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 x-axis and is bounded on the left and right, respectively, by the planes that are perpendicular to the x-axis at x=a and x=b.  We are finding the volume V of the solid, assuming that its cross-sectional 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 cross-sectional 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

1.     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 x-axis at x=a and x=b.  If, for each x in [a, b] the cross-sectional area of S perpendicular to the x-axis 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 y-axis at y=c and y=d.  If, for each y in [c, d], the cross-sectional area of S perpendicular to the y-axis 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 cross-sectional 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 y-axis passes through apex and is perpendicular to the base, and the x-axis passes through the base and is parallel to a side of the base.
   At any ‘y’ in the interval [0, h] on the y-axis, the cross section perpendicular to the y-axis 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 non-negative on [a, b] and let R be the region that is bounded by y=f(x), below by the x-axis, 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 x-axis is given by
                                    V= _a∫^b π[f(x)]²dx
                                      = _a∫^b π y² dx
                                       = π _a∫^by² dx                              

Example: Find the volume of a paraboloid of revolution formed by revolving the parabola y²=8x about the x-axis from x=0 to x=6
Solution: The equation of the parabola is y²=8x,
Hence volume = ₀∫⁶ πy²dx
                     = π₀∫⁶ 8x dx
                     = 8π[x²/2]₀⁶
                     = 8π x ½ [6²-0²]
                     =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 r₁, outer radius r₂, and height h can be written as
                         V = (area of cross section).height
Let f be continuous and non-negative on [a, b] and let R be the region that is bounded above by y=f(x) below by the x-axis, 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 y-axis is given by
                       V = _a∫^b 2πxf(x)dx

1.                          
     
   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=x² is revolved about the y-axis.
   Solution: V= ₀∫¹2πx(x-x²)dx = 2π ₀∫¹ (x²-x³)dx
                                    = 2π [(1/3) – (1/4)]
                                    = π/6 cubic units.

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Reference Links:

http://www.intmath.com/applications-integration/2-area-under-curve.php

http://www.cliffsnotes.com/study_guide/Volumes-of-Solids-with-Known-Cross-Sections.topicArticleId-39909,articleId-39906.html

http://en.wikipedia.org/wiki/Solid_of_revolution

http://en.wikipedia.org/wiki/Shell_integration

http://en.wikipedia.org/wiki/Riemann_sum