]> Haese & Harris - Chapter 28 Notes

28: Volumes of Revolution

A: Solids of Revolution

There are many shapes that are created by rotating a simple shape in space. It turns out that the volume of such objects can be found using calculus.

Here's the idea: take a small interval along the axis of rotation, and use that to create a rectangle (much like we did when beginning integrals). Now, spin that rectangle around the axis to create a very small cylinder. Knowing the volume of that cylinder, and being able to add up a lot of those cylinders means that we can find the volume of a solid that is generated by rotating a function around an axis.

Let's consider a function (it doesn't matter which—call it $f (x)$ ) from $x = 0$ to $x = 4$ . Here's the graph:

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When this is rotated around the x-axis, the resulting shape is…

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How do we get the volume of this thing? By slicing it up into manageable parts, and taking the limit.

OK, let's consider a really tiny sliver along the x-axis and make a little rectangle…like this:

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If we take this little rectangle, and spin it around the x-axis, we get a little cylinder (really short cylinders are called disks):

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The volume of a cylinder/disk is simple to calculate: $V = π r^{2} h$ . The radius is just the function value ( $f (x)$ ), and the height is tiny (let's call it $Δ x$ ).

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So—the volume of this small cylinder is $π {(f (x))}^{2} Δ x$ . Now, I want to add up the volumes of a bunch of these disks (specifically, from $x = 0$ to $x = 4$ ) and take the limit as $Δ x$ approaches zero. That should give the volume of this object! $V = lim_{Δ x \to 0} \sum_{x = 0}^{4} π {(f (x))}^{2} Δ x$

Now wait just a minute…this looks familiar. Haven't we done this before?

Yes! That looks like an integral!

And it is.

The volume of the object is $V = \int_{0}^{4} π {(f (x))}^{2} d x$ . Of course, you don't always have to start at zero and stop at four…you can use any reasonable limits for the integration.

This example assumes a rotation around the x-axis. If you want to rotate around the y-axis, then you probably want to rewrite the function so that x is a function of y.

Examples

[1.] Find the volume generated by rotating the region bounded by $y = x^{2}$ , $x = 1$ and $y = 0$ about the x-axis.

The region:

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The solid:

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The volume is $\int_{0}^{1} π {(x^{2})}^{2} d x = \int_{0}^{1} π x^{4} d x = \frac{π}{5} x^{5}]_{0}^{1} = \frac{π}{5}$ .

[2.] Find the volume generated by rotating the region bounded by $x = y - y^{2}$ and $x = 0$ about the y-axis.

The region:

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The solid:

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The volume is $\int_{0}^{1} π (y - y^{2}) d y = π \int_{0}^{1} (y^{2} - 2 y^{3} + y^{4}) d y = π {(\frac{1}{3} y^{3} - \frac{1}{2} y^{4} + \frac{1}{5} y^{5})}_{0}^{1} = π (\frac{1}{3} - \frac{1}{2} + \frac{1}{5}) = \frac{π}{30}$ .

[3.] Find the volume generated by rotating the region bounded by $\frac{1}{\sqrt{x^{2} + 1}}$ , $x = 0$ , $x = 1$ and $y = 0$ about the x-axis.

The region:

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The solid:

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The volume is $\int_{0}^{1} π {(\frac{1}{\sqrt{x^{2} + 1}})}^{2} d x = π \int_{0}^{1} \frac{1}{x^{2} + 1} d x = π \cdot {tan}^{- 1} (x)]_{0}^{1} = π ({tan}^{- 1} (1) - {tan}^{- 1} (0)) = \frac{π^{2}}{4}$ .

B: Volumes for Two Defining Functions

Sometimes, the area that you want to rotate doesn't come out solid—it has a hole of some type. This arises when the area that is rotated is the area between two functions, rather than the area beneath just one function.

Consider these functions:

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When that very unusual area is rotated, it looks like this:

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(it is difficult to see, perhaps, that this thing is empty inside…try!)

We can still consider a very small section along the axis to create a rectangle. However, when this is rotated around the axis, you get what most would recognize as a ring (often called a washer)!

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The volume of this ring is not difficult to find…

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It's just the volume of the disk with the larger radius, minus the volume of the disk with the smaller radius. $V = π {(f (x))}^{2} Δ x - π {(g (x))}^{2} Δ x = π [{(f (x))}^{2} - {(g (x))}^{2}] Δ x$

Again, we want to sum these up and take the limit as $Δ x$ approaches zero…and again, this results in an integral!

$V = \int_{a}^{b} π [{(f (x))}^{2} - {(g (x))}^{2}] d x$

The other way that a solid can be defined by two functions is that the boundary isn't constant. Consider this area between $y = sin (x)$ , $y = cos (x)$ , and $y = 0$ :

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(I can't get my computer to draw a picture of the solid)

This must be separated into regions if you want to rotate it about the x-axis…unless you want to learn about a technique called shells…

Examples

[4.] Find the volume generated by rotating the region bounded by $x = y^{2}$ and $y = x^{2}$ about the x-axis.

The region:

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The solid:

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The volume is $\int_{0}^{1} π ({(\sqrt{x})}^{2} - {(x^{2})}^{2}) d x = π \int_{0}^{1} (x - x^{4}) d x = π {(\frac{1}{2} x^{2} - \frac{1}{5} x^{5})}_{0}^{1} = π (\frac{1}{2} - \frac{1}{5}) = \frac{3 π}{10}$ .

[5.] Find the volume generated by rotating the region bounded by $y = x$ and $y = \sqrt{x}$ about the y-axis.

The region:

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The solid:

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The volume is $\int_{0}^{1} π (y - y^{2}) d y = π \int_{0}^{1} (y - y^{2}) d y = π {(\frac{1}{2} y^{2} - \frac{1}{3} y^{3})}_{0}^{1} = π (\frac{1}{2} - \frac{1}{3}) = \frac{π}{6}$ .

[6.] Find the volume generated by rotating the region bounded by $y = sec (x)$ , $x + y = 4$ , $x = - \frac{π}{4}$ and $x = \frac{π}{4}$ about the x-axis.

The region:

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The solid:

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The volume is $\int_{- \frac{π}{4}}^{\frac{π}{4}} π ({(sec (x))}^{2} - {(4 - x)}^{2}) d x = π \int_{- \frac{π}{4}}^{\frac{π}{4}} ({sec}^{2} (x) - (16 - 8 x + x^{2})) d x$ . That looks pretty ugly…I'll let my calculator work this one. The volume is 73.7.

Extra: Cross Sections

In general, if we know the cross sectional area of a solid at any point along an axis (typically the x-axis), then we can find the volume of the object by integrating the area function along the axis.

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If, for any value of x in the interval $[a, b]$ the cross sectional area $A (x)$ is known (or can be found), then the volume of the object is $\int_{a}^{b} A (x) d x$ .

For many problems, we will know a function that traces the base of the solid, and it will be up to us to use a little geometry to find the cross sectional area. For example, consider the area between the y-axis, $y = 4$ and $y = x^{2}$ .

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Now, let's tip that over into three dimensions before we create a solid.

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To create the solid, let's construct a square for each value of x.

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All that remains is to write and evaluate the integral. The volume is $\int_{0}^{2} {(4 - x^{2})}^{2} d x$ , which works out to 17.1.

Examples

[7.] The base of a solid is the region enclosed by $y = x^{2}$ and $y = 1$ . Cross sections parallel to the y-axis are equilateral triangles. Find the volume of the solid.

The region:

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The solid:

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The volume is $\int_{0}^{1} \frac{\sqrt{3}}{4} {(2 \sqrt{y})}^{2} d y = \sqrt{3} \int_{0}^{1} y d y = \sqrt{3} {(\frac{1}{2} y^{2})}_{0}^{1} = \frac{\sqrt{3}}{2}$ .

[8.] The base of a solid is the triangular region with vertices $(0, 0)$ , $(3, 0)$ and $(0, 2)$ . Cross sections perpendicular to the x-axis are semicircles. Find the volume of the solid.

The region:

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The solid:

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The volume is $\int_{0}^{2} \frac{π}{2} {(\frac{1}{2} (- \frac{2}{3} x + 2))}^{2} d x$ . Again, since this looks quite nasty, I'll allow my calculator to do the dirty work, giving a result of 1.51.

Extra Ideas

What if you wanted to rotate a region around some line other than the x- or y-axis?

How can you use integration to find the length of a curve?

What if you wanted to find the surface area of a solid generated by rotating a region?

Page last validated 2010-08-15