After you are comfortable with the rules for integration, it will become time for you to begin applying those calculus integrals. One way in which you will do this is by finding volumes of figures using the definite integral. Before we begin this article, you must understand that the definite integral is the area under a curve between the axis of integration and the curve.

You ought to also be comfortable with the various calculus formulas for integration. Please review the rules, u-substitution, and integration by parts if you would like.

How it works

As you are likely aware, the definite integral is nothing more than a Riemann sum of a function, using rectangles with infinitely small widths. For instance, the integral:

calculates the area between the x-axis and the function y=x

^{2}by adding together all the infinitely "skinny" rectangles created between the axis and the function. In other words, we are able to calculate the area under the curve by adding together what are virtually line segments.

The principle for finding volumes is quite similar. Let us say we wanted to find the volume of the figure that would be created if we assembled a bunch of squares of height 5 (infinitely "skinny" rectangular prisms) on the space between y=x

^{2}and the axis. It should make sense that, to find the volume, we must:

1. Determine the area of one square, and

2. Add all the areas together

1. Determine the area of one square

The question is: how do we find the area of one square? Area is nothing more than height times width. In this case, our height is certainly 5, but what is our width? It will vary with the function as we go along the x-axis. In fact, the width is the function. So, the area of each square along the interval from 0 to 2 is 5x

^{2}.

However, we are looking for volume. Instead of considering these as squares, let's go back to thinking of them as infinitely "skinny" rectangular prisms. Volume is just area times depth. Depth, in this case, is dx.

2. Add all the areas together

We now have the formula for the area of any square along the interval from 0 to 2 between the x-axis and the function y=x^2: 5x^2 dx. We must now add all these squares together to get a definite volume of the figure created.

This is just a three-dimensional application of the definite integral. We will add the areas of the squares together by using:

Following the integral power rule you are left with:

which, when evaluated from 0 to 2, is:

40/3 - 0 = 40/3

Therefore, we conclude that the volume of the figure created by using squares of height 5 lined up between the x-axis and y=x^2 from 0 to 2 is 40/3.

Conclusion

I hope you now have a decent understanding of how calculus formulas can find volumes of all kinds of figures. With these skills, you are setting yourself up for success in your calculus course. Good luck with your studies!

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