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.

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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*S_a^b(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*S_a^b((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*S_a^b(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.

(Calculus:)

A solid of revolution is a solid figure made by rotating a curve about an axis, in a way reminiscent of the way a lathe works. The general method for determining the volume of these things was worked out several centuries ago by Riemann, who is generally credited with the development of the subfield of mathematics that this sort of thing falls under: differential geometry.

The simplest solid of revolution is a cylinder. In this example, the curve that generates the volume is a line parallel to the axis of revolution, which, while sounding the United States' next enemy combatant or a really cool rock band, is usually something boring like the x-axis or y-axis on the Cartesian plane. No calculus is needed to find the volume of a cylinder. Even in ancient times it was known that the volume of a prism is equal to the height times the area of the base, and in this case the base is a circle with area πr².

The next simplest solid of revolution is a cone. In this case, the curve is a straight line that crosses the axis of revolution. This formula too was known to the ancients, who generalized (correctly) that the volume of a pyramid is one third the volume of the prism with the same height and base.

The limit of ancient knowledge in this area is due to everyone's favorite last great Greek mathematician, Pappus of Alexandria, who said that the volume of a real-world solid of revolution was equal to the area of a radial cross-section (that is, the area between the curve being rotated and the axis of revolution) times the circumference of the circle traversed by the centroid of said cross-section. Pretty awesome for someone who hadn't heard of calculus and didn't have Mathematica to draw 3-D pictures for him. Pappus' theorem gives the formula for a cone, cylinder, sphere, and torus without resorting to calculus. Luckily, we can do better.

If we consider the x-axis to be the axis of revolution, we can describe a volume of revolution by a function y=f(x) over an interval (a, b). The area under the curve given by the usual definite integral ∫_a^bf(x) dx is the area of what we have called the radial cross-section. Thinking back to the Riemann sums that give rise to this integral, the area under the curve is estimated by successively better approximations of the area by rectangles. In finding the volume of a solid of revolution, we replace each of these rectangles with a cylinder whose height will be made successively smaller and whose radius will be given by the function at hand. Accordingly, as in the limit where the width of the rectangles approaches zero the usual Riemann sum approaches the actual area, in the limit where the height of the cylinders approaches zero the sum of volumes of cylinders approaches the actual volume.

This suggests that we should partition (a, b) with a finite sequence of points x(i), with x(0) = a and the last point, x(n) = b. Then if we sum up the volumes of cylinders with radius f(x(i)) and height x(i+1) - x(i), we find a sum of the form

V = Σ_i π (f(x(i)))² (x(i+1) - x(i))

And passing to the limit where the difference between points in the partition approaches zero, we have an integral of the form

V = ∫_a^b π (f(x))² dx