The first step is to find the lower and upper bounds, the points where the two functions intersect:

Knowing these, the solid generated will have the following volume:

The disc method can be used to find the volume of this solid of revolution. To use the disc method, we must set up and evaluate an integral of the form

where a and b are the endpoints of the interval, and R(x) is the distance between the function and the rotation axis. 

First find R(x):

Next set up and evaluate the integral:

Simplifying, we find that the volume of the solid is

This problem can be solved using the Disk Method and the equation 

.

Using our equation to formulate this equation we get the following.

Applying the power rule of integrals which states 

we get,

.

Looking at the graph, we see that the curves intersect at the point  and  is on top.

Using the general equation 

 where  is on top.

This gives us that the volume is  

The volume can be solved by 

 where  and  from 0 to 1 where the curves intersect.

Then, we formulate 

We take the region and think about revolving rectangles of height  and width  about the line . This forms discs of volume . We then sum all of these discs on the interval and take the limit as the number of discs on the interval approaches infinity to arrive at the definite integral 

Recall that volume is the definite integral of surface area over a given region. 

Given our surface area formula, we can determine the volume in one step. 

To solve this problem, treat the function

As the radius of a disc with infinitesimal thickness . This disc would then have a volume defined the surface area:

And a thickness or height:

The volume of the solid generated would be the sum of these discs, i.e. the integral:

Since the functions are being rotated around the y-axis, they should be rewritten. It may help to write them in terms of y:

And

Here, function designations  have been used solely to eliminate the ambiguity of using x for both functions.

Now, there are two points of intersection for these functions:

These will be the bounds for the integration to follow.

Finally, treat these two functions as the radii for two regions: a disk from the larger radius, and a hole from the smaller radius, allowing for the formation of a ring with a cross-sectional area:

These disks have infinitesimal thickness , and so the volume of the solid can be found by adding all of these disks together; i.e. by taking the integral:

The formula for the volume of the region revolved around the x-axis on the interval  is given as

where .

As such,

 .

When taking the integral, we will use the inverse power rule which states,

 .

Applying this rule we get

 .

And by the corollary of the First Fundamental Theorem of Calculus

.

As such, the volume is

 cubic units.