Finding volumes of strange shapes can be very easy if those shapes are
A great many objects we use and engineer are actually manufactured in this way, by "turning" a piece of material on a lathe to make a shape with a round profile in one of its dimensions. Think of a bolt or screw, or a pipe. Here are some other examples of objects with at least some circular symmetry.
The cartoon below illustrates how solid of revolution is generated from a plane curve. The curve f(x) (black) is shown in the left-hand panel. If we revolve f(x) around the y-axis 1/4 of the way around the blue circle, we get the shape in the second panel. Think of it as one fourth of the full "goblet"
shape on the far right. Likewise, rotations of 180˚ and 360˚ give us half a goblet and the full shape. Put a stem on it and it's a nice wine glass. In this section we'll learn how to calculate volumes of figures obtained be this kind of revolution of a 2-D curve.
It's best to illustrate this technique by example, then we'll recap the method at the end.
As a first example , let's calculate the volume of a sphere of radius R. Of course, we already know the answer,
so we'll be able to check our result.
Any sphere can be approximated by a stack of appropriately-sized disks, as shown on the left. The volume of each disk is πr2Δx, where r is the radius of the specific disk and Δx is its height.
Then as we do with calculus, we let the width of each disk approach zero, replacing Δx with the infinitesimal width dx, and summing over an infinite number of disks, to get the integral
There are two crucial steps to the problem. The first is getting the integrand correct, and we're part way there - we have to get A(x). The second is the limits of integration. Here it's pretty easy to see that for a sphere of radius R conveniently centered at (0, 0), the limits are -R to R.
From the geometry shown on the left, we just use the Pythagorean theorem to show that r2 = R2 - x2, the area as a function of x. Note that R is a parameter of this problem. We'd just plug in whatever value of R we need for the sphere we're using.
Now our integral and its solution are reasonably straightforward:
Sometimes it's easier to set up these problems so that we integrate along the y-axis instead of x. One example is the volume enclosed by revolving the parabola y = x2 about the y-axis between x=0 and x=4, as shown on the right.
The disks are horizontal this time, but the geometry is very simple. The radius of each disk is x, but because we're integrating along the y-axis, we express x in terms of y. We actually want x2 (the radius squared) so this couldn't be easier: y = x2
Now the easy integral is (below):
... which is about 25 cubic units.
In this example, we revolve the quadratic function f(x) = (x-3)2 + 1 around the x axis and calculate the volume of the resulting figure between x = 0 and x = 5.
Because this is a volume of revolution, we can place disks of width dx perpendicular to the x-axis, and integrate along x.
The radius of each disk is f(x), so the volume of each is π [f(x)]2or π[(x-3)2 + 1]2. If we integrate between the limits of x=0 and x=5, we get:
Volume of revolution by the disk method
where A(x) is the area of a disk and x is the coordinate along which the integration is performed, and dx is the width of the disk along that coordinate.
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