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
We can conclude the result in the following way,
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.
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 x ½ [62-02]
= 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.
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