindrical coordinates (3D)

Polar in

x y

plus

z

x = r cos θ, y = r sin θ, z = z

. Differential:

d V = r d z d r d θ

. Best for solids with circular cross-sections (cylinders, paraboloids about

z

-axis).

Spherical coordinates

x = ρ sin ϕ cos θ, y = ρ sin ϕ sin θ, z = ρ cos ϕ

. Here

ρ \geq 0

is distance from origin,

θ \in [0, 2 π)

is the azimuthal angle around

z

-axis,

ϕ \in [0, π]

is the polar angle from

+ z

. Differential:

d V = ρ^{2} sin ϕ d ρ d θ d ϕ

Strategy tips

$ϕ = 0$ along $+ z$ , $ϕ = π /2$ in $x y$ -plane, $ϕ = π$ along $- z$ .
Spheres centered at origin become $ρ \leq a$ — bounds are constants!

Choosing the best system

**Cartesian**: rectangular boxes, simple regions in

x, y, z

. **Cylindrical**: circular symmetry about an axis, paraboloids, cylinders. **Spherical**: spheres, cones, regions described by distance from origin. The **best choice gives constant bounds** wherever possible.

Sphere of radius $a$ in each system

Cartesian:

- a \leq x \leq a, - a^{2} - x^{2} \leq y \leq a^{2} - x^{2}, - a^{2} - x^{2} - y^{2} \leq z \leq a^{2} - x^{2} - y^{2}

— ugly. Cylindrical:

0 \leq θ \leq 2 π, 0 \leq r \leq a, - a^{2} - r^{2} \leq z \leq a^{2} - r^{2}

— better. **Spherical:

0 \leq ρ \leq a, 0 \leq θ \leq 2 π, 0 \leq ϕ \leq π

— all constants. The clear winner.**

Worked examples

Example 1

Find the area of one leaf of the four-leafed rose

r = sin (2 θ)

1
One leaf occurs for $0 \leq θ \leq π /2$ (where $sin (2 θ) \geq 0$ ).
2
Polar area formula.
$A = \frac{1}{2} \int_{0}^{π /2} sin^{2} (2 θ) d θ = \frac{1}{2} \int_{0}^{π /2} \frac{1 - cos ( 4 θ )}{2} d θ$
3
Compute.
$= \frac{1}{4} [θ - \frac{s i n ( 4 θ )}{4}]_{0}^{π /2} = \frac{π}{8}$

Answer

A = π /8

Example 2

Set up and evaluate the volume of the solid above the cone

z = x^{2} + y^{2}

and below the sphere

x^{2} + y^{2} + z^{2} = 4

using spherical coordinates.

1
Cone $z = x^{2} + y^{2}$ in spherical: $ρ cos ϕ = ρ sin ϕ \Rightarrow tan ϕ = 1 \Rightarrow ϕ = π /4$ .
2
Sphere $ρ = 2$ . Region: $0 \leq ρ \leq 2, 0 \leq θ \leq 2 π, 0 \leq ϕ \leq π /4$ .
3
Volume integral.
$V = \int_{0}^{π /4} \int_{0}^{2 π} \int_{0}^{2} ρ^{2} sin ϕ d ρ d θ d ϕ$
4
Evaluate.
$= \frac{8}{3} \int_{0}^{π /4} sin ϕ d ϕ \cdot 2 π = \frac{16 π}{3} (1 - cos (π /4)) = \frac{16 π}{3} (1 - \frac{2}{2})$

Answer

V = \frac{16 π}{3} (1 - \frac{2}{2})

Example 3

Convert

\iint_{D} x^{2} d A

over the disk

x^{2} + y^{2} \leq 4

to polar and evaluate.

1
Polar setup.
$\int_{0}^{2 π} \int_{0}^{2} r^{2} cos^{2} θ \cdot r d r d θ = \int_{0}^{2 π} cos^{2} θ d θ \cdot \int_{0}^{2} r^{3} d r$
2
Compute.
$= π \cdot 4 = 4 π$

Answer

4 π

Interactive visualizations

Loading visualization…

r2.00

θ (rad)0.79

x = r cos θ = 2.00 \cdot cos (0.79) = 1.414

y = r sin θ = 2.00 \cdot sin (0.79) = 1.414

Inverse:

r = x^{2} + y^{2}, tan θ = y / x

Polar conversion. Slide

r

and

θ

to see the corresponding

(x, y)

Loading visualization…

r2.00

θ1.05

z1.50

(x, y, z) = (r cos θ, r sin θ, z) = (1.00, 1.73, 1.50)

Cylindrical: circle in

x y

at constant

z

, plus a vertical lift. Note green segment (

z

) and blue segment (radial in

x y

Loading visualization…

ρ2.00

θ0.79

φ1.05

x = ρ sin ϕ cos θ = 1.22

y = ρ sin ϕ sin θ = 1.22

z = ρ cos ϕ = 1.00

Spherical:

ρ, θ, ϕ

. The sphere shrinks/grows with

ρ

Formulas in this topic

Polar conversion

x = r cos θ, y = r sin θ, d A = r d r d θ

2D conversions and area element.

Cylindrical conversion

x = r cos θ, y = r sin θ, z = z, d V = r d z d r d θ

3D, axial symmetry.

Spherical conversion

x = ρ sin ϕ cos θ, y = ρ sin ϕ sin θ, z = ρ cos ϕ

3D, radial.

Spherical volume element

d V = ρ^{2} sin ϕ d ρ d θ d ϕ

Critical Jacobian.

Polar coordinates (2D)

Cylindrical coordinates (3D)

Spherical coordinates

Choosing the best system

Sphere of radius $a$ in each system

Worked examples

Interactive visualizations

Formulas in this topic

Related topics