Triple Integrals in Spherical Coordinates

Learning Objectives

  • Evaluate a triple integral by changing to spherical coordinates.

Review of Spherical Coordinates

In three-dimensional space [latex]{\mathbb{R}^{3}}[/latex] in the spherical coordinate system, we specify a point [latex]P[/latex] by its distance [latex]{\rho}[/latex] from the origin, the polar angle [latex]{\theta}[/latex] from the positive [latex]x[/latex]-axis (same as in the cylindrical coordinate system), and the angle [latex]{\varphi}[/latex] from the positive [latex]z[/latex]-axis and the line [latex]OP[/latex] (Figure 1). Note that [latex]{\rho} \ {\geq} \ {0}[/latex] and [latex]{0} \ {\leq} \ {\varphi} \ {\leq} \ {\pi}[/latex]. (Refer to Cylindrical and Spherical Coordinates for a review.) Spherical coordinates are useful for triple integrals over regions that are symmetric with respect to the origin.

A depiction of the spherical coordinate system: a point (x, y, z) is shown, which is equal to (rho, theta, phi) in spherical coordinates. Rho serves as the spherical radius, theta serves as the angle from the x axis in the xy plane, and phi serves as the angle from the z axis.

Figure 1. The spherical coordinate system locates points with two angles and a distance from the origin.

Recall the relationships that connect rectangular coordinates with spherical coordinates.

From spherical coordinates to rectangular coordinates:

[latex]{{x} = {\rho} \ {\sin} \ {\varphi} \ {\cos} \ {\theta}}, {{y} = {\rho} \ {\sin} \ {\varphi} \ {\sin} \ {\theta}}, \ {\text{and}} \ {{z} = {\rho} \ {\cos} \ {\varphi}}.[/latex]

From rectangular coordinates to spherical coordinates:

[latex]{\rho}^{2} = {x^2} + {y^2} + {z^2}, \ {\tan}{\theta} = {\frac{y}{x}}, {\varphi} = {\arccos}{\left ( {\frac{z}{\sqrt{{x^2} + {y^2} + {z^2}}}} \right )}.[/latex]

Other relationships that are important to know for conversions are

  • [latex]{r} = {\rho} \ {\sin} \ {\varphi}[/latex]
  • [latex]{\theta} = {\theta}[/latex]                                                     These equations are used to convert from spherical coordinates to cylindrical coordinates
  • [latex]{z} = {\rho} \ {\cos} \ {\varphi}[/latex]

and

  • [latex]{\rho} = {\sqrt{{r^2}+{z^2}}}[/latex]
  • [latex]{\theta} = {\theta}[/latex]                                                     These equations are used to convert from cylindrical coordinates to spherical coordinates.
  • [latex]{\varphi} = {\arccos}{\left ( {\frac{z}{\sqrt{{r^2}+{z^2}}}} \right )}[/latex]

The following figure shows a few solid regions that are convenient to express in spherical coordinates.

This figure consists of four figures. In the first, a sphere is shown with the note Sphere rho = c (constant). In the second, a half plane is drawn from the z axis with the note Half plane theta = c (constant). In the last two figures, a half cone is drawn in each with the note Half cone phi = c (constant). In the first of these, the cone opens up and it is marked 0 < c < pi/2. In the second of these, the cone opens down and it is marked pi/2 < c < pi.

Figure 2. Spherical coordinates are especially convenient for working with solids bounded by these types of surfaces. (The letter [latex]c[/latex] indicates a constant.)

Integration in Spherical Coordinates

We now establish a triple integral in the spherical coordinate system, as we did before in the cylindrical coordinate system. Let the function [latex]{f}{({\rho},{\theta},{\varphi})}[/latex] be continuous in a bounded spherical box, [latex]{B} = {\left \{ {({\rho},{\theta},{\varphi})}{\mid}{a} \ {\leq} \ {\rho} \ {\leq} \ {{b},{\alpha}} \ {\leq} \ {\theta} \ {\leq} \ {{\beta},{\gamma}} \ {\leq} \ {\varphi} \ {\leq} \ {\psi} \right \}}[/latex]. We then divide each interval into [latex]l[/latex], [latex]m[/latex], and [latex]n[/latex] subdivisions such that [latex]{{\Delta}{\rho} = {\frac{{b}-{a}}{l}}}, \ {{\Delta}{\theta} = {\frac{{\beta}-{\alpha}}{m}}}, \ {{\Delta}{\varphi} = {\frac{{\psi}-{\gamma}}{n}}}[/latex].

