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[since 4(r 2 x 2 ) is an even function] r 2 (r) r3 3 (0 0) = 8 2 3 r 3 = 16 3 r 3
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33.1 Find the volume of the solid generated by revolving the given region about the given axis. (a) The region under the parabola y = x 2 , above the x-axis, between x = 0 and x = 1; about the x-axis. (b) The same region as in part (a), but about the y-axis. The region is shown in Fig. 33-10.
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APPLICATIONS OF INTEGRATION II: VOLUME
[CHAP. 33
Fig. 33-10
(a) Use the disk formula, V = (b) Use the cylindrical shell formula, V = 2
1 0 1 0
(x 2 )2 dx =
x 4 dx =
x5 5
x(x 2 ) dx = 2
x 3 dx = 2
x4 4
= 2
33.2 Let R be the region between y = x 2 and y = x (see Fig. 33-11). Find the volume of the solid obtained by revolving R around: (a) the x-axis; (b) the y-axis.
Fig. 33-11
The curves intersect at (0, 0) and (1, 1). (a) By the washer formula, V = (b) (Method 1)
(x 2 (x 2 )2 ) dx =
(x 2 x 4 ) dx =
x5 x3 3 5
1 1 3 5
2 15
Use the washer formula along the y-axis, V =
(( y)2 y2 ) dy =
(y y2 ) dy =
y3 y2 2 3
1 1 2 3
CHAP. 33]
APPLICATIONS OF INTEGRATION II: VOLUME
(Method 2) We can integrate along the x-axis and use the difference of two cylindrical shell formulas, V = 2 = 2
x(x) dx
x(x 2 ) dx
= 2
(x 2 x 3 ) dx 1 12 = 6
x4 x3 3 4
= 2
1 1 3 4
(0 0) = 2
The formula used in method 2 can be formulated as follows: V = 2
x(g(x) f (x)) dx
difference of cylindrical shells
where V is the volume of the solid obtained by revolving about the y-axis the region bounded above by y = g(x), below by y = f (x), and lying between x = a and x = b, with 0 a < b.
33.3 Find the volume of the solid whose base is a circle of radius r and such that every cross section perpendicular to a particular xed diameter D is an equilateral triangle.
Let the center of the circular base be the origin, and let the x-axis be the diameter D. The area of the cross section at x is A(x) = hs/2 (see Fig. 33-12). Now, in the horizontal right triangle, x2 + and in the vertical right triangle, h2 + s 2 = s2 2 s2 = s2 h2 + 4 s2 h2 = 3 4 s h = 3 = 3 r 2 x2 2 s 2 = r2 2 or s = 2 r 2 x2
Hence, A(x) =
3(r 2 x 2 ) an even function and the cross-section formula gives V= 3
r r
(r 2 x 2 ) dx = 2 3
r3 = 2 3 r3 3
x3 (r 2 x 2 ) dx = 2 3 r 2 x 3 0 2 3 4 3 3 =2 3 r = r 3 3
Fig. 33-12
APPLICATIONS OF INTEGRATION II: VOLUME
[CHAP. 33
33.4 Establish the disk formula V =
( f (x))2 dx.
We assume as valid the expression r 2 h for the volume of a cylinder of radius r and height h. Divide the interval [a, b] into n equal subintervals, each of length x = (b a)/n (see Fig. 33-13). Consider the volume Vi obtained by revolving the region Ri above the ith subinterval about the x-axis. If mi and Mi denote the absolute minimum and the absolute maximum of f on the ith subinterval, it is plain that Vi must lie between the volume of a cylinder of radius mi and height x and the volume of a cylinder of radius Mi and height x,
2 mi
x Vi Mi2
2 mi
Vi Mi2 x
The intermediate-value theorem for the continuous function ( f (x))2 guarantees the existence of some point xi in the ith subinterval such that
Vi Hence, x
= ( f (xi ))2 n
Vi = ( f (xi ))2 f (xi ) 2
Vi =
Since this relation holds (for suitable numbers xi ) for arbitrary n, it must hold in the limit as n , n
V = lim
( f (xi ))2
( f (x))2 dx
which is the disk formula. The name derives from the use of cylindrical disks (of thickness
x) to approximate the Vi .
33.5 Establish the cylindrical shell formula V = 2
xf (x) dx.
Divide the interval [a, b] into n equal subintervals, each of length x. Let Ri be the region above the ith subinterval (see Fig. 33-14). Let xi be the midpoint of the ith subinterval, xi = (xi 1 + xi )/2. Now the solid obtained by revolving the region Ri about the y-axis is approximately the solid obtained by revolving the rectangle with base x and height yi = f (xi ). The latter solid is a cylindrical shell; that is, it is the difference between
Fig. 33-13
Fig. 33-14
CHAP. 33]
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