Spherical Coordinates

Learning Objectives

Convert from spherical to rectangular coordinates.
Convert from rectangular to spherical coordinates.

In the Cartesian coordinate system, the location of a point in space is described using an ordered triple in which each coordinate represents a distance. In the cylindrical coordinate system, location of a point in space is described using two distances ([latex]r[/latex] and[latex]z[/latex]) and an angle measure ([latex]\theta[/latex]). In the spherical coordinate system, we again use an ordered triple to describe the location of a point in space. In this case, the triple describes one distance and two angles. Spherical coordinates make it simple to describe a sphere, just as cylindrical coordinates make it easy to describe a cylinder. Grid lines for spherical coordinates are based on angle measures, like those for polar coordinates.

DEFINITION

In the spherical coordinate system, a point [latex]P[/latex] in space (Figure 1) is represented by the ordered triple [latex](\rho,\theta,\varphi)[/latex] where

[latex]\rho[/latex] (the Greek letter rho) is the distance between [latex]P[/latex] and the origin [latex](\rho\ne0)[/latex];
[latex]\theta[/latex] is the same angle used to describe the location in cylindrical coordinates;
[latex]\varphi[/latex] (the Greek letter phi) is the angle formed by the positive [latex]z[/latex]-axis and line segment [latex]\overline{OP}[/latex], where [latex]O[/latex] is the origin and [latex]0\leq\varphi\leq\pi[/latex].

This figure is the first quadrant of the 3-dimensional coordinate system. It has a point labeled “(x, y, z) = (rho, theta, phi).” There is a line segment from the origin to the point. It is labeled “rho.” The angle between this line segment and the z-axis is phi. There is a line segment in the x y-plane from the origin to the shadow of the point. This segment is labeled “r.” The angle between the x-axis and r is theta.

Figure 1. The relationship among spherical, rectangular, and cylindrical coordinates.

By convention, the origin is represented as [latex](0, 0, 0)[/latex] in spherical coordinates.

THEOREM: converting among Spherical, cylindrical, and rectangular coordinates

Rectangular coordinates [latex](x, y, z)[/latex] and spherical coordinates [latex](\rho,\theta,\varphi)[/latex] of a point are related as follows:

[latex]\begin{aligned} x&=\rho\sin\varphi\cos\theta \quad \text{These equations are used to convert from spherical coordinates to rectangular coordinates.} \\ y&=\rho\sin\varphi\sin\theta \\ z&=\rho\cos\varphi \\ &\text{and} \\ \rho^2&=x^2+y^2+z^2 \quad \text{These equations are used to convert from rectangular coordinates to spherical coordinates.} \\ \tan\theta&=\frac{y}x \\ \varphi&=\arccos\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right) \end{aligned}[/latex]

If a point has cylindrical coordinates [latex](r,\theta,z)[/latex], then these equations define the relationship between cylindrical and spherical coordinates.

[latex]\begin{aligned} r&=\rho\sin\varphi \quad \text{These equations are used to convert from spherical coordinates to cylindrical coordinates} \\ \theta&=\theta \\ z&=\rho\cos\varphi \\ &\text{and} \\ \rho&=\sqrt{r^2+z^2} \quad \text{These equations are used to convert from cylindrical coordinates to spherical coordinates} \\ \theta&=\theta \\ \varphi&=\arccos\left(\frac{z}{\sqrt{r^2+z^2}}\right) \end{aligned}[/latex]

The formulas to convert from spherical coordinates to rectangular coordinates may seem complex, but they are straightforward applications of trigonometry. Looking at Figure 2, it is easy to see that [latex]r=\rho\cos\varphi[/latex]. Then, looking at the triangle in the [latex]xy[/latex]-plane with [latex]r[/latex] as its hypotenuse, we have [latex]x=r\cos\theta=\rho\sin\varphi\cos\theta[/latex]. The derivation of the formula for [latex]y[/latex] is similar. Figure 9 in Cylindrical Coordinates also shows that [latex]\rho^2=r^2+z^2=x^2+y^2+z^2[/latex] and [latex]z=\rho\cos\varphi[/latex]. Solving this last equation for [latex]\varphi[/latex] and then substituting [latex]\rho=\sqrt{r^2+z^2}[/latex] (from the first equation) yields [latex]\varphi=\arccos\left(\frac{z}{\sqrt{r^2+z^2}}\right)[/latex]. Also, note that, as before, we must be careful when using the formula [latex]\tan\theta=\frac{y}x[/latex] to choose the correct value of [latex]\theta[/latex].

This figure is the first quadrant of the 3-dimensional coordinate system. It has a point labeled “(x, y, z) = (r, theta, z) = (rho, theta, phi).” There is a line segment from the origin to the point. It is labeled “rho.” The angle between this line segment and the z-axis is phi. There is a line segment in the x y-plane from the origin to the shadow of the point. This segment is labeled “r.” The angle between the x-axis and r is theta.The distance from r to the point is labeled “z.”

Figure 2. The equations that convert from one system to another are derived from right-triangle relationships.

As we did with cylindrical coordinates, let’s consider the surfaces that are generated when each of the coordinates is held constant. Let [latex]c[/latex] be a constant, and consider surfaces of the form [latex]\rho=c[/latex]. Points on these surfaces are at a fixed distance from the origin and form a sphere. The coordinate [latex]\theta[/latex] in the spherical coordinate system is the same as in the cylindrical coordinate system, so surfaces of the form [latex]\theta=c[/latex] are half-planes, as before. Last, consider surfaces of the form [latex]\varphi=c[/latex]. The points on these surfaces are at a fixed angle from the [latex]z[/latex]-axis and form a half-cone (Figure 3).

This figure has three images. The first image is a sphere centered in the 3-dimensional coordinate system. The second figure is a vertical plane with an edge on the z-axis in the 3-dimensional coordinate system. The third image is an elliptical cone with the center at the origin of the 3-dimensional coordinate system.

Figure 3. In spherical coordinates, surfaces of the form [latex]\rho=c[/latex] are spheres of radius [latex]\rho[/latex] (a), surfaces of the form [latex]\theta=c[/latex] are half-planes at an angle [latex]\theta[/latex] from the [latex]x[/latex]-axis (b), and surfaces of the form [latex]\varphi=c[/latex] are half-cones at an angle [latex]\varphi[/latex] from the [latex]z[/latex]-axis (c).

Example: converting from spherical coordinates

Plot the point with spherical coordinates [latex]\left(8,\frac{\pi}3,\frac{\pi}6\right)[/latex] and express its location in both rectangular and cylindrical coordinates.

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Plot the point with spherical coordinates [latex]\left(2,-\frac{5\pi}6,\frac{\pi}6\right)[/latex] and describe its location in both rectangular and cylindrical coordinates.

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Watch the following video to see the worked solution to the above Try IT.

You can view the transcript for “CP 2.58” here (opens in new window).

Example: converting from rectangular coordinates

Convert the rectangular coordinates [latex](-1,1,\sqrt6)[/latex] to both spherical and cylindrical coordinates.

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Example: identifying surfaces in the spherical coordinate system

Describe the surfaces with the given spherical equations.

[latex]\theta=\frac{\pi}3[/latex]
[latex]\varphi=\frac{5\pi}6[/latex]
[latex]\rho=6[/latex]
[latex]\rho=\sin\theta\sin\varphi[/latex]

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The variable [latex]\theta[/latex] represents the measure of the same angle in both the cylindrical and spherical coordinate systems. Points with coordinates [latex](\rho,\frac{\pi}3,\varphi)[/latex] lie on the plane that forms angle [latex]\theta=\frac{\pi}3[/latex] with the positive [latex]x[/latex]-axis. Because [latex]\rho>0[/latex], the surface described by equation [latex]\theta=\frac{\pi}3[/latex] is the half-plane shown in Figure 6.

Figure 6. The surface described by equation [latex]\theta=\frac{\pi}3[/latex] is a half-plane.
Equation [latex]\varphi=\frac{5\pi}6[/latex] describes all points in the spherical coordinate system that lie on a line from the origin forming an angle measuring [latex]\frac{5\pi}6[/latex] rad with the positive [latex]z[/latex]-axis. These points form a half-cone (Figure 7). Because there is only one value for [latex]\varphi[/latex] that is measured from the positive [latex]z[/latex]-axis, we do not get the full cone (with two pieces).

Figure 7. The equation [latex]\varphi=\frac{5\pi}6[/latex] describes a cone.

To find the equation in rectangular coordinates, use equation [latex]\varphi=\arccos\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)[/latex].

