Multiple Integration

# 34 Triple Integrals in Cylindrical and Spherical Coordinates

### Learning Objectives

- Evaluate a triple integral by changing to cylindrical coordinates.
- Evaluate a triple integral by changing to spherical coordinates.

Earlier in this chapter we showed how to convert a double integral in rectangular coordinates into a double integral in polar coordinates in order to deal more conveniently with problems involving circular symmetry. A similar situation occurs with triple integrals, but here we need to distinguish between cylindrical symmetry and spherical symmetry. In this section we convert triple integrals in rectangular coordinates into a triple integral in either cylindrical or spherical coordinates.

Also recall the chapter opener, which showed the opera house l’Hemisphèric in Valencia, Spain. It has four sections with one of the sections being a theater in a five-story-high sphere (ball) under an oval roof as long as a football field. Inside is an IMAX screen that changes the sphere into a planetarium with a sky full of twinkling stars. Using triple integrals in spherical coordinates, we can find the volumes of different geometric shapes like these.

### Review of Cylindrical Coordinates

As we have seen earlier, in two-dimensional space a point with rectangular coordinates can be identified with in polar coordinates and vice versa, where and are the relationships between the variables.

In three-dimensional space a point with rectangular coordinates can be identified with cylindrical coordinates and vice versa. We can use these same conversion relationships, adding as the vertical distance to the point from the -plane as shown in the following figure.

To convert from rectangular to cylindrical coordinates, we use the conversion and To convert from cylindrical to rectangular coordinates, we use and The -coordinate remains the same in both cases.

In the two-dimensional plane with a rectangular coordinate system, when we say (constant) we mean an unbounded vertical line parallel to the -axis and when (constant) we mean an unbounded horizontal line parallel to the -axis. With the polar coordinate system, when we say (constant), we mean a circle of radius units and when (constant) we mean an infinite ray making an angle with the positive -axis.

Similarly, in three-dimensional space with rectangular coordinates the equations and where and are constants, represent unbounded planes parallel to the -plane, -plane and -plane, respectively. With cylindrical coordinates by and where and are constants, we mean an unbounded vertical cylinder with the -axis as its radial axis; a plane making a constant angle with the -plane; and an unbounded horizontal plane parallel to the -plane, respectively. This means that the circular cylinder in rectangular coordinates can be represented simply as in cylindrical coordinates. (Refer to Cylindrical and Spherical Coordinates for more review.)

### Integration in Cylindrical Coordinates

Triple integrals can often be more readily evaluated by using cylindrical coordinates instead of rectangular coordinates. Some common equations of surfaces in rectangular coordinates along with corresponding equations in cylindrical coordinates are listed in (Figure). These equations will become handy as we proceed with solving problems using triple integrals.

Circular cylinder | Circular cone | Sphere | Paraboloid | |
---|---|---|---|---|

Rectangular | ||||

Cylindrical |

As before, we start with the simplest bounded region in to describe in cylindrical coordinates, in the form of a cylindrical box, ((Figure)). Suppose we divide each interval into subdivisions such that and Then we can state the following definition for a triple integral in cylindrical coordinates.

Consider the cylindrical box (expressed in cylindrical coordinates)

If the function is continuous on and if is any sample point in the cylindrical subbox ((Figure)), then we can define the triple integral in cylindrical coordinates as the limit of a triple Riemann sum, provided the following limit exists:

Note that if is the function in rectangular coordinates and the box is expressed in rectangular coordinates, then the triple integral is equal to the triple integral and we have

As mentioned in the preceding section, all the properties of a double integral work well in triple integrals, whether in rectangular coordinates or cylindrical coordinates. They also hold for iterated integrals. To reiterate, in cylindrical coordinates, Fubini’s theorem takes the following form:

Suppose that is continuous on a rectangular box which when described in cylindrical coordinates looks like

Then and

The iterated integral may be replaced equivalently by any one of the other five iterated integrals obtained by integrating with respect to the three variables in other orders.

Cylindrical coordinate systems work well for solids that are symmetric around an axis, such as cylinders and cones. Let us look at some examples before we define the triple integral in cylindrical coordinates on general cylindrical regions.

Evaluate the triple integral where the cylindrical box is

As stated in Fubini’s theorem, we can write the triple integral as the iterated integral

The evaluation of the iterated integral is straightforward. Each variable in the integral is independent of the others, so we can integrate each variable separately and multiply the results together. This makes the computation much easier:

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Evaluate the triple integral

Follow the same steps as in the previous example.

If the cylindrical region over which we have to integrate is a general solid, we look at the projections onto the coordinate planes. Hence the triple integral of a continuous function over a general solid region in where is the projection of onto the -plane, is

In particular, if then we have

Similar formulas exist for projections onto the other coordinate planes. We can use polar coordinates in those planes if necessary.

Consider the region inside the right circular cylinder with equation bounded below by the -plane and bounded above by the sphere with radius centered at the origin ((Figure)). Set up a triple integral over this region with a function in cylindrical coordinates.

First, identify that the equation for the sphere is We can see that the limits for are from to Then the limits for are from to Finally, the limits for are from to Hence the region is

Therefore, the triple integral is

Consider the region inside the right circular cylinder with equation bounded below by the -plane and bounded above by Set up a triple integral with a function in cylindrical coordinates.

Analyze the region, and draw a sketch.

Let be the region bounded below by the cone and above by the paraboloid ((Figure)). Set up a triple integral in cylindrical coordinates to find the volume of the region, using the following orders of integration:

- The cone is of radius 1 where it meets the paraboloid. Since and (assuming is nonnegative), we have Solving, we have Since we have Therefore So the intersection of these two surfaces is a circle of radius in the plane The cone is the lower bound for and the paraboloid is the upper bound. The projection of the region onto the -plane is the circle of radius centered at the origin.

Thus, we can describe the region as

Hence the integral for the volume is

- We can also write the cone surface as and the paraboloid as The lower bound for is zero, but the upper bound is sometimes the cone and the other times it is the paraboloid. The plane divides the region into two regions. Then the region can be described as

Now the integral for the volume becomes

Redo the previous example with the order of integration

and

Note that is independent of and

Let *E* be the region bounded below by the -plane, above by the sphere and on the sides by the cylinder ((Figure)). Set up a triple integral in cylindrical coordinates to find the volume of the region using the following orders of integration, and in each case find the volume and check that the answers are the same:

- Note that the equation for the sphere is

and the equation for the cylinder is

Thus, we have for the region

Hence the integral for the volume is

- Since the sphere is which is and the cylinder is which is we have that is, Thus we have two regions, since the sphere and the cylinder intersect at in the -plane

and

Hence the integral for the volume is

Redo the previous example with the order of integration

and

A figure can be helpful. Note that is independent of and

### Review of Spherical Coordinates

In three-dimensional space in the spherical coordinate system, we specify a point by its distance from the origin, the polar angle from the positive (same as in the cylindrical coordinate system), and the angle from the positive and the line ((Figure)). Note that and (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.

Recall the relationships that connect rectangular coordinates with spherical coordinates.

From spherical coordinates to rectangular coordinates:

From rectangular coordinates to spherical coordinates:

Other relationships that are important to know for conversions are

and

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

### 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 be continuous in a bounded spherical box, We then divide each interval into subdivisions such that

Now we can illustrate the following theorem for triple integrals in spherical coordinates with being any sample point in the spherical subbox For the volume element of the subbox in spherical coordinates, we have as shown in the following figure.

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

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.

If is continuous on a spherical solid box then

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.

Evaluate the iterated triple integral

As before, in this case the variables in the iterated integral are actually independent of each other and hence we can integrate each piece and multiply:

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 and mean the increments in volume and area, respectively. The variables and are used as the variables for integration to express the integrals.

The triple integral of a continuous function over a general solid region

in where is the projection of onto the -plane, is

In particular, if then we have

Similar formulas occur for projections onto the other coordinate planes.

Set up an integral for the volume of the region bounded by the cone and the hemisphere (see the figure below).

Using the conversion formulas from rectangular coordinates to spherical coordinates, we have:

For the cone: or or or

For the sphere: or or or

Thus, the triple integral for the volume is

Set up a triple integral for the volume of the solid region bounded above by the sphere and bounded below by the cone

Follow the steps of the previous example.

Let be the region bounded below by the cone and above by the sphere ((Figure)). Set up a triple integral in spherical coordinates and find the volume of the region using the following orders of integration:

- Use the conversion formulas to write the equations of the sphere and cone in spherical coordinates.