Now we can illustrate the following theorem for triple integrals in spherical coordinates with [latex]{\left ( {{\rho}^{*}_{ijk}}, {{\theta}^{*}_{ijk}}, {{\varphi}^{*}_{ijk}} \right )}[/latex] being any sample point in the spherical subbox [latex]{B}_{ijk}[/latex]. For the volume element of the subbox [latex]{\Delta}{V}[/latex] in spherical coordinates, we have [latex]{\Delta}{V} = {({\Delta}{\rho})}{({\rho}{\Delta}{\varphi})}{({\rho} \ {\sin} \ {\varphi}{\Delta}{\theta})}[/latex], as shown in the following figure.

In the spherical coordinate space, a box is projected onto the polar coordinate plane. On the polar coordinate plane, the projection has area rho sin phi Delta theta. On the z axis, a distance Delta rho is indicated, and from these boundaries, angles are made that project through the edges of the box. There is also a blown up version of the box that shows it has sides Delta rho, rho Delta phi, and rho sin phi Delta theta, with overall volume Delta V = rho squared sin phi Delta rho Delta phi Delta theta.

Figure 3. The volume element of a box in spherical coordinates.

definition


The triple integral in spherical coordinates is the limit of a triple Riemann sum,

[latex]\displaystyle\lim_{l,m,n\to\infty}\displaystyle\sum^l_{i=1}\displaystyle\sum^m_{j=1}\displaystyle\sum^n_{k=1}f(\rho^*_{ijk},\theta^*_{ijk},\varphi^*_{ijk})(\rho^*_{ijk})^2\sin\varphi\Delta\rho\Delta\theta\Delta\varphi[/latex]

provided the limit exists.

As with the other multiple integrals we have examined, all the properties work similarly for a triple integral in the spherical coordinate system, and so do the iterated integrals. Fubini’s theorem takes the following form.

theorem: fubini’s theorem for spherical coordinates

If [latex]{f}{({\rho},{\theta},{\varphi})}[/latex] is continuous on a spherical solid box [latex]{B} = {[{a},{b}]} \ {\times} \ {[{\alpha},{\beta}]} \ {\times} {[{\gamma},{\psi}]}[/latex], then

[latex]\underset{B}{\displaystyle\iiint}{f}{({\rho},{\theta},{\varphi})} \ {\rho}^{2}{\sin} \ {\varphi} \ {{d}{\rho}} \ {{d}{\varphi}} \ {{d}{\theta}} = {\displaystyle\int^{{\varphi}={\psi}}_{{\varphi}={\gamma}}} \ {\displaystyle\int^{{\theta}={\beta}}_{{\theta}={\alpha}}} \ {\displaystyle\int^{{\rho}={b}}_{{\rho}={a}}} \ {f}{({\rho},{\theta},{\varphi})} \ {\rho}^{2}{\sin} \ {\varphi} \ {{d}{\rho}} \ {{d}{\varphi}} \ {{d}{\theta}}.[/latex]

This iterated integral may be replaced by other iterated integrals by integrating with respect to the three variables in other orders.

As stated before, spherical coordinate systems work well for solids that are symmetric around a point, such as spheres and cones. Let us look at some examples before we consider triple integrals in spherical coordinates on general spherical regions.

Example: evaluating a triple integral in spherical coordinates

Evaluate the iterated triple integral

[latex]{\displaystyle\int^{{\theta}={{2}{\pi}}}_{{\theta}={0}}} \ {\displaystyle\int^{{\varphi}={{\pi}/{2}}}_{{\varphi}={0}}} \ {\displaystyle\int^{{\rho}={1}}_{{p}={0}}} \ {\rho}^{2}{\sin} \ {\varphi} \ {{d}{\rho}} \ {{d}{\varphi}} \ {{d}{\theta}}[/latex].

The concept of triple integration in spherical coordinates can be extended to integration over a general solid, using the projections onto the coordinate planes. Note that [latex]dV[/latex] and [latex]dA[/latex] mean the increments in volume and area, respectively. The variables [latex]V[/latex] and [latex]A[/latex] are used as the variables for integration to express the integrals.

The triple integral of a continuous function [latex]{f}{({\rho},{\theta},{\varphi})}[/latex] over a general solid region

[latex]{E} = {\left \{ {({\rho},{\theta},{\varphi})}{\mid}{({\rho},{\theta})} \ {\in} \ {{D},{{u_1} \ {({\rho},{\theta}}})} \ {\leq} \ {\varphi} \ {\leq} \ {{u_2} \ {({\rho},{\theta})}} \right \}}[/latex]

in [latex]{\mathbb{R}}^{3}[/latex], where [latex]D[/latex] is the projection of [latex]E[/latex] onto the [latex]{\rho}{\theta}[/latex]-plane, is

[latex]\underset{E}{\displaystyle\iiint}{f}{({\rho},{\theta},{\varphi})}{dV} = {\iint\limits_{D}}{\left [ \ {\int\limits^{{u_2}{({\rho},{\theta})}}_{{u_1}{({\rho},{\theta})}}} \ {f}{({\rho},{\theta},{\varphi})}{d}{\varphi} \right ]}{{d}{A}}.[/latex]

In particular, [latex]{D} = {\left \{ {({\rho},{\theta})}{\mid}{g_1} \ {(\theta)} \ {\leq} \ {\rho} \ {\leq} \ {{g_2} \ {(\theta)},{\alpha}} \ {\leq} \ {\theta} \ {\leq} \ {\beta} \right \}}[/latex], then we have

[latex]\underset{E}{\displaystyle\iiint}{f}{({\rho},{\theta},{\varphi})}{dV} = {\displaystyle\int^{\beta}_{\alpha}} \ {\displaystyle\int^{{g_2}{({\theta})}}_{{g_1}{({\theta})}}} \ {\displaystyle\int^{{u_2}{({\rho},{\theta})}}_{{u_1}{({\rho},{\theta})}}} \ {f}{({\rho},{\theta},{\varphi})}{{\rho}^{2}}{\sin} \ {\varphi} \ {{d}{\varphi}} \ {{d}{\rho}} \ {{d}{\theta}}.[/latex]

Similar formulas occur for projections onto the other coordinate planes.

Example: setting up a triple integral in spherical coordinates

Set up an integral for the volume of the region bounded by the cone [latex]{z} = {\sqrt{{3}{({x^2}+{y^2})}}}[/latex] and the hemisphere [latex]{z} = {\sqrt{{4}-{x^2}-{y^2}}}[/latex] (see the figure below).

A hemisphere with equation z = the square root of (4 minus x squared minus y squared) in the upper half plane, and within it, a cone with equation z = the square root of (3 times (x squared + y squared)) that is pointing down, with vertex at the origin.

Figure 4. A region bounded below by a cone and above by a hemisphere.

try it

Set up a triple integral for the volume of the solid region bounded above by the sphere [latex]{\rho} = {2}[/latex] and bounded below by the cone [latex]{\varphi} = {{\pi}/{3}}[/latex].

Example: interchanging order of integration in spherical coordinates

Let [latex]E[/latex] be the region bounded below by the cone [latex]{z} = {\sqrt{{x^2}+{y^2}}}[/latex] and above by the sphere [latex]{z} = {x^2} + {y^2} + {z^2}[/latex] (Figure 5). Set up a triple integral in spherical coordinates and find the volume of the region using the following orders of integration:

  1. [latex]{{d}{\rho}} \ {{d}{\phi}} \ {{d}{\theta}}[/latex],
  2. [latex]{{d}{\varphi}} \ {{d}{\rho}} \ {{d}{\theta}}[/latex].
A sphere with equation z = x squared + y squared + z squared, and within it, a cone with equation z = the square root of (x squared + y squared) that is pointing down, with vertex at the origin.

