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)

=

(2Πx)(Δxf (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)^{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 = Πf (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