The application of integrals to the computation of areas in the plane can be extended to the computation of certain volumes in space, namely those of solids of revolution. A solid of revolution arises from revolving the region below the graph of a function f (x) about the x- or y-axis of the plane. A cone arises in this way from a triangular region, a sphere from a semicircular region, and a cylinder from a rectangular region. These are just a few of the possibilities for solids of revolution.
There are two primary methods for finding the volume of a solid of revolution. The shell method is applied to a solid obtained by revolving the region below the graph of a function f (x) from a to b about the y-axis. It approximates the solid with a number of thin cylindrical shells, obtained by revolving about the y-axis the thin rectangular regions used to approximate the corresponding region in the plane. This is illustrated in the figure below.
The volume of a thin cylindrical shell of radius x, thickness Δx, and height f (x) is equal to
|Π(x + )2f (x) - Π(x - )2f (x)||=||Π(2xΔx)f (x)|
Here by "cylindrical shell" we mean the region between two concentric cylinders whose radii differ only very slightly; precisely speaking, this formula is not correct for any positive thickness, but approaches the correct value as the thickness Δx shrinks to zero. Since we will ultimately consider such a limit, this formula will yield the correct volume in our application.
If we sum together the volumes of a family of such cylindrical shells, covering the entire interval from a to b, and take the limit as Δx→ 0 (and consequently as the number of cylindrical shells approaches infinity), we end up with the integral
|Vol = 2Πxf (x)dx = 2Πxf (x)dx|
The disk method for finding volumes applies to a solid obtained by revolving the region below the graph of a function f (x) from a to b about the x-axis. Here the solid is approximated by a number of very thin disks, standing sideways with the x-axis through their centers. These disks are obtained by revolving about the x-axis the thin rectangular regions used to approximate the area of the corresponding region in the plane. This is illustrated in the figure below.
The volume of such a disk is (exactly) the area of the base times the height; hence, if the corresponding rectangle has width Δx and height f (x), the volume is equal to Πf (x)2Δx. Taking the sum of the volumes of all the disks (covering the entire interval from a to b) and taking the limit as Δx→ 0 gives the integral
|Vol = Πf (x)2dx = Πf (x)2dx|
The disk method is a special case of a more general method called the cross-sectional area method. In the disc method, the quantity we end up integrating, from a to b, is Πf (x)2, the cross-sectional area of the solid when sliced by a plane through x perpendicular to the x-axis. Even when the cross-section is not a disk (as it is in the case of more general solids of revolution), there may still be a function A(x) that gives the area of the cross section obtained by slicing the solid with the plane through x and perpendicular to the x-axis. The volume of the solid is then given by
|Vol = A(x)dx|