For the sphere:

For the cone:

Hence the integral for the volume of the solid region becomes

- Consider the -plane. Note that the ranges for and (from part a.) are

The curve meets the line at the point Thus, to change the order of integration, we need to use two pieces:

Hence the integral for the volume of the solid region becomes

In each case, the integration results in

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.

Convert the following integral into cylindrical coordinates:

The ranges of the variables are

The first two inequalities describe the right half of a circle of radius Therefore, the ranges for and are

The limits of are hence

Convert the following integral into spherical coordinates:

The ranges of the variables are

The first two ranges of variables describe a quarter disk in the first quadrant of the -plane. Hence the range for is

The lower bound is the upper half of a cone and the upper bound is the upper half of a sphere. Therefore, we have which is

For the ranges of we need to find where the cone and the sphere intersect, so solve the equation

This gives

Putting this together, we obtain

Use rectangular, cylindrical, and spherical coordinates to set up triple integrals for finding the volume of the region inside the sphere but outside the cylinder

Rectangular:

Cylindrical:

Spherical:

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.

Find the volume of the spherical planetarium in l’Hemisphèric in Valencia, Spain, which is five stories tall and has a radius of approximately ft, using the equation

We calculate the volume of the ball in the first octant, where and using spherical coordinates, and then multiply the result by for symmetry. Since we consider the region as the first octant in the integral, the ranges of the variables are

Therefore,

This exactly matches with what we knew. So for a sphere with a radius of approximately ft, the volume is

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

Find the volume of the ellipsoid

We again use symmetry and evaluate the volume of the ellipsoid using spherical coordinates. As before, we use the first octant and and then multiply the result by

In this case the ranges of the variables are

Also, we need to change the rectangular to spherical coordinates in this way:

Then the volume of the ellipsoid becomes

Find the volume of the space inside the ellipsoid and outside the sphere

This problem is directly related to the l’Hemisphèric structure. The volume of space inside the ellipsoid and outside the sphere might be useful to find the expense of heating or cooling that space. We can use the preceding two examples for the volume of the sphere and ellipsoid and then substract.

First we find the volume of the ellipsoid using and in the result from (Figure). Hence the volume of the ellipsoid is

From (Figure), the volume of the sphere is

Therefore, the volume of the space inside the ellipsoid and outside the sphere is approximately

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 balloons participating each year.

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 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 feet and the radius of the small end of the frustum is feet. A graph of our balloon model and a cross-sectional diagram showing the dimensions are shown in the following figure.

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.

- Find the volume of the balloon in two ways.
- 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.)
- Verify the answer using the formulas for the volume of a sphere, and for the volume of a cone,

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

- 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 seconds. This action affects the temperature in a -foot-wide column feet high, directly above the burner. A cross section of the balloon depicting this column in shown in the following figure.

Assume that after the pilot activates the burner for seconds, the temperature of the air in the column described above*increases*according to the formula

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

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

- Find the average temperature of the air in the balloon after the pilot has activated the burner for seconds.

### Key Concepts

- To evaluate a triple integral in cylindrical coordinates, use the iterated integral

- To evaluate a triple integral in spherical coordinates, use the iterated integral

### Key Equations

**Triple integral in cylindrical coordinates**

**Triple integral in spherical coordinates**

In the following exercises, evaluate the triple integrals over the solid

- Let be a cylindrical shell with inner radius outer radius and height where and Assume that a function defined on can be expressed in cylindrical coordinates as where and are differentiable functions. If and where and are antiderivatives of and respectively, show that

- Use the previous result to show that where is a cylindrical shell with inner radius outer radius and height

- Let be a cylindrical shell with inner radius outer radius and height where and Assume that a function defined on can be expressed in cylindrical coordinates as where are differentiable functions. If where is an antiderivative of show that

where and are antiderivatives of and respectively. - Use the previous result to show that where is a cylindrical shell with inner radius outer radius and height

In the following exercises, the boundaries of the solid are given in cylindrical coordinates.

- Express the region in cylindrical coordinates.
- Convert the integral to cylindrical coordinates.

*E* is bounded by the right circular cylinder the -plane, and the sphere

a. b.

is bounded by the right circular cylinder the -plane, and the sphere

is located in the first octant and is bounded by the circular paraboloid the cylinder and the plane

a. b.

is located in the first octant outside the circular paraboloid and inside the cylinder and is bounded also by the planes and

In the following exercises, the function and region are given.

- Express the region and the function in cylindrical coordinates.
- Convert the integral into cylindrical coordinates and evaluate it.

a. b.

a. b.

In the following exercises, find the volume of the solid whose boundaries are given in rectangular coordinates.

is above the -plane, inside the cylinder and below the plane

is below the plane and inside the paraboloid

is bounded by the circular cone and

is located above the -plane, below outside the one-sheeted hyperboloid and inside the cylinder

is located inside the cylinder and between the circular paraboloids and

is located inside the sphere above the -plane, and inside the circular cone

is located outside the circular cone and between the planes and

is located outside the circular cone above the -plane, below the circular paraboloid, and between the planes

**[T]** Use a computer algebra system (CAS) to graph the solid whose volume is given by the iterated integral in cylindrical coordinates Find the volume of the solid. Round your answer to four decimal places.

**[T]** Use a CAS to graph the solid whose volume is given by the iterated integral in cylindrical coordinates Find the volume of the solid Round your answer to four decimal places.

Convert the integral into an integral in cylindrical coordinates.

Convert the integral into an integral in cylindrical coordinates.

In the following exercises, evaluate the triple integral over the solid

is bounded above by the half-sphere with and below by the cone

is bounded above by the half-sphere with and below by the cone

Show that if is a continuous function on the spherical box then

- A function is said to have spherical symmetry if it depends on the distance to the origin only, that is, it can be expressed in spherical coordinates as where Show that

where is the region between the upper concentric hemispheres of radii and centered at the origin, with and a spherical function defined on - Use the previous result to show that where

- Let be the region between the upper concentric hemispheres of radii
*a*and*b*centered at the origin and situated in the first octant, where Consider*F*a function defined on*B*whose form in spherical coordinates is Show that if and then

where is an antiderivative of and is an antiderivative of - Use the previous result to show that where is the region between the upper concentric hemispheres of radii and centered at the origin and situated in the first octant.

In the following exercises, the function and region are given.

- Express the region and function in cylindrical coordinates.
- Convert the integral into cylindrical coordinates and evaluate it.

a. b.

a. b.

In the following exercises, find the volume of the solid whose boundaries are given in rectangular coordinates.

Use spherical coordinates to find the volume of the solid situated outside the sphere and inside the sphere with

Use spherical coordinates to find the volume of the ball that is situated between the cones

Convert the integral into an integral in spherical coordinates.

Convert the integral into an integral in spherical coordinates.

Convert the integral into an integral in spherical coordinates and evaluate it.

**[T]** Use a CAS to graph the solid whose volume is given by the iterated integral in spherical coordinates Find the volume of the solid. Round your answer to three decimal places.

**[T]** Use a CAS to graph the solid whose volume is given by the iterated integral in spherical coordinates as Find the volume of the solid. Round your answer to three decimal places.

**[T]** Use a CAS to evaluate the integral where lies above the paraboloid and below the plane

**[T]**

- Evaluate the integral where is bounded by the spheres and
- Use a CAS to find an approximation of the previous integral. Round your answer to two decimal places.

Express the volume of the solid inside the sphere and outside the cylinder as triple integrals in cylindrical coordinates and spherical coordinates, respectively.

Express the volume of the solid inside the sphere and outside the cylinder that is located in the first octant as triple integrals in cylindrical coordinates and spherical coordinates, respectively.

The power emitted by an antenna has a power density per unit volume given in spherical coordinates by

where is a constant with units in watts. The total power within a sphere of radius meters is defined as Find the total power

watts

Use the preceding exercise to find the total power within a sphere of radius 5 meters when the power density per unit volume is given by

A charge cloud contained in a sphere of radius *r* centimeters centered at the origin has its charge density given by where The total charge contained in is given by Find the total charge

Use the preceding exercise to find the total charge cloud contained in the unit sphere if the charge density is

### Glossary

- triple integral in cylindrical coordinates
- the limit of a triple Riemann sum, provided the following limit exists:

- triple integral in spherical coordinates
- the limit of a triple Riemann sum, provided the following limit exists: