Shell Method
The shell method is a technique for finding the volumes of solids of revolutions. It considers vertical slices of the region being integrated rather than horizontal ones, so it can greatly simplify certain problems where the vertical slices are more easily described.
Summary
The shell method is a method of finding volumes by decomposing a solid of revolution into cylindrical shells. Consider a region in the plane that is divided into thin vertical strips. If each vertical strip is revolved about the -axis, then the vertical strip generates a disk, as we showed in the disk method. However, if this thin vertical strip is revolved about the -axis, we obtain a different object of revolution, one that looks like a cylindrical shell, or an empty tin can with the top and bottom removed. The resulting volume of the cylindrical shell is the surface area of the cylinder times the thickness of the cylinder wall, or
The shell method calculates the volume of the full solid of revolution by summing the volumes of these thin cylindrical shells as the thickness goes to in the limit:
Notice that the axis of rotation and the variable of integration are on a different axis, i.e., even though we are integrating about the -axis, we are actually rotating about the -axis.
Basic Examples
Use the shell method to find the volume generated by revolving the region bounded by , , , and about the -axis.
rectangle slices
(The shell method corresponds to rotating the red rectangles about the -axis.)
In this case, the radius of the cylindrical shell is and the height is . Since the endpoints of the -interval of the solid are and , the volume of the solid is
Use the shell method to find the volume generated by revolving the region bounded by and about the -axis.
image
(The shell method corresponds to rotating the red rectangles about the -axis.)
These 2 graphs intersect at the points and .
The radius of the cylindrical shell is and the height is . Hence, the volume of the solid is
Using the disk method, we saw how to find the volume of a sphere of radius . Use the shell method to find the volume of a sphere of radius with a vertical hole of radius bored through the center of the sphere.
The cylindrical shell radius is and the cylindrical shell height is Then the volume of the cylindrical shell is
Now, since a vertical hole of radius is bored through the center, the endpoints of the -interval of the solid are and , which gives
When to use the shell method?
It is sometimes difficult to distinguish between using the disk method or the shell method. Many people do not like the shell method, because they do not understand what is happening. As such, they try and use the disk method almost exclusively.
However, there are instances when the shell method is much simpler:
- When the function is rotated around the -axis.
- (rotated around the -axis) When the graph is not a function on , but is a function on .
- When is hard to integrate but is easy to integrate (especially by parts).
Let's look at some examples.
1. When the function is rotated about the -axis:
The shell method will yield a direct answer, but the disk method requires us to figure out how to evaluate the corresponding volume.
Determine the volume of the solid obtained when the region bounded by , the line , and the -axis is rotated about the -axis.
image
(The shell method corresponds to using the red rectangles, while the disk method corresponds to using the blue rectangles.)
By the shell method, as ranges from 0 to 1, the corresponding radius is and the height is . Hence, the volume is
If we wanted to use the disk method, we will take the cylinder obtained by rotating the unit square, and then subtract off the region bounded by and the -axis, rotated about the -axis. This gives us
2. When the graph is not a function on , but is a function on :
The shell method will yield a direct answer, but the disk method requires us to figure out all the corresponding functions.
Determine the volume of the solid obtained when the region bounded by is rotated about the -axis.
image
(The shell method corresponds to using the red rectangles, while the disk method corresponds to using the blue rectangles minus the yellow rectangles.)
When . When . As ranges from 0 to 1, ranges from 1 to 3.
Notice that we have a function in terms of . Using the shell method, as ranges from 1 to 3, the corresponding radius is , and the height of the shell is . Hence, the volume is
If we wanted to use the disk method, we will have to take the volume generated by the "top curve" and then subtract away the volume generated by the "bottom curve." We will have to calculate the functions as follows: , or , so . So, the "top curve" is and the "bottom curve" is .
Then, the volume would be equal to
Using the shell method to find the volume of revolution will allow you to approach problems like this:
Intermediate Examples
The shell method can also be applied to compute volumes of revolution around other axes of rotation.
Use the shell method to find the volume generated by revolving the region bounded by , , and about the line .
Since the region is revolved about the line , we consider cylindrical shells with center axis line . Then the cylindrical shell radius is , the cylindrical shell height is , and the cylindrical shell volume is
Since the endpoints are given by and , we have
As humans haven't been able to reach the sun, we have to work with mathematical models of the interior of the sun based on observations that are made from afar. The standard model of the sun assumes that the density of heated gas ( in grams per cm) throughout the interior follows the best-fit formula
where represents the core, and represents the surface.
Given that the physical radius of the sun is centimeters, what is the total mass of the sun (in kg)?
This problem is part of Calvin's set Fun In Multiple Dimensions.
Hint: We are integrating along the radius. What is the corresponding area element?