7.3 #30 part (c): Use the disk or the shell method to find the volume ...

7.3 #30 part (c): Use the disk or the shell method to find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line

Solution: I had made a mistake on what the inner radius of the washer should be. Refer to the drawing: The drawing I had done in class was the same, but for The distance from the line to the function is actually

. So the integral should be:

I had just put

. That was the mistake.

To verify, I did it with the shell method as well. In this case, two parts are needed. The function and the line

intersect at

, and putting the function in terms of

gives

. The lower

cylindrical shell can be found without integration, using the formula for the volume of a cylinder, so the first part is

. Notice that this is the volume of one large

cylinder with radius 10 and height 4 minus a smaller cylinder with radius other part (above

), the radius of each shell is

and height 4. For the

and the height of each shell is

(right function minus left function). So the volume of this part is

Adding this to the volume of the other part gives

The disk method is possible as well. For this to work, you would need to find the volume of the solid generated by revolving the region bounded by the function

and the lines

, and

. You would also need to find the volume of the cylinder that would result from revolving the rectangular area below (and above the -axis) between and around the line . Then, you would subtract the first volume from the volume of the cylinder. The volume of the cylinder is . The volume of the other region is

Subtracting this volume from the volume of the cylinder gives