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the cylinders obtained by rotating the rectangles with the same height f (xi ) and with bases [0, xi 1 ] and [0, xi ]. Hence, it has volume 2 2 2 2 xi f (xi ) xi 1 f (xi ) = f (xi )(xi xi 1 ) = f (xi )(xi + xi 1 )(xi xi 1 ) = f (xi )(2xi )( x) = 2 xi f (xi ) x
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Thus, the total V is approximated by
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which in turn approximates the de nite integral 2
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33.6 Establish the cross-section formula V =
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Divide the interval [a, b] into n equal subintervals [xi 1 , xi ], each of length x. Choose a point xi in [xi 1 , xi ]. If n is large, making x small, the piece of the solid between xi 1 and xi will be very nearly a (noncircular) disk, of thickness x and base area A(xi ) (see Fig. 33-15). This disk has volume A(xi ) x. Thus, n b
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A(xi )
A(x) dx
Fig. 33-15 33.7 (Solids of Revolution about Lines Parallel to a Coordinate Axis) If a region is revolved about a line parallel to a coordinate axis, we translate the line (and the region along with it) so that it goes over into the coordinate axis. The functions de ning the boundary of the region have to be recalculated. The volume obtained by revolving the new region around the new line is equal to the desired volume. (a) Consider the region R bounded above by the parabola y = x 2 , below by the x-axis, and lying between x = 0 and x = 1 [see Fig. 33-16(a)]. Find the volume obtained by revolving R around the horizontal line y = 1. (b) Find the volume obtained by revolving the region R of part (a) about the vertical line x = 2.
(a) Move R vertically upward by one unit to form a new region R . The line y = 1 moves up to become the x-axis. R is bounded above by y = x 2 + 1 and below by the line y = 1. The volume we want is obtained by revolving R
APPLICATIONS OF INTEGRATION II: VOLUME
[CHAP. 33
Fig. 33-16
about the x-axis. The washer formula applies, V = =
((x 2 + 1)2 12 ) dx =
(x 4 + 2x 2 ) dx 13 15
2 x5 + x3 5 3
1 2 + 5 3
(b) Move R two units to the right to form a new region R # [see Fig. 33-16(b)]. The line x = 2 moves over to become the y-axis. R # is bounded above by y = (x 2)2 and below by the x-axis and lies between x = 2 and x = 3. The volume we want is obtained by revolving R # about the y-axis. The cylindrical shell formula applies, V = 2 = 2
x(x 2)2 dx = 2
(x 3 4x 2 + 4x) dx = 2
1 4 4 3 x x + 2x 2 4 3 = 2 11 12
1 4 4 3 (3) (3) + 2(3)2 4 3
1 4 4 3 (2) (2) + 2(2)2 4 3
11 = 6
Supplementary Problems
Strategy: In calculating the volume of a solid of revolution we usually apply either the disk formula (or the washer formula) or the cylindrical shells formula (or the difference of cylindrical shells formula). To decide which formula to use: (1) Decide along which axis you are going to integrate. This depends on the shape and position of the region R that is revolved. (2) (i) Use the disk formula (or the washer formula) if the region R is revolved perpendicular to the axis of intergration. (ii) Use the cylindrical shells formula (or the difference of cylindrical shells formula) if the region R is revolved parallel to the axis of integration.
33.8 Find the volume of the solid generated by revolving the given region about the given axis. (a) The region above the curve y = x 3 , under the line y = 1, and between x = 0 and x = 1; about the x-axis. (b) The region of part (a); about the y-axis. (c) The region below the line y = 2x, above the x-axis, and between x = 0 and x = 1; about the y-axis. (d) The region between the parabolas y = x 2 and x = y2 ; about either the x-axis or the y-axis.
CHAP. 33]
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