# Subtract the Hole from the Whole.

Sometimes I think textbooks are too rigorous. Behind every Riemann sum is a definite integral. So, authors routinely show how to solve an application of integration problem by developing the method starting from the Riemann sum and proceeding to an integral that give the result that is summarized in a “formula.” There is nothing wrong with that except that often the formula is all the students remember and are lost when faced with a similar situation that the formula does not handle. .

Volume of solid figure problems are developed from the idea that if a solid figure has a regular cross-section (that is, when cut perpendicular to a line, each face is similar – in the technical sense – to all the others). They are all squares, or equilateral triangles, or whatever. The last shape considered is usually a “washer”, that is, an annulus or two concentric circles. This is formed by revolving the region between two curves around a line. Authors develop a formula for such volumes: $\displaystyle \pi \int_{a}^{b}{{{\left( R\left( x \right) \right)}^{2}}-{{\left( r\left( x \right) \right)}^{2}}dx}$.

Now there is nothing wrong with that, but I like to give the students their chance to show off. They can usually figure out the answer without Riemann sums. Here is my suggestion. After students have had some practice with circular cross-sections (“Disk” method”) I give them a series of three volumes to find.

Example 1: The curve $f\left( x \right)=\sin \left( \pi x \right)$ on the interval [0, ½] is revolved around the x-axis to form a solid figure. Find the volume of this figure. Solution: $\displaystyle V=\int_{0}^{1}{\pi {{\left( \sin \left( \pi x \right) \right)}^{2}}dx}=\frac{\pi }{4}$

Example 2: The curve $g\left( x \right)=8{{x}^{3}}$ on the interval [0, ½] is revolved around the x-axis to form a solid figure. Find the volume of this figure. Solution: $\displaystyle V=\int_{0}^{1/2}{\pi {{\left( 8{{x}^{3}} \right)}^{2}}dx=}\frac{\pi }{14}$

These they find easy. Then, leaving the first two examples in plain view, I give them:

Example 3: The region in the first quadrant between the graphs of $f\left( x \right)=\sin \left( \pi x \right)$ and $g\left( x \right)=8{{x}^{3}}$ is revolved around the x-axis. Find the volume of the resulting figure. A little thinking and (rarely) a hint and they have it. $\displaystyle V=\frac{\pi }{4}-\frac{\pi }{14}$

What did the do? Easy, they subtracted the hole from the whole. We discuss this and why they think it is correct. We try one or two others. And now they are set to do any “washer” method problem without another formula to memorize.

Extensions:

1. In symbols, when rotation around a horizontal line, if R(x) is the distance from the curve farthest from the line of rotation and r(x) the distance from the closer curve to the line of rotation the result can be summarized in the formula $\displaystyle V = \int_{a}^{b}{\pi {{\left( R\left( x \right) \right)}^{2}}dx}-\int_{a}^{b}{\pi {{\left( r\left( x \right) \right)}^{2}}dx}$.

Notice, that I like to keep the $\pi$ inside the integral sign so that each integrand looks like the formula for the area of a circle. What the students need to know is to subtract the volume hole from the outside volume. With that                idea and the disk method they can do any volume by washers problem.

2. You should show the students how this equation above can be rearranged into the formula in their books, $\displaystyle V = \pi \int_{a}^{b}{{{\left( R\left( x \right) \right)}^{2}}-{{\left( r\left( x \right) \right)}^{2}}dx}$.

This is so that they understand that the formulas are the same, and not think you’ve forgotten to tell them something important. It is also a good exercise in working with the notation. (see MPAC 5 – Notational fluency)

3. Next discuss what ${{\left( \pi R\left( x \right) \right)}^{2}}-\pi {{\left( r\left( x \right) \right)}^{2}}$ is the area of and how it relates to this problem. See if the students can understand what the textbook is doing; what shape the book is using.. Discuss the Riemann sum approach. (MPAC 1 Reasoning with definitions and theorems, and MPAC 5 Notational fluency)

4. With the idea of subtracting the “hole” try a problem like this. Example 4: The region in the first quadrant between x-axis and the graphs of $f\left( x \right)=\sqrt{x}$ and $g\left( x \right)=\sqrt{2x-4}$ is revolved around the x-axis. Find the volume of the resulting figure. Solution: $\displaystyle V=\int_{0}^{4}{\pi {{\left( \sqrt{x} \right)}^{2}}dx}-\int_{2}^{4}{\pi {{\left( \sqrt{2x-4} \right)}^{2}}dx}=4\pi$

(Notice the limits of integration.)

Traditionally, this is done by the method of cylindrical shells, but you don’t need that. You could divide the region into two parts with a vertical line at x = 2 and use disks on the left and washers on the right, but you don’t need to do that either. Just subtract the hole from the whole.

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