[latex]\begin{aligned} \frac{5\pi}6&=\arccos\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right) \\ \cos\frac{5\pi}6&=\frac{z}{\sqrt{x^2+y^2+z^2}} \\ -\frac{\sqrt3}2&=\frac{z}{\sqrt{x^2+y^2+z^2}} \\ \frac34&=\frac{z}{\sqrt{x^2+y^2+z^2}} \\ \frac{3x^2}4+\frac{3y^2}4+\frac{3z^2}4&=z^2 \\ \frac{3x^2}4+\frac{3y^2}4-\frac{z^2}4&=0. \end{aligned}[/latex]

This is the equation of a cone centered on the [latex]z[/latex]-axis.
Equation [latex]\rho=6[/latex] describes the set of all points [latex]6[/latex] units away from the origin—a sphere with radius [latex]6[/latex] (Figure 8).

Figure 8. Equation [latex]\rho=6[/latex] describes a sphere with radius [latex]6[/latex].
To identify this surface, convert the equation from spherical to rectangular coordinates, using equations [latex]y=\rho\sin\varphi\sin\theta[/latex] and [latex]\rho^2=x^2+y^2+z^2[/latex]:
[latex]\begin{aligned} \rho&=\sin\theta\sin\varphi \\ \rho^2&=\rho\sin\theta\sin\varphi&\text{Multiply both sides of the equation by }\varphi \\ x^2+y^2+z^2&=y&\text{Substitute rectangular variables using the equations above.} \\ x^2+y^2-y+z^2&=0&\text{Subtract }y\text{ from both sides of the equation.} \\ x^2+y^2-y+\frac14+z^2&=\frac14&\text{Complete the square.} \\ x^2+\left(y-\frac12\right)^2+z^2&=\frac14&\text{Rewrite the middle terms as a perfect square.} \end{aligned}[/latex]

The equation describes a sphere centered at point [latex] (0,\frac{1}{2},0)[/latex] with radius [latex]\frac{1}{2}[/latex].

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Describe the surfaces defined by the following equations.

[latex]\rho=13[/latex]
[latex]\theta=\frac{2\pi}3[/latex]
[latex]\varphi=\frac{\pi}4[/latex]

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Spherical coordinates are useful in analyzing systems that have some degree of symmetry about a point, such as the volume of the space inside a domed stadium or wind speeds in a planet’s atmosphere. A sphere that has Cartesian equation [latex]x^{2}+y^{2}+z^{2}=c^{2}[/latex] has the simple equation [latex]\rho=c[/latex] in spherical coordinates.

In geography, latitude and longitude are used to describe locations on Earth’s surface, as shown in Figure 9. Although the shape of Earth is not a perfect sphere, we use spherical coordinates to communicate the locations of points on Earth. Let’s assume Earth has the shape of a sphere with radius 4000 mi. We express angle measures in degrees rather than radians because latitude and longitude are measured in degrees.

This figure is an image of the Earth. It has the prime meridian labeled, which is a circle on the surface circumnavigating the Earth vertically through the poles. The equator is also labeled which is a horizontal circle circumnavigating the Earth. Three vectors extend out from the center of Earth. Two of them extend to the equator and indicate a measurement of longitude. Two of them extend to a vertical polar circle and indicate a measurement of latitude.

Figure 9. In the latitude–longitude system, angles describe the location of a point on Earth relative to the equator and the prime meridian.

Let the center of Earth be the center of the sphere, with the ray from the center through the North Pole representing the positive [latex]z[/latex]-axis. The prime meridian represents the trace of the surface as it intersects the [latex]xz[/latex]-plane. The equator is the trace of the sphere intersecting the [latex]xy[/latex]-plane.

Example: converting latitude and longitude to spherical coordinates

The latitude of Columbus, Ohio, is [latex]40^{\small\circ}[/latex] N and the longitude is [latex]83^{\small\circ}[/latex] W, which means that Columbus is [latex]40^{\small\circ}[/latex] north of the equator. Imagine a ray from the center of Earth through Columbus and a ray from the center of Earth through the equator directly south of Columbus. The measure of the angle formed by the rays is [latex]40^{\small\circ}[/latex]. In the same way, measuring from the prime meridian, Columbus lies [latex]83^{\small\circ}[/latex] to the west. Express the location of Columbus in spherical coordinates.

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Sydney, Australia is at [latex]34^{\small\circ}[/latex]S and [latex]151^{\small\circ}[/latex]E. Express Sydney’s location in spherical coordinates.

