Multiple Integration

# 32 Double Integrals in Polar Coordinates

### Learning Objectives

• Recognize the format of a double integral over a polar rectangular region.
• Evaluate a double integral in polar coordinates by using an iterated integral.
• Recognize the format of a double integral over a general polar region.
• Use double integrals in polar coordinates to calculate areas and volumes.

Double integrals are sometimes much easier to evaluate if we change rectangular coordinates to polar coordinates. However, before we describe how to make this change, we need to establish the concept of a double integral in a polar rectangular region.

### Polar Rectangular Regions of Integration

When we defined the double integral for a continuous function in rectangular coordinates—say, over a region in the -plane—we divided into subrectangles with sides parallel to the coordinate axes. These sides have either constant -values and/or constant -values. In polar coordinates, the shape we work with is a polar rectangle, whose sides have constant -values and/or constant -values. This means we can describe a polar rectangle as in (Figure)(a), with

In this section, we are looking to integrate over polar rectangles. Consider a function over a polar rectangle We divide the interval into subintervals of length and divide the interval into subintervals of width This means that the circles and rays for and divide the polar rectangle into smaller polar subrectangles ((Figure)(b)).

(a) A polar rectangle (b) divided into subrectangles (c) Close-up of a subrectangle.

As before, we need to find the area of the polar subrectangle and the “polar” volume of the thin box above Recall that, in a circle of radius the length of an arc subtended by a central angle of radians is Notice that the polar rectangle looks a lot like a trapezoid with parallel sides and and with a width Hence the area of the polar subrectangle is

Simplifying and letting we have Therefore, the polar volume of the thin box above ((Figure)) is

Finding the volume of the thin box above polar rectangle

Using the same idea for all the subrectangles and summing the volumes of the rectangular boxes, we obtain a double Riemann sum as

As we have seen before, we obtain a better approximation to the polar volume of the solid above the region when we let and become larger. Hence, we define the polar volume as the limit of the double Riemann sum,

This becomes the expression for the double integral.

Definition

The double integral of the function over the polar rectangular region in the -plane is defined as

Again, just as in Double Integrals over Rectangular Regions, the double integral over a polar rectangular region can be expressed as an iterated integral in polar coordinates. Hence,

Notice that the expression for is replaced by when working in polar coordinates. Another way to look at the polar double integral is to change the double integral in rectangular coordinates by substitution. When the function is given in terms of and using changes it to

Note that all the properties listed in Double Integrals over Rectangular Regions for the double integral in rectangular coordinates hold true for the double integral in polar coordinates as well, so we can use them without hesitation.

Sketching a Polar Rectangular Region

Sketch the polar rectangular region

As we can see from (Figure), and are circles of radius and covers the entire top half of the plane. Hence the region looks like a semicircular band.

The polar region lies between two semicircles.

Now that we have sketched a polar rectangular region, let us demonstrate how to evaluate a double integral over this region by using polar coordinates.

Evaluating a Double Integral over a Polar Rectangular Region

Evaluate the integral over the region

First we sketch a figure similar to (Figure) but with outer radius From the figure we can see that we have

Sketch the region and evaluate

Hint

Evaluating a Double Integral by Converting from Rectangular Coordinates

Evaluate the integral where is the unit circle on the -plane.

The region is a unit circle, so we can describe it as

Using the conversion and we have

Evaluating a Double Integral by Converting from Rectangular Coordinates

Evaluate the integral where

We can see that is an annular region that can be converted to polar coordinates and described as (see the following graph).

The annular region of integration

Hence, using the conversion and we have

Evaluate the integral where is the circle of radius on the -plane.

Hint

Follow the steps in the previous example.

### General Polar Regions of Integration

To evaluate the double integral of a continuous function by iterated integrals over general polar regions, we consider two types of regions, analogous to Type I and Type II as discussed for rectangular coordinates in Double Integrals over General Regions. It is more common to write polar equations as than so we describe a general polar region as (see the following figure).

A general polar region between and

Double Integrals over General Polar Regions

If is continuous on a general polar region as described above, then

Evaluating a Double Integral over a General Polar Region

Evaluate the integral where is the region bounded by the polar axis and the upper half of the cardioid

We can describe the region as as shown in the following figure.

The region is the top half of a cardioid.

Hence, we have

Evaluate the integral

Hint

Graph the region and follow the steps in the previous example.

### Polar Areas and Volumes

As in rectangular coordinates, if a solid is bounded by the surface as well as by the surfaces and we can find the volume of by double integration, as

If the base of the solid can be described as then the double integral for the volume becomes

We illustrate this idea with some examples.

Finding a Volume Using a Double Integral

Find the volume of the solid that lies under the paraboloid and above the unit circle on the -plane (see the following figure).

The paraboloid .

By the method of double integration, we can see that the volume is the iterated integral of the form where

This integration was shown before in (Figure), so the volume is cubic units.

Finding a Volume Using Double Integration

Find the volume of the solid that lies under the paraboloid and above the disk on the -plane. See the paraboloid in (Figure) intersecting the cylinder above the -plane.

Finding the volume of a solid with a paraboloid cap and a circular base.

First change the disk to polar coordinates. Expanding the square term, we have Then simplify to get which in polar coordinates becomes and then either or Similarly, the equation of the paraboloid changes to Therefore we can describe the disk on the -plane as the region

Hence the volume of the solid bounded above by the paraboloid and below by is

Notice in the next example that integration is not always easy with polar coordinates. Complexity of integration depends on the function and also on the region over which we need to perform the integration. If the region has a more natural expression in polar coordinates or if has a simpler antiderivative in polar coordinates, then the change in polar coordinates is appropriate; otherwise, use rectangular coordinates.

Finding a Volume Using a Double Integral

Find the volume of the region that lies under the paraboloid and above the triangle enclosed by the lines and in the -plane ((Figure)).

First examine the region over which we need to set up the double integral and the accompanying paraboloid.

Finding the volume of a solid under a paraboloid and above a given triangle.

The region is Converting the lines and in the -plane to functions of and we have and respectively. Graphing the region on the -plane, we see that it looks like Now converting the equation of the surface gives Therefore, the volume of the solid is given by the double integral

As you can see, this integral is very complicated. So, we can instead evaluate this double integral in rectangular coordinates as

Evaluating gives

To answer the question of how the formulas for the volumes of different standard solids such as a sphere, a cone, or a cylinder are found, we want to demonstrate an example and find the volume of an arbitrary cone.

Finding a Volume Using a Double Integral

Use polar coordinates to find the volume inside the cone and above the

The region for the integration is the base of the cone, which appears to be a circle on the (see the following figure).

Finding the volume of a solid inside the cone and above the -plane.

We find the equation of the circle by setting

This means the radius of the circle is so for the integration we have and Substituting and in the equation we have Therefore, the volume of the cone is

cubic units.

Analysis

Note that if we were to find the volume of an arbitrary cone with radius units and height units, then the equation of the cone would be

We can still use (Figure) and set up the integral as

Evaluating the integral, we get

Use polar coordinates to find an iterated integral for finding the volume of the solid enclosed by the paraboloids and

cubic units

Hint

Sketching the graphs can help.

As with rectangular coordinates, we can also use polar coordinates to find areas of certain regions using a double integral. As before, we need to understand the region whose area we want to compute. Sketching a graph and identifying the region can be helpful to realize the limits of integration. Generally, the area formula in double integration will look like

Finding an Area Using a Double Integral in Polar Coordinates

Evaluate the area bounded by the curve

Sketching the graph of the function reveals that it is a polar rose with eight petals (see the following figure).

Finding the area of a polar rose with eight petals.

Using symmetry, we can see that we need to find the area of one petal and then multiply it by Notice that the values of for which the graph passes through the origin are the zeros of the function and these are odd multiples of Thus, one of the petals corresponds to the values of in the interval Therefore, the area bounded by the curve is

Finding Area Between Two Polar Curves

Find the area enclosed by the circle and the cardioid

First and foremost, sketch the graphs of the region ((Figure)).

Finding the area enclosed by both a circle and a cardioid.

We can from see the symmetry of the graph that we need to find the points of intersection. Setting the two equations equal to each other gives

One of the points of intersection is The area above the polar axis consists of two parts, with one part defined by the cardioid from to and the other part defined by the circle from to By symmetry, the total area is twice the area above the polar axis. Thus, we have

Evaluating each piece separately, we find that the area is

Find the area enclosed inside the cardioid and outside the cardioid

Hint

Sketch the graph, and solve for the points of intersection.

Evaluating an Improper Double Integral in Polar Coordinates

Evaluate the integral

This is an improper integral because we are integrating over an unbounded region In polar coordinates, the entire plane can be seen as

Using the changes of variables from rectangular coordinates to polar coordinates, we have

Evaluate the integral

Hint

Convert to the polar coordinate system.

### Key Concepts

• To apply a double integral to a situation with circular symmetry, it is often convenient to use a double integral in polar coordinates. We can apply these double integrals over a polar rectangular region or a general polar region, using an iterated integral similar to those used with rectangular double integrals.
• The area in polar coordinates becomes
• Use and to convert an integral in rectangular coordinates to an integral in polar coordinates.
• Use and to convert an integral in polar coordinates to an integral in rectangular coordinates, if needed.
• To find the volume in polar coordinates bounded above by a surface over a region on the -plane, use a double integral in polar coordinates.

### Key Equations

• Double integral over a polar rectangular region
• Double integral over a general polar region

In the following exercises, express the region in polar coordinates.

is the region of the disk of radius centered at the origin that lies in the first quadrant.

is the region between the circles of radius and radius centered at the origin that lies in the second quadrant.

is the region bounded by the -axis and

is the region bounded by the -axis and

In the following exercises, the graph of the polar rectangular region is given. Express in polar coordinates.

In the following graph, the region is situated below and is bounded by and

In the following graph, the region is bounded by and

In the following exercises, evaluate the double integral over the polar rectangular region

where

where

where

where

In the following exercises, the integrals have been converted to polar coordinates. Verify that the identities are true and choose the easiest way to evaluate the integrals, in rectangular or polar coordinates.

In the following exercises, convert the integrals to polar coordinates and evaluate them.

Evaluate the integral where is the region bounded by the polar axis and the upper half of the cardioid

Find the area of the region bounded by the polar axis and the upper half of the cardioid

Evaluate the integral where is the region bounded by the part of the four-leaved rose situated in the first quadrant (see the following figure).

Find the total area of the region enclosed by the four-leaved rose (see the figure in the previous exercise).

Find the area of the region which is the region bounded by and

Find the area of the region which is the region inside the disk and to the right of the line

Determine the average value of the function over the region bounded by the polar curve where (see the following graph).

Determine the average value of the function over the region bounded by the polar curve where (see the following graph).

Find the volume of the solid situated in the first octant and bounded by the paraboloid and the planes and

Find the volume of the solid bounded by the paraboloid and the plane

1. Find the volume of the solid bounded by the cylinder and the planes and
2. Find the volume of the solid outside the double cone inside the cylinder and above the plane
3. Find the volume of the solid inside the cone and below the plane by subtracting the volumes of the solids and
1. Find the volume of the solid inside the unit sphere and above the plane
2. Find the volume of the solid inside the double cone and above the plane
3. Find the volume of the solid outside the double cone and inside the sphere

a. b. c.

For the following two exercises, consider a spherical ring, which is a sphere with a cylindrical hole cut so that the axis of the cylinder passes through the center of the sphere (see the following figure).

If the sphere has radius and the cylinder has radius find the volume of the spherical ring.

A cylindrical hole of diameter cm is bored through a sphere of radius cm such that the axis of the cylinder passes through the center of the sphere. Find the volume of the resulting spherical ring.

Find the volume of the solid that lies under the double cone inside the cylinder and above the plane

Find the volume of the solid that lies under the paraboloid inside the cylinder and above the plane

Find the volume of the solid that lies under the plane and above the disk

Find the volume of the solid that lies under the plane and above the unit disk

A radial function is a function whose value at each point depends only on the distance between that point and the origin of the system of coordinates; that is, where Show that if is a continuous radial function, then where and with and

Use the information from the preceding exercise to calculate the integral where is the unit disk.

Let be a continuous radial function defined on the annular region where and is a differentiable function. Show that

Apply the preceding exercise to calculate the integral where is the annular region between the circles of radii and situated in the third quadrant.

Let be a continuous function that can be expressed in polar coordinates as a function of only; that is, where with and Show that where is an antiderivative of

Apply the preceding exercise to calculate the integral where

Let be a continuous function that can be expressed in polar coordinates as a function of only; that is, where with and Show that where and are antiderivatives of and respectively.

Evaluate where

A spherical cap is the region of a sphere that lies above or below a given plane.

1. Show that the volume of the spherical cap in the figure below is
2. A spherical segment is the solid defined by intersecting a sphere with two parallel planes. If the distance between the planes is show that the volume of the spherical segment in the figure below is

In statistics, the joint density for two independent, normally distributed events with a mean and a standard distribution is defined by Consider the Cartesian coordinates of a ball in the resting position after it was released from a position on the z-axis toward the -plane. Assume that the coordinates of the ball are independently normally distributed with a mean and a standard deviation of (in feet). The probability that the ball will stop no more than feet from the origin is given by where is the disk of radius a centered at the origin. Show that

The double improper integral may be defined as the limit value of the double integrals over disks of radii a centered at the origin, as a increases without bound; that is,

1. Use polar coordinates to show that
2. Show that by using the relation

### Glossary

polar rectangle
the region enclosed between the circles and and the angles and it is described as