To end this series of posts on visualizing solid figures, we will look at a problem that relates to how volumes of solid figures are formed. It has 5 parts which will be presented first. Then the solution will be given.

Consider the region in the first quadrant bounded by the graphs of the parabola and the line both for .

This region is revolved around the *y*-axis to form a solid figure.

- Use the disk/washer method to find the volume of this figure.
- Use the method of cylindrical shells to find the volume of this figure.
- Use the disk/washer method to find a number
*j*, such that when*x*=*j*the volume of the figure is one-half that of the original figure. - Use the method of cylindrical shells to find a number
*k*, such that when*x*=*k*the volume of the figure is one-half that of the original figure. - The answers to parts a) and b) should be the same, but the answers to parts c) and d) are different. Explain why they are different.

**Solutions:**

1. The volume by the disk/washer method is

2. The volume by the method of cylindrical shells is

3. The value is be found by solving the equation for *j*:

, so

4. The value is be found by solving the equation for *k*:

and

5. The reason the values are not the same is this. Think of the revolved parabola as a bowl. If you pour water into the bowl until it is half full, the bowl looks like the figure below. The water is pooled in the bottom of the bowl as you would expect. This is what happens when using the disk/washer method. The washers stack up starting in the bottom of the bowl until it is half full.

On the other hand, the method of cylindrical shells sort of wraps the water in layers (the shells) around the *y*-axis. Picture the water being sprayed on the *y*-axes and frozen there. Each new layer (shell) increases the amount and you end up half of the total volume arranged as a cylinder with a rounded (paraboloid shaped) bottom as shown in the figure below. Both bowls contain the same amount of water, arranged differently.