# Rotational Volume Using Shell Method

• Michele Nunes
In summary, the conversation discusses using the shell method to find the volume of a solid generated by revolving a plane region around the y-axis, with equations y = 4 - x2 and y = 0. The individual is confused about why the integral 2π∫[from -2 to 2] (4x-x3)dx yields 0 when plugged into a calculator, and questions whether rotating the region from 0 to 2 and -2 to 2 would give the same result. The expert explains that the wrong answer is obtained because the shell method was applied incorrectly, and suggests using a = 0 as the lower limit of integration.
Michele Nunes

## Homework Statement

Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis.
y = 4 - x2
y = 0

## The Attempt at a Solution

Okay I understand that the region is symmetric about the y-axis, however I still don't understand why the integral 2π∫[from -2 to 2] (4x-x3)dx comes out to be 0 when I plug it into my calculator. I know that you can just do the integral from 0 to 2 and then multiply the whole thing by 2 and it comes out correctly but why does doing it from -2 to 2 come out as 0? Aren't both methods trying to calculate the same thing? Why do both get different answers? Wouldn't rotating just the region from 0 to 2 about the y-axis and rotating the region from -2 to 2 about the y-axis give you the same cylindrical solid?

Michele Nunes said:

## Homework Statement

Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis.
y = 4 - x2
y = 0

## The Attempt at a Solution

Okay I understand that the region is symmetric about the y-axis, however I still don't understand why the integral 2π∫[from -2 to 2] (4x-x3)dx comes out to be 0 when I plug it into my calculator. I know that you can just do the integral from 0 to 2 and then multiply the whole thing by 2 and it comes out correctly but why does doing it from -2 to 2 come out as 0? Aren't both methods trying to calculate the same thing? Why do both get different answers? Wouldn't rotating just the region from 0 to 2 about the y-axis and rotating the region from -2 to 2 about the y-axis give you the same cylindrical solid?
You get the wrong answer for the volume using

$$V=\int_{-2}^2(4-x^2) ⋅ xdx$$

because you have applied the shell method incorrectly. Since the region y = 4x - x2 is symmetrical about the axis of rotation, in this case the y-axis, you should have made the lower limit of integration a = 0 instead of a = -2.

http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithCylinder.aspx

## 1. What is the shell method and how is it used in calculating rotational volume?

The shell method is a mathematical technique used to calculate the volume of a solid of revolution, or a three-dimensional object formed by rotating a two-dimensional shape around an axis. It involves slicing the solid into thin cylindrical shells and using the formula V = 2πrh to find the volume of each shell, then adding up the volumes of all the shells to get the total volume of the solid.

## 2. What are the differences between the shell method and the disk method?

The disk method and the shell method are both used to calculate rotational volumes, but they differ in the shape of the slices used. The disk method involves slicing the solid into thin circular disks, while the shell method slices the solid into thin cylindrical shells. In general, the shell method is more useful for solids with holes or irregular shapes, while the disk method is more suitable for solids with simple, circular cross-sections.

## 3. How do you determine the limits of integration for the shell method?

The limits of integration for the shell method depend on the axis of rotation and the shape of the solid. For a solid rotated around the x-axis, the limits will be in terms of x, and for a solid rotated around the y-axis, the limits will be in terms of y. To determine the limits, you can use the intersection points of the solid with the axis of rotation as the endpoints of your integral.

## 4. Can the shell method be used for solids with varying cross-sectional areas?

Yes, the shell method can be used for solids with varying cross-sectional areas. In this case, the radius (r) in the formula V = 2πrh will change as you move along the axis of rotation, but the height (h) will remain constant. The integral for the shell method will then involve a function for the radius, and the limits of integration will depend on the intersection points of the solid with the axis of rotation.

## 5. Are there any other methods for calculating rotational volume?

Yes, besides the shell method and the disk method, there are other methods for calculating rotational volume, such as the washer method and the cylindrical shell method. The washer method is similar to the disk method, but it takes into account solids with holes or voids. The cylindrical shell method is similar to the shell method, but it involves slicing the solid into thin rectangular strips instead of cylindrical shells. The method you choose will depend on the shape of the solid and the axis of rotation.

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