Chapter 8 Applications of Integration VolumesofSolidsofRevolution
Solidsof
Revolution
Rotationabout:
x‐axis
y‐axis
Disk
Method
Washer
Method(1)
2
CylindricalShell
Method(2)
2
or
2
2
2
or
or
2
AreaCrossSection
Method(4)
or
2
Differenceof
ShellsMethod(2)(3)
2
Notes: 1. The Washer Method is an extension of the Disk Method. 2.
is the radius of the cylindrical shell. In cases where there is a gap between the axis of revolution and the functions being revolved, is the distance between the axis of revolution and either or , as appropriate. 3. The Difference of Shells Method is an extension of the Cylindrical Shell Method. 4. The function is the area of the cross section being integrated. Chapter 8 Applications of Integration DiskandWasherMethods
The formulas for the Disk Method and Washer Method for calculating volumes of revolution are provided above. Below, we present an approach that can be used to calculate volumes of revolution using these methods. Under the Disk Method, we integrate the area of the region between a curve and its axis of revolution to obtain volume. Since each cross‐section of the resulting object will be a circle, we use the formula as our starting point. The resulting formula is: or
The Washer Method is simply a dual application of the Disk Method. Consider the illustration at right. If we want the area of the shaded region, we subtract the area of the smaller circle from the area of the larger circle. The same occurs with the Washer Method; since we integrate cross‐sectional area to find volume, so to obtain the volume of revolution of a region between the two curves we integrate the difference in the areas between the two curves. Below is a set of steps that can be used to determine the volume of revolution of a region between two curves. The approach is illustrated based on the following example: Example: Find the volume that results from revolving the region between the curves and about the line 2√ 6. Steps 1. Graph the equations provided and any other information given in the problem (illustrated below). Then, isolate the section of the graph that we want to work with (illustrated at right). The disks we will use are shown as green and orange vertical lines. The dashed objects are reflections of the curves and disks over the axis of revolution; these give us an idea of what the central cross‐section of the 3 shape will look like after revolution. You do not need to draw these. Integration Interval Chapter 8 Applications of Integration 2. Identify whether there is a gap between the region to be revolved and the axis of revolution. In the example, the axis of revolution is 6, so there is clearly a gap between a) the red and blue curves, and b) the axis of revolution. Therefore, we will use the Washer Method. 3. Set up the integral form to be used. radius
a. Disk Method: radius
or smallradius
bigradius
b. Washer Method: bigradius
smallradius
or 4. Identify the variable of integration (i.e., are we using be perpendicular to the axis of revolution. or ?). The disks used must a. If we are revolving around an axis, use the variable of that axis. b. If the axis of revolution is a line of the form, or , use the opposite variable from the one that occurs in the equation of the axis. In the example, the axis of revolution is 6, so we will integrate with respect to . Note: The expressions used in the integration must be in terms of the variable of integration. So, for example, if the variable of integration is and the equation of a curve is given as , we must invert this to the form before integrating. 5. Identify the limits of integration. In the example, the curves intersect at 4. This results in an equation for volume in the form: bigradius
smallradius
0 and 6. Substitute the expressions for the big and small radii inside the integral. In the example, we have the following: a. bigradius
b. smallradius
6
6
2√ This results in the following: √
~
.
Note that this matches the value calculated using the Difference of Shells Method below. Chapter 8 Applications of Integration CylindricalShellMethods
The formulas for the Cylindrical Shell Method and Difference of Shells Method for calculating volumes of revolution are provided above. Below, we present an approach that can be used to calculate volumes of revolution using these methods. Under the Cylindrical Shell Method, we integrate the volume of a shell across the appropriate values of or . We use the formula for the volume of a cylinder as our starting point (i.e., 2
, where is typically the function provided). The resulting formula is: 2
or
2
The Difference of Shells Method is essentially a dual application of the Cylindrical Shell Method. We want the volume of the cylinder whose height is the difference between two functions (see illustration at right). Below is a set of steps that can be used to determine the volume of revolution of a region between two curves. The approach is illustrated based on the following example: Example: Find the volume that results from revolving the region between the curves and about the line 2√ 6. Steps 1. Graph the equations provided and any other information given in the problem (illustrated below left). Then, isolate the section of the graph that we want to work with (illustrated below right). Also shown are reflections of the curves over the axis of revolution (dashed curves); this allows us to see the “other side” of the cylindrical shells we will use. A typical shell is shown as a green cylinder. Integration Interval Chapter 8 Applications of Integration 2. Identify whether the integration involves one or two curves. a. One curve: Use the Cylindrical Shell Method. b. Two curves: Use the Difference of Shells Method. This is the case in the example. 3. Set up the integral form to be used. Let be the radius of the shell. 2
a. Cylindrical Shell Method: 2
b. Difference of Shells Method: 2
2
or . differenceofshellheights
differenceofshellheights
or . 4. Identify the variable of integration (i.e., are we using be parallel to the axis of revolution. or ?). The shells used must a. If we are revolving around an axis, use the opposite variable of that axis. b. If the axis of revolution is a line of the form, or , use the same variable as the one that occurs in the equation of the axis. In the example, the axis of revolution is 6, so we will integrate with respect to . 2
differenceofshellheights
5. Identify the limits of integration. In the example, the curves intersect at 4. This results in an equation for volume in the form: 2
differenceofshellheights 0 and 6. Substitute the expressions for and the difference of shell heights into the integral. In the example, we need to convert each equation to the form because is the variable of integration: a.
2
so 2√ so The difference of shell heights, then, is 2
1
4
2
. b. The radius of a shell is the difference between the line in the interval, so the radius is 6
. 6 and the value of This results in the following: ~
.
Note that this matches the value calculated using the Washer Method above.
Chapter 8 Applications of Integration AreaCross‐SectionMethod
Some problems require us to determine volume of a solid using its base and cross‐sectional area across that base. These are not problems based on revolution of a shape, so we use a more basic formula (that does not involve π): or
Below is a set of steps that can be used to determine volume for this type of problem. The approach is illustrated using the following example: Example: Find the volume of a solid with a base of 2√sin over the interval 0, if the cross‐sections perpendicular to the ‐axis are equilateral triangles whose bases stretch from the ‐axis to the curve. Steps 1. Graph the curve of the base over the interval specified. 2. Determine the variable of integration. This will always be the variable whose axis is perpendicular to the cross‐sections specified. In the example, the variable of integration is . 3. Determine the limits of integration. This is typically the interval provided in the problem. In the example, this is the interval 0, . 4. Draw the cross‐section you are provided in the problem. In the example, we are working with equilateral triangles with base equal to the function 2√sin . 5. Determine the area of the cross‐section in terms of the appropriate variable. We need the area of an equilateral triangle for this example. This area can be developed from basic principles using the illustration at right, or from the formula: In the example: √
√
√
, where is the length of the base of the triangle. 2√sin
√3 sin 6. Integrate the area of the cross‐section using the limits determined in Step 3. √3sin √3cos
0
√ ~ .