Figure 5. A region bounded below by a cone and above by a sphere.

Before we end this section, we present a couple of examples that can illustrate the conversion from rectangular coordinates to cylindrical coordinates and from rectangular coordinates to spherical coordinates.

Example: converting from rectangular coordinates to cylindrical coordinates

Convert the following integral into cylindrical coordinates:

[latex]{\displaystyle\int^{{y} = {1}}_{{y} = {-1}}}\displaystyle\int^{{x} = {\sqrt{{1} - {{y}^{2}}}}}_{{x} = {0}} \ {\displaystyle\int^{{z} = {\sqrt{{x^2}+{y^2}}}}_{{z} = {{x}^{2}}+{{y}^{2}}}} \ {{x}{y}{z}} \ {{d}{z}} \ {{d}{x}} \ {{d}{y}}.[/latex]

Example: converting from rectangular coordinates to spherical coordinates

Convert the following integral into spherical coordinates:

[latex]{\displaystyle\int^{{y} = {3}}_{{y} = {0}}}\displaystyle\int^{{x} = {\sqrt{{9} - {{y}^{2}}}}}_{{x} = {0}}\displaystyle\int^{{z} = {\sqrt{{18} - {{x}^{2}} - {{y}^{2}}}}}_{{z} = {\sqrt{{x}^{2} + {{y}^{2}}}}} \ {({{x}^{2}} + {{y}^{2}} + {{z}^{2}})} \ {{d}{z}} \ {{d}{x}} \ {{d}{y}}.[/latex]

try it

Use rectangular, cylindrical, and spherical coordinates to set up triple integrals for finding the volume of the region inside the sphere [latex]x^2+y^2+z^2=4[/latex] but outside the cylinder [latex]x^2+y^2=1[/latex].

A sphere with equation x squared + y squared + z squared = 4, and within it, a cylinder with equation x squared + y squared = 1.

Figure 6.

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 5.32” here (opens in new window).

Now that we are familiar with the spherical coordinate system, let’s find the volume of some known geometric figures, such as spheres and ellipsoids.

Example: chapter opener: finding the volume of L’hemisphÈric

Find the volume of the spherical planetarium in l’Hemisphèric in Valencia, Spain, which is five stories tall and has a radius of approximately [latex]50[/latex] ft, using the equation [latex]x^2+y^2+z^2=r^2[/latex].

A picture of l’Hemisphèric, which is a giant glass structure that is in the shape of an ellipsoid.

Figure 7. (credit: modification of work by Javier Yaya Tur, Wikimedia Commons)

For the next example we find the volume of an ellipsoid.

Example: finding the volume of an ellipsoid

Find the volume of the ellipsoid [latex]\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1[/latex].

Example: finding the volume of the space inside an ellipsoid and outside a sphere

Find the volume of the space inside the ellipsoid [latex]\frac{x^2}{75^2}+\frac{y^2}{80^2}+\frac{z^2}{90^2}=1[/latex] and outside the sphere [latex]x^2+y^2+z^2=50^2[/latex].

Activity: hot air balloons

Hot air ballooning is a relaxing, peaceful pastime that many people enjoy. Many balloonist gatherings take place around the world, such as the Albuquerque International Balloon Fiesta. The Albuquerque event is the largest hot air balloon festival in the world, with over [latex]500[/latex] balloons participating each year.

A picture of many hot air balloons.

Figure 8. Balloons lift off at the 2001 Albuquerque International Balloon Fiesta. (credit: David Herrera, Flickr)

As the name implies, hot air balloons use hot air to generate lift. (Hot air is less dense than cooler air, so the balloon floats as long as the hot air stays hot.) The heat is generated by a propane burner suspended below the opening of the basket. Once the balloon takes off, the pilot controls the altitude of the balloon, either by using the burner to heat the air and ascend or by using a vent near the top of the balloon to release heated air and descend. The pilot has very little control over where the balloon goes, however—balloons are at the mercy of the winds. The uncertainty over where we will end up is one of the reasons balloonists are attracted to the sport.