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Cylindrical and spherical coordinates give us the flexibility to select a coordinate system appropriate to the problem at hand. A thoughtful choice of coordinate system can make a problem much easier to solve, whereas a poor choice can lead to unnecessarily complex calculations. In the following example, we examine several different problems and discuss how to select the best coordinate system for each one.

Example: choosing the best coordinate system

In each of the following situations, we determine which coordinate system is most appropriate and describe how we would orient the coordinate axes. There could be more than one right answer for how the axes should be oriented, but we select an orientation that makes sense in the context of the problem. Note: There is not enough information to set up or solve these problems; we simply select the coordinate system (Figure 10).

Find the center of gravity of a bowling ball.
Determine the velocity of a submarine subjected to an ocean current.
Calculate the pressure in a conical water tank.
Find the volume of oil flowing through a pipeline.
Determine the amount of leather required to make a football.

This figure has 5 images. The first image shows bowling balls. The second image is a submarine traveling on an ocean surface. The third image is a traffic cone. The fourth image is a pipeline across some barren land. The fifth image is a football.

Figure 10. (credit: (a) modification of work by scl hua, Wikimedia, (b) modification of work by DVIDSHUB, Flickr, (c) modification of work by Michael Malak, Wikimedia, (d) modification of work by Sean Mack, Wikimedia, (e) modification of work by Elvert Barnes, Flickr)

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Clearly, a bowling ball is a sphere, so spherical coordinates would probably work best here. The origin should be located at the physical center of the ball. There is no obvious choice for how the [latex]x[/latex]-, [latex]y[/latex]– and [latex]z[/latex]-axes should be oriented. Bowling balls normally have a weight block in the center. One possible choice is to align the [latex]z[/latex]-axis with the axis of symmetry of the weight block.
A submarine generally moves in a straight line. There is no rotational or spherical symmetry that applies in this situation, so rectangular coordinates are a good choice. The [latex]z[/latex]-axis should probably point upward. The [latex]x[/latex]– and [latex]y[/latex]-axes could be aligned to point east and north, respectively. The origin should be some convenient physical location, such as the starting position of the submarine or the location of a particular port.
A cone has several kinds of symmetry. In cylindrical coordinates, a cone can be represented by equation [latex]z=kr[/latex], where [latex]k[/latex] is a constant. In spherical coordinates, we have seen that surfaces of the form [latex]\varphi=c[/latex] are half-cones. Last, in rectangular coordinates, elliptic cones are quadric surfaces and can be represented by equations of the form [latex]z^2=\frac{x^2}{a^2}+\frac{y^2}{b^2}[/latex]. In this case, we could choose any of the three. However, the equation for the surface is more complicated in rectangular coordinates than in the other two systems, so we might want to avoid that choice. In addition, we are talking about a water tank, and the depth of the water might come into play at some point in our calculations, so it might be nice to have a component that represents height and depth directly. Based on this reasoning, cylindrical coordinates might be the best choice. Choose the [latex]z[/latex]-axis to align with the axis of the cone. The orientation of the other two axes is arbitrary. The origin should be the bottom point of the cone.
A pipeline is a cylinder, so cylindrical coordinates would be best the best choice. In this case, however, we would likely choose to orient our [latex]z[/latex]-axis with the center axis of the pipeline. The [latex]x[/latex]-axis could be chosen to point straight downward or to some other logical direction. The origin should be chosen based on the problem statement. Note that this puts the [latex]z[/latex]-axis in a horizontal orientation, which is a little different from what we usually do. It may make sense to choose an unusual orientation for the axes if it makes sense for the problem.
A football has rotational symmetry about a central axis, so cylindrical coordinates would work best. The [latex]z[/latex]-axis should align with the axis of the ball. The origin could be the center of the ball or perhaps one of the ends. The position of the [latex]x[/latex]-axis is arbitrary.

Watch the following video to see the worked solution to Example: Choosing the Best Coordinate System.

You can view the transcript for “Ex 2.67” here (opens in new window).

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Which coordinate system is most appropriate for creating a star map, as viewed from Earth (see the following figure)?

This figure is a circle with a star chart in the middle.

Figure 11. Star map as viewed from Earth.

How should we orient the coordinate axes?

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Module 2: Vectors in Space

Learning Objectives

Example: converting from spherical coordinates

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Example: converting from rectangular coordinates

Example: identifying surfaces in the spherical coordinate system

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Example: converting latitude and longitude to spherical coordinates

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Example: choosing the best coordinate system

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