Math Core

Polar coordinates made disks and circular regions easy in the plane. In space, the same idea gives two new coordinate systems. Cylindrical coordinates suit solids with an axis of symmetry, like cylinders, cones and paraboloids. Spherical coordinates suit solids built from spheres and cones centered at the origin.

Cylindrical coordinates

Cylindrical coordinates are polar coordinates in the $xy$ -plane with $z$ left alone. A point is written $(r, \theta, z)$ , where

x = r\cos\theta, \qquad y = r\sin\theta, \qquad z = z, \qquad r^2 = x^2 + y^2.

Simple equations describe common surfaces. $r = 2$ is a cylinder of radius $2$ around the $z$ -axis, $\theta = \frac{\pi}{4}$ is a vertical half-plane, and $z = r^2$ is the paraboloid $z = x^2 + y^2$ .

To find the volume element, take a small polar area element $r\,dr\,d\theta$ and give it height $dz$ .

Triple integrals in cylindrical coordinates

If $E$ is the solid $u_1(r, \theta) \le z \le u_2(r, \theta)$ over a polar region $D$ , then

\iiint_E f\,dV = \iint_D \int_{u_1}^{u_2} f(r\cos\theta,\ r\sin\theta,\ z)\,r\,dz\,dr\,d\theta.

The volume element is $dV = r\,dz\,dr\,d\theta$ .

Worked example: An integral over a solid cylinder

Evaluate $\displaystyle\iiint_E \left(x^2 + y^2\right)dV$ , where $E$ is the solid cylinder $x^2 + y^2 \le 1$ , $0 \le z \le 2$ .

Solution. In cylindrical coordinates, $E$ is $0 \le \theta \le 2\pi$ , $0 \le r \le 1$ , $0 \le z \le 2$ , and $x^2 + y^2 = r^2$ :

\int_0^{2\pi} \int_0^1 \int_0^2 r^2 \cdot r\,dz\,dr\,d\theta = 2\pi \cdot \frac{1}{4} \cdot 2 = \pi.

A vertical slice through the solid of the next example: between the bowl z = r² and the dome z = 8 − r². They meet where r = 2.Open in grapher →

Worked example: Between two paraboloids

Find the volume of the solid between the paraboloids $z = x^2 + y^2$ and $z = 8 - x^2 - y^2$ .

Solution. In cylindrical coordinates the surfaces are $z = r^2$ (below) and $z = 8 - r^2$ (above). They meet where $r^2 = 8 - r^2$ , so $r = 2$ . The shadow is the disk $r \le 2$ :

V = \int_0^{2\pi} \int_0^2 \int_{r^2}^{8 - r^2} r\,dz\,dr\,d\theta = 2\pi\int_0^2 \left(8r - 2r^3\right)dr = 2\pi\left[4r^2 - \frac{r^4}{2}\right]_0^2 = 2\pi(16 - 8) = 16\pi.

Spherical coordinates

A point in spherical coordinates is $(\rho, \theta, \phi)$ :

$\rho \ge 0$ is the distance from the origin, so $\rho^2 = x^2 + y^2 + z^2$ .
$\theta$ is the same angle as in cylindrical coordinates.
$\phi$ is the angle down from the positive $z$ -axis, with $0 \le \phi \le \pi$ .

The distance from the $z$ -axis is $r = \rho\sin\phi$ , and the height is $z = \rho\cos\phi$ . Combining with polar coordinates,

x = \rho\sin\phi\cos\theta, \qquad y = \rho\sin\phi\sin\theta, \qquad z = \rho\cos\phi.

In these coordinates, $\rho = a$ is a sphere of radius $a$ , $\phi = \frac{\pi}{2}$ is the $xy$ -plane, and $\phi = c$ (for $0 < c < \frac{\pi}{2}$ ) is a cone opening upward.

A small spherical "box" has sides $d\rho$ (outward), $\rho\,d\phi$ (along a meridian) and $\rho\sin\phi\,d\theta$ (along a circle of radius $\rho\sin\phi$ ). Multiplying gives the volume element.

Triple integrals in spherical coordinates

\iiint_E f\,dV = \iiint f(\rho\sin\phi\cos\theta,\ \rho\sin\phi\sin\theta,\ \rho\cos\phi)\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta.

The volume element is $dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ .

Common mistake

Don't forget the extra factors. In cylindrical coordinates $dV = r\,dz\,dr\,d\theta$ , not $dz\,dr\,d\theta$ . In spherical coordinates $dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ . Also, $\phi$ runs only from $0$ to $\pi$ , not to $2\pi$ : it is $\theta$ that goes all the way around.

Worked example: Volume of a ball

Find the volume of the ball $x^2 + y^2 + z^2 \le a^2$ .

Solution. The ball is $0 \le \rho \le a$ , $0 \le \phi \le \pi$ , $0 \le \theta \le 2\pi$ . The limits are constants, so the integral splits:

V = \int_0^{2\pi} \int_0^{\pi} \int_0^a \rho^2\sin\phi\,d\rho\,d\phi\,d\theta = 2\pi \cdot \Big[-\cos\phi\Big]_0^{\pi} \cdot \frac{a^3}{3} = 2\pi \cdot 2 \cdot \frac{a^3}{3} = \frac{4}{3}\pi a^3.

Worked example: An ice cream cone

Find the volume of the solid that lies above the cone $z = \sqrt{x^2 + y^2}$ and inside the sphere $x^2 + y^2 + z^2 = 4$ .

Solution. On the cone, $z = r$ , so $\rho\cos\phi = \rho\sin\phi$ , which means $\tan\phi = 1$ and $\phi = \frac{\pi}{4}$ . Above the cone means closer to the $z$ -axis, so $0 \le \phi \le \frac{\pi}{4}$ . The sphere is $\rho = 2$ :

V = \int_0^{2\pi} \int_0^{\pi/4} \int_0^2 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta = 2\pi\left(1 - \frac{\sqrt{2}}{2}\right)\frac{8}{3} = \frac{8\pi}{3}\left(2 - \sqrt{2}\right) \approx 4.91.

Tip

Choose coordinates by the shape and the integrand. Expressions like $x^2 + y^2$ and circular shadows suggest cylindrical. Expressions like $x^2 + y^2 + z^2$ , spheres, and cones through the origin suggest spherical.

Practice

Practice 1

Convert the cylindrical point $(r, \theta, z) = \left(2, \frac{\pi}{3}, 5\right)$ to rectangular coordinates $(x, y, z)$ .

Enter a point like (2, -3)

Practice 2

Convert the rectangular point $\left(1, 1, \sqrt{2}\right)$ to spherical coordinates $(\rho, \theta, \phi)$ .

Enter a point like (2, -3)

Practice 3

Which equation describes the cone $z = \sqrt{3\left(x^2 + y^2\right)}$ in spherical coordinates?

Practice 4

Find the volume of the solid under the paraboloid $z = 4 - x^2 - y^2$ and above the $xy$ -plane.