Application of Integrals: Finding the Volume


Elements of Volume


When we were working on Riemann sums to approximate integrals, we took small "elements of area", in this case, rectangles, and summed them all together to get the total area under the curve. A similar idea is used when trying to calculate the volume of solids.

Instead of taking small elements of area like a rectangle, we need to map out a small element of volume, which is equal to the cross-sectional area multiplied by our interval Δx. (That's right, just like a Riemann sum.) Then we'll sum them all up to find the total volume of the solid.


There are several ways to come up with this element of volume, but only two are really important to this class:

The "Disc" Method

In this method, the volume is defined as a "disc", or a cross-section of the volume. Obviously, this works best with shapes where the cross-sectional area is very easily defined, such as a circle.

This method can be used with any shape in which the cross-section can be simply and uniformly defined along an axis, but the primary case in which the disc method is used is with shapes that the cross-section will always be the shape of a circle, such as a volume of revolution. This volume is created when the area under the curve of some function f(x) is revolved around either the x or y axis, creating a volume.

Example 1: f(x) = √x, 1 ≤ x ≤4


As you can see in the above example, the area under the curve of √x from [1,4] is revolved around the x-axis, creating a volume with a circular cross-section. So, let's go ahead and solve this problem using the disc method.

First, let's establish what the disc method is all about: Cut a cross-section of the volume in such a way that it gives you a nice circle shape (in this example, the cutting plane is perpendicular to the x-axis), come up with the equation for the area A(x) of that cross-section, and then integrate that area along the axis perpendicular to the cross-section (in this case, its the x-axis).

The area of a circle is A = πr2. So, what is r? In this case, r is the value of the function f(x), which is √x. So, the area of the cross-section is A(x) = π(√x)2 = πx.

Finally, the volume of the solid is equal to:



The "Shell" Method

The shell method is a method that can be used when the disc method becomes difficult to use, such as with very complex shapes. In certain cases, it makes your life much easier.

However, there are tighter restrictions on when you can use the "shell" method:
  1. It can only be used if the shape is SYMMETRICAL around some axis.
  2. It can only be used if the shape is CIRCULAR around the axis perpendicular to its symmetry.
For instance, a cylinder is a great candidate for the shell method. It is symmetrical from the side, and it is circular around the axis perpendicular to its side (the top). Additionally, volumes of revolution always fulfill these requirements. (Why?)

The shell method is similar to the disc method in that we're defining an element of volume and then summing them up with a Riemann sum (an integral). But, instead of discs, we're defining thin, cylindrical shells.

Here's an animation showing how these shells are used to sum up the volume of a revolved volume:
Copyright Lila F. Roberts, Georgia College & State University

The shells are analogous to the rectangles from normal Riemann sums. The surface area of the shell is multiplied by the interval Δx. The surface area of a cylinder is equal to its circumference times its height, or AC = 2πrh. Then the shells are summed from the innermost radius to the outermost radius.


Example 2: What is the volume of the solid created when the area between the curves f(x) = √x and g(x) = x2 from 0 ≤ x ≤ 1 is revolved around the y-axis?
Copyright Lila F. Roberts, Georgia College & State University

As the animation shows, the small little area creates a "bowl"-shaped solid. Let's determine the element of volume using the shell method.

First, let's establish that we're going to map these shells starting from the inside (x = 0) to the outside (x = 1), so the radius r is just the value of x (r = x). Now, we need to determine the height of each cylinder, h. It's not as easy as just saying that it's the value of the function f(x), because this area is bounded by two functions: f(x) = √x on top, and g(x) = x2 on bottom. So, the height is going to be the difference between these functions. h = f(x) - g(x) = √x - x2.

Now let's write our equation out: