Introduction
We have already learned to find the area of a plane region bounded by two curves which is obtained by integrating the length of a general cross section over an appropriate interval. Here we will see that the same basic principle can be used to find volumes of certain three dimensional solids.
Let S be a solid that extends along the xaxis and is bounded on the left and right, respectively, by the planes that are perpendicular to the xaxis at x=a and x=b. We are finding the volume V of the solid, assuming that its crosssectional area A(x) is known at each x in the interval [a, b].
To solve this problem we divide the interval [a, b] into n subintervals, which has the effect of dividing the solid into n slabs [fig (ii)]
If we assume that the width of the k^{th} slab is ∆x_{k}, then the volume of the slab can be approximated by the volume of a right cylinder of width (height) ∆x_{k} and crosssectional area A(x_{k}*), where x_{k}* is a number in the k^{th} subinterval. Adding these approximations yields the following Riemann sum that approximates the volume V:
V ≈ ΣA(x_{k}*)∆x_{k}
Taking the limit as n increases and the widths of the subintervals approach zero yields the definite integral
 V = lim ΣA(x_{k}*)∆x_{k} = _{a}∫^{b} A(x)dx
_{max}∆x_{k} 0
We can conclude the result in the following way,
Volume formula
Let S be a solid bounded by two parallel planes perpendicular to the xaxis at x=a and x=b. If, for each x in [a, b] the crosssectional area of S perpendicular to the xaxis is A(x), then the volume of the solid is,
V= _{a}∫^{b} A(x) dx provided A(x) is integrable.
Volume Formula
Let S be a solid bounded by two parallel planes perpendicular to the yaxis at y=c and y=d. If, for each y in [c, d], the crosssectional area of S perpendicular to the yaxis is A(y), then the volume of the solid is,
V = _{c}∫^{d} A(y) dy provided A(y) is integrable.
In words, these formulas states that, “The volume of a solid can be obtained by integrating the crosssectional area from one end of the solid to the other”.
Example: Find the formula for the volume of a right pyramid whose altitude is h and whose base is a square with sides of length a.
Solution: We introduce a rectangular coordinate system in which the yaxis passes through apex and is perpendicular to the base, and the xaxis passes through the base and is parallel to a side of the base.
At any ‘y’ in the interval [0, h] on the yaxis, the cross section perpendicular to the yaxis is a square. If ‘s’ denotes the length of a side of this square, then by similar triangles.
Solids of revolution
A solid of revolution is a solid that is generated by revolving a plane region about a line that lies in the same plane as the region; the line is called the axis of revolution.
Volume of a solid of revolution
Let f be continuous and nonnegative on [a, b] and let R be the region that is bounded by y=f(x), below by the xaxis, and on the sides by the lines x=a and x=b, then the volume of the solid or revolution that is generated by revolving the region R about the xaxis is given by
V= _{a}∫^{b} π[f(x)]^{2}dx
= _{a}∫^{b} π y^{2} dx
= π _{a}∫^{b}y^{2} dx
Example: Find the volume of a paraboloid of revolution formed by revolving the parabola y^{2}=8x about the xaxis from x=0 to x=6
Solution: The equation of the parabola is y^{2}=8x,
Hence volume = _{0}∫^{6} πy^{2}dx
= π_{0}∫^{6} 8x dx
= 8π[x^{2}/2]_{0}^{6}
= 8π x ½ [6^{2}0^{2}]
=4πx36
= 144πcubic units.
Volume by cylindrical shells
A cylindrical shell is a solid enclosed by two concentric right circular cylinders. The volume V of a cylindrical shell with inner radius r_{1}, outer radius r_{2}, and height h can be written as
V = (area of cross section).height
Let f be continuous and nonnegative on [a, b] and let R be the region that is bounded above by y=f(x) below by the xaxis, and on the sides by the lines x=a and x=b. Then the volume V of the solid revolution that is generated by revolving the region R about the yaxis is given by
V = _{a}∫^{b} 2πxf(x)dx

Example: Use cylindrical shells to find the volume of the solid generated when the region R in the first quadrant enclosed between y=x and y=x^{2} is revolved about the yaxis.
Solution: V= _{0}∫^{1}2πx(xx^{2})dx = 2π _{0}∫^{1} (x^{2}x^{3})dx
= 2π [(1/3) – (1/4)]
= π/6 cubic units.
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Reference Links:
http://www.intmath.com/applicationsintegration/2areaundercurve.php
http://www.cliffsnotes.com/study_guide/VolumesofSolidswithKnownCrossSections.topicArticleId39909,articleId39906.html
http://en.wikipedia.org/wiki/Solid_of_revolution