In this project we use triple integrals to learn more about hot air balloons. We model the balloon in two pieces. The top of the balloon is modeled by a half sphere of radius [latex]28[/latex] feet. The bottom of the balloon is modeled by a frustum of a cone (think of an ice cream cone with the pointy end cut off). The radius of the large end of the frustum is [latex]28[/latex] feet and the radius of the small end of the frustum is [latex]6[/latex] feet. A graph of our balloon model and a cross-sectional diagram showing the dimensions are shown in the following figure.

This figure consists of two parts, a and b. Figure a shows a representation of a hot air balloon in xyz space as a half sphere on top of a frustrum of a cone. Figure b shows the dimensions, namely, the radius of the half sphere is 28 ft, the distance from the bottom to the top of the frustrum is 44 ft, and the diameter of the circle at the top of the frustrum is 12 ft.

Figure 9. (a) Use a half sphere to model the top part of the balloon and a frustum of a cone to model the bottom part of the balloon. (b) A cross section of the balloon showing its dimensions.

We first want to find the volume of the balloon. If we look at the top part and the bottom part of the balloon separately, we see that they are geometric solids with known volume formulas. However, it is still worthwhile to set up and evaluate the integrals we would need to find the volume. If we calculate the volume using integration, we can use the known volume formulas to check our answers. This will help ensure that we have the integrals set up correctly for the later, more complicated stages of the project.

  1. Find the volume of the balloon in two ways.
    1. Use triple integrals to calculate the volume. Consider each part of the balloon separately. (Consider using spherical coordinates for the top part and cylindrical coordinates for the bottom part.)
    2. Verify the answer using the formulas for the volume of a sphere, [latex]V=\frac43\pi{r}^3[/latex], and for the volume of a cone, [latex]V=\frac13\pi{r}^2h[/latex].

    In reality, calculating the temperature at a point inside the balloon is a tremendously complicated endeavor. In fact, an entire branch of physics (thermodynamics) is devoted to studying heat and temperature. For the purposes of this project, however, we are going to make some simplifying assumptions about how temperature varies from point to point within the balloon. Assume that just prior to liftoff, the temperature (in degrees Fahrenheit) of the air inside the balloon varies according to the function

    [latex]\large{T_0(r,\theta,z)=\frac{z-r}{10}+210}[/latex]

  2. What is the average temperature of the air in the balloon just prior to liftoff? (Again, look at each part of the balloon separately, and do not forget to convert the function into spherical coordinates when looking at the top part of the balloon.)
    Now the pilot activates the burner for [latex]10[/latex] seconds. This action affects the temperature in a [latex]12[/latex]-foot-wide column [latex]20[/latex] feet high, directly above the burner. A cross section of the balloon depicting this column in shown in the following figure.

    This figure shows the dimensions of the balloon and the hot air, namely, the radius of the half sphere is 28 ft, the distance from the bottom to the top of the frustrum is 44 ft, the diameter of the circle at the top of the frustrum is 12 ft, and the inner column of hot air has height 20 ft and diameter 12 ft.

    Figure 10. Activating the burner heats the air in a [latex]20[/latex]-foot-high, [latex]12[/latex]-foot-wide column directly above the burner.

    Assume that after the pilot activates the burner for [latex]10[/latex] seconds, the temperature of the air in the column described above increases according to the formula

    [latex]\large{H(r,\theta,z)=-2z-48.}[/latex]


    Then the temperature of the air in the column is given by

    [latex]\large{T_1(r,\theta,z)=\frac{z-r}{10}+210+(-2z-48)}[/latex],


    while the temperature in the remainder of the balloon is still given by

    [latex]\large{T_0(r,\theta,z)=\frac{z-r}{10}+210}[/latex].

  3. Find the average temperature of the air in the balloon after the pilot has activated the burner for [latex]10[/latex] seconds.