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 kth slab is ∆xk, then the volume of the slab can be approximated by the volume of a right cylinder of width (height) ∆xk and cross-sectional area A(xk*), where xk* is a number in the kth subinterval. Adding these approximations yields the following Riemann sum that approximates the volume V:
V ≈ ΣA(xk*)∆xk
Taking the limit as n increases and the widths of the subintervals approach zero yields the definite integral
- V = lim ΣA(xk*)∆xk = a∫b A(x)dx
max∆xk 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
-
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.
Now try it yourself! Should you still need any help, click here to schedule live online session with e Tutor!
About eAge Tutoring:
eAgeTutor.com is the premium online tutoring provider. Using materials developed by highly qualified educators and leading content developers, a team of top-notch software experts, and a group of passionate educators, eAgeTutor works to ensure the success and satisfaction of all of its students.
Contact us today to learn more about our tutoring programs and discuss how we can help make the dreams of the student in your life come true!
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