display | more...
Solids of revolution are one of the many applications of integration. A solid of revolution is generated when a region bounded by functions is rotated about an axis. Generally, this will be either the x or y axis, but it can be any axis. The interesting and useful parts about solids of revolution are finding their volumes and surface areas; for example, volume is important for when the displacement method of finding volume is inconvenient, while surface area is important for finding the minimum size of a package for an odd-shaped, fragile object.

There are several methods for finding the volume of a solid of revolution, the three that I currently know of being the disc method, the washer method, and the shell method.

The Disc Method

The disc method is fairly simple, providing you know basic integration. Use this method when there are no "gaps" in your solid; i.e. a semisphere A graphing utility can be extremely helpful, especially for functions that are not easily integrable.

  1. Sketch the graph of the function. This will help you visualize how the solid is to be generated, as well as how to determine the interval of the integral.
  2. Determine where the two functions that bind the region intersect.
  3. Determine the radius function, R(x). R(x) = f(x) - g(x), where f(x) is either the top-most (revolution about the x axis or an axis parallel to the x axis) or right-most (revolution about the y axis or an axis parallel to the y axis), and g(x) is either the bottom-most or left-most.
  4. Determine the interval of the integral. Use the information gained from finding the intersects.
  5. Integrate by using the formula V = pi*Sab(R(x))^2dx. This will give you your volume.

The Washer Method

The washer method is almost identical to the disc method, but it is used for solids with holes in them.

  1. Follow steps 1, 2, 3, and 4.
  2. Along with R(x), you will also need to find r(x), the inner radius of the solid. To find this, assume the solid is bound by f(x), g(x), and h(x). Assume that f(x) and h(x) are the top/right-most and bottom/left-most, respectively. g(x) is a function between f(x) and h(x). r(x) = g(x) - h(x)
  3. Integrate by using the formula V = pi*Sab((R(x))^2-(r(x))^2)dx. You now have the volume of the solid using the washer method.

The Shell Method

The shell method is an alternative to the washer method, which may be easier to calculate at times. It does, however, differ from both the disc method and the washer method.

  1. Sketch the graph of the function.
  2. Determine about which axis you are going to revolve the region, then, draw a small rectangle inside the region parallel to the axis of revolution.
  3. Find the radius of the shell that will be formed when the region is revolved about the axis. To find the radius when revolving a solid about the x axis, subtract the bottom-most curve from the top-most curve. To find the radius when revolving the solid about the y axis, subtract the left-most curve from the right-most curve. When revolving about the x axis, the radius is y; when revolving about the y axis, the radius is x.
  4. Find the height of the rectangle. For solids rotated about a horizontal axis, this is accomplished by subtracting the left-most curve from the right-most curve. For solids rotated about a verticle axis, it is bottom-most from top-most.
  5. Integrate by using the formula V = 2pi*Sab(r(x)*h(x))dx, where r(x) is the radius of the shell, and h(x) is the height of the shell/rectangle. This formula is for solids rotated about verticle axes. Substitute y for x if you are rotating the solid about horizontal axes.

    If you're a math major/math professor/math teacher, please correct me if I have done something horribly wrong. Most of this has been done from memory, and, being that I'm none of the people I have mentioned above, I would not doubt that there is some flaw in my work.

OK, here's the shell method from somebody who aced AP calculus.

Shell method of integrating volumes of solids of revolution

The shell method is useful if y(r) is easier to integrate than r(y) (for instance, if the volume is rotated about the y-axis). Imagine a cylinder made of concentric shells (a scroll comes to mind). The volume of such a shell is

V = π(R2 - r2)h
Approximating the volume by concentric shells:
V = ∑(π((r + Δr)2 - r2)h(r))
= ∑(π(r2 + 2rΔr + Δr2 - r2)h(r))
= ∑(π(2rΔr + Δr^2)h(r))
Taking the limit as Δr goes to 0:
= 2πr h(r) ∂r
which seems intuitive, as 2πr h(r) is the area of such a shell.

Log in or register to write something here or to contact authors.