Applications of the Integral
Volumes of Solids
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 + )2 f (x) - Π(x - )2 f (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 = 2Π xf (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)2 Δx . 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)2 dx = Π f (x)2 dx|
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|