* Spherical coordinates* represent points in using three numbers: .

is the distance from to the point.

is "the polar coordinate " --- that is, project the ray from the origin to the point down to a ray in the x-y plane. Measure the angle from the positive x-axis to in the usual way.

is the angle from the positive z-axis to the ray from the origin to the point.

The conversion equations are

Note also that

However, if you're converting the coordinates of a single point from one coordinate system to another, the best thing is to draw a picture and use trigonometry.

* Example.* A point has rectangular coordinates
. Find its spherical coordinates.

First, . Since and , by Pythagoras.

, so .

The spherical coordinates are .

Note: If a point lies below the x-y-plane, will be greater than . In that case, you can't use or "as is" to give , since those inverse trig functions only produce angles less than or equal to .

* Example.* A point has cylindrical coordinates
.
Find its spherical coordinates.

. Since and , by Pythagoras. Finally, the radius lies below the x-y-plane, so .

The spherical coordinates are .

* Example.* Let r be the polar coordinate radius.
Express r in terms of spherical coordinates.

I can take (so ) and , so I can avoid taking absolute values in the last square root step.

* Example.* Sketch the region in space described
by the following spherical coordinate inequalities:

The region lies inside the sphere of radius 1 but above the cone . Note that the latter is

The spherical conversion equations are

They define a function

The Jacobian of f is

The absolute value is . Hence, when you go from rectangular coordinates to spherical coordinates, the differentials convert by

Therefore, in order to convert a triple integral from rectangular coordinates to spherical coordinates, you should do the following:

1. Convert the limits of integration by describing the region of integration by inequalities in spherical coordinates.

2. Convert the integrand using the spherical conversion formulas:

3. Convert the differentials by

* Example.* Let R be the interior of the sphere
. Compute

R is a sphere of radius 2 centered at the origin. The interior of the sphere is described by the inequalities

Moreover, since ,

Therefore, the integral becomes

When should you consider converting a triple integral to spherical coordinates? Here are two rough guidelines:

(a) Consider converting to spherical coordinates when the region of
integration involves graphs that "look nice" in spherical.
For example, * spheres* and *
cones* often produce regions that can be described by simple
inequalities in spherical coordinates.

(b) Consider converting to spherical coordinates when the integrand involves terms like ( ).

* Example.* Let R be the region defined by the
inequalities and . Compute

is a sphere of radius 2 centered at the origin, so refers to the interior of the sphere.

is a cone opening at a angle to the z-axis. Hence, refers to the region lying above the cone.

Together, the inequalities specify the region inside the sphere but above the cone. (It's shaped like an ice-cream cone.)

Evidently, the region "goes all the way around" the z-axis,
so . To determine the
inequalities for , think of a searchlight beam
emanating from the origin. The beam *enters* the ice-cream
cone at the origin ( ) and *leaves* the
ice-cream cone through the top of the ice-cream, which is our sphere
( ). Hence, .

is the angle measured "downward" from the z-axis. Think of an umbrella held upside-down. If you open the umbrella, through what range of angles will the ribs sweep as they pass through the region? Since the cone makes a ( ) angle with the z-axis, it follows that .

The inequalities in spherical coordinates which describe the region are

Finally, since , the integral is

Copyright 2018 by Bruce Ikenaga