# How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region #y = 1 + x^2#, #y = 0#, #x = 0#, #x = 2# rotated about the x-axis?

##### 1 Answer

See explanation below.

#### Explanation:

Here is the region to be revolved about the

If we must use shells, we will need to take our representative slices horizontally and use thickness

We need to rewrite the boundaries as functions of

Note that, in the region

For the 'height' of the shell (which is lying on its side), we will use the

For

(We could do an integral for this, but we learned the volume of a cylinder in geometry class.)

From

The radius of the representative shell is

We integrate:

Adding the two volumes, we get:

**Note**

If we had been allowed to use disks, we would integrate:

# = pi [x+(2x^3)/3 + x^5/5]_0^2#

# = pi [2+16/3+32/5] = pi [(30+80+96)/15] = 206/15 pi#

Of course, we get the same answer, but we only need to do one calculation and the integral is (I think) simpler.