This volume can be found with the washer method, which is done using the equation

where the region to be revolved around the x axis has the upper bound of f(x), the lower bound of g(x), and exists on the interval [a, b].  Applying this equation:

This volume can be found using the shell method by implementing the formula

Where the region to be rotated about the y-axis is between f(x) and g(x) on the interval .  

Additionally, one can notice that here, because f(x)=-g(x), the region between the two function is the same as twice the region betwee f(x) and the x-axis, simplifying our problem.  Using this symmetry and the above formula:

In order to evaluate an integral by parts, utilize the equation:

For the integral given, we can define u and dv:

From this, we can evalute du and v:

Combining these elements usng the equation above, we can begin to evaluate the integral:

The final term of this answer needs to be further, again by integration by parts:

Doing the integration:

Substituting this in back for the previous partial solution:

To solve this, first factor out one cosine and write the rest in terms of sine:

From here, we can do a substitution where u=sin(x) and du=cos(x)dx:

Finally, re-substitute sin(x) for u:

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

where 

As such

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

Applying this rule term by term we get

And by the corollary of the Fundamental Theorem of Calculus 

As such the volume is

 units cubed

Since we are revolving a function of  around the y-axis, we will use the method of cylindrical shells to find the volume.

Using the formula for cylindrical shells, we have

.

Find the volume of the solid region swept out by the area bounded by the the functions   and  about the -axis. 

The cross sectional area of the solid can be written as a function of  by first finding the cross-sectional areas swept out by each of the individual functions as they rotate about the x-axis, and then subtracting the inner area. 

Visualize a line drawn perpendicularly to the x-axis at some point meeting the function . As we rotate about the x-axis, we sweep out a circular cross-sectional area with a radius equal to the function value  at the corresponding value of . The area of this disk is therefore:

Similarly for the function , the cross-sectional area is: 

The plot of the two functions helps visualize the geometry. The parabola is clearly the "inner" radius, and therefore we must subtract the areas  from  to find the cross-sectional area of the solid:

Now we must find the limits of integration by finding the intersection points. 

Solve for 

The other solution is . Now simply integrate the cross sectional area over the interval  to find the volume of the solid, 

 To simplify further, write the terms with a common denominator and factor. 

The volume of a solid of revolution for a given function is found using the disk method. Answer is obtained upon evaluating the following integral: