Applications of Integration: Area and Volume

This section covers:

One very useful application of Integration is finding the area and volume of “curved” figures, that we couldn’t typically get without using Calculus. Since we already know that can use the integral to get the area between the x– and y-axis and a function, we can also get the volume of this figure by rotating the figure around either one of the axes.

Area Between Curves

Since we know how to get the area under a curve here in the Definite Integrals section, we can also get the area between two curves by subtracting the bottom curve from the top curve everywhere where the top curve is higher than the bottom curve. The cool thing about this is it even works if one of the curves is below the x-axis, as long as the higher curve always stays above the lower curve in the integration interval.

Note that we may need to find out where the two curves intersect (and where they intersect the x-axis) to get the limits of integration. And sometimes we have to divide up the integral if the functions cross over each other in the integration interval.

Here is the formal definition of the area between two curves:

Let’s try some problems:

Notice this next problem, where it’s much easier to find the area with respect to y, since we don’t have to divide up the graph.  When we integrate with respect to y, we will have horizontal rectangles (parallel to the x-axis) instead of vertical rectangles (perpendicular to the x-axis), since we’ll use “dy” instead of “dx”.  And if we have the functions in terms of x, we need to use Inverse Functions to get them in terms of y.

Here are more problems where we take the area with respect to y:

Volumes of Solids by Cross Sections

Now that we know how to get areas under and between curves, we can use this method to get the volume of a three-dimensional solid, either with cross sections, or by rotating a curve around a given axis.  Think about it; every day engineers are busy at work trying to figure out how much material they’ll need for certain pieces of metal, for example, and they are using calculus to figure this stuff out!

Let’s first talk about getting the volume of solids by cross-sections of certain shapes.  When doing these problems, think of the bottom of the solid being flat on your horizontal paper, and the 3-D part of it coming up from the paper.  Cross sections might be squares, rectangles, triangles, semi-circles, trapezoids, or other shapes. We’ll have to use some geometry to get these areas.

Cross sections can either be perpendicular to the x-axis or y-axis; in our examples, they will be perpendicular to the x-axis, which is what is we are used to.

So given the cross sectional area A(x) in interval [a, b], and cross sections are perpendicular to the x-axis, the volume of this solid is $$\text{Volume = }\int\limits_{a}^{b}{{A\left( x \right)}}\,dx$$

Here are examples of volumes of cross sections between curves. Slices of the volume are shown to better see how the volume is obtained:

Volumes of Solids: The Disk Method

Now let’s talk about getting a volume by revolving a function or curve around a given axis to obtain a solid of revolution.

Since we know now how to get the area of a region using integration, we can get the volume of a solid by rotating the area around a line, which results in a right cylinder, or disk. (Remember that the formula for the volume of a cylinder is $$\pi {{r}^{2}}\cdot \text{height}$$). Note that the radius is the distance from the axis of revolution to the function, and the “height” of each disk, or slice is “dx”:

Let’s try some problems:

Volumes of Solids: The Washer Method

The washer method is similar to the disk method, but it covers solids of revolution that have “holes”, where we have inner and outer functions, thus inner and outer radii.

So now we have two revolving solids and we basically subtract the area of the inner solid from the area of the outer one.  Note that for this to work, the middle function must be completely inside (or touching) the outer function over the integration interval.

Let’s try some problems:

Volumes of Solids: The Shell Method

The shell method for finding volume of a solid of revolution uses integration along an axis perpendicular to the axis of revolution instead of parallel, as we’ve seen with the disk and washer methods.  The nice thing about the shell method is that you can integrate around the y-axis and not have to take the inverse of functions.  Also, the rotational solid can have a hole in it (or not), so it’s a little more robust.  It’s not intuitive though, since it deals with an infinite number of “surface areas” of rectangles in the shapes of cylinders (shells).

Here are the equations for the shell method:

Since I believe the shell method is no longer required the Calculus AP tests (at least for the AB test), I will not be providing examples and pictures of this method.  Please let me know if you want it discussed further.

Learn these rules and practice, practice, practice!

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On to Integration by Parts — you are ready!