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An argument for the disk formula is sketched in Problem 33.4.
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APPLICATIONS OF INTEGRATION II: VOLUME
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[CHAP. 33
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Fig. 33-2 If we interchange the roles of x and y and revolve the area under the graph of x = g(y) about the y-axis, then the same reasoning leads to the disk formula V =
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EXAMPLE Applying the disk formula to Fig. 33-2(a), we obtain
V = =
h o h r 2 x dx = h o
r2 2 x dx h2
h r 2 x3 r2 = 2 h2 3 h o
r2h h3 0 = 3 3
which is the standard formula for the volume of a cone with height h and radius of base r.
Now let f and g be two functions such that 0 g(x) f (x) for a x b, and revolve the region R between the curves y = f (x) and y = g(x) about the x-axis (see Fig. 33-3). The resulting solid of revolution has a volume V which is the difference between the volume of the solid of revolution generated by the region under y = f (x) and the volume of the solid of revolution generated by the region under y = g(x). Hence, by Theorem 33.1, V =
( f (x))2 (g(x))2 dx
washer formula1
Fig. 33-3
1 So
termed because the cross section obtained by revolving a vertical segment has the shape of a plumber s washer.
CHAP. 33]
APPLICATIONS OF INTEGRATION II: VOLUME
x and y = x (see Fig. 33-4). The curves obviously intersect in the points (0, 0) and (1, 1). The bowl-shaped solid of revolution generated by revolving R about the x-axis has volume
EXAMPLE Consider the region R bounded by the curves y =
V = =
1 (( x)2 x 2 ) dx =
(x x 2 ) dx =
x3 x2 2 3
1 1 2 3
0 = 6
Fig. 33-4 Cylindrical Shell Method Let f be a continuous function such that f (x) 0 for a x b, where a 0. As usual, let R be the region under the curve y = f (x), above the x-axis, and between x = a and x = b (see Fig. 33-5). Now, however, revolve R about the y-axis. The resulting solid of revolution has volume V = 2
xf (x) dx = 2
xy dx
cylindrical shell formula
For the basic idea behind this formula and its name, see Problem 33.5.
Fig. 33-5 EXAMPLE Consider the function f (x) = r 2 x2 for 0 x r. The graph of f is the part of the circle x2 + y2 = r 2 that lies
in the rst quadrant. Revolution about the y-axis of the region R under the graph of f (see Fig. 33-6) produces a solid hemisphere of radius r. By the cylindrical shell formula, V = 2
x r 2 x 2 dx
APPLICATIONS OF INTEGRATION II: VOLUME
[CHAP. 33
Fig. 33-6
To evaluate V substitute u = r 2 x 2 . Then du = 2x dx, and the limits of integration x = 0 and x = r become u = r 2 and u = 0, respectively,
0 r2 0 r 1 du = u1/2 du = u1/2 du 2 2 0 r
V = 2 =
u1/2
2 2 3/2 r 2 2 u = (r 2 )3/2 = r 3 3 3 3 0
(This result is more easily obtained by the disk formula V =
x 2 dy. Try it.)
33.2 VOLUME BASED ON CROSS SECTIONS Assume that a solid (not necessarily a solid of revolution) lies entirely between the plane perpendicular to the x-axis at x = a and the plane perpendicular to the x-axis at x = b. For a x b, let the plane perpendicular to the x-axis at that value of x intersect the solid in a region of area A(x), as indicated in Fig. 33-7. Then the volume V of the solid is given by V=
A(x) dx
cross-section formula
For a derivation, see Problem 33.6.
Fig. 33-7
CHAP. 33]
APPLICATIONS OF INTEGRATION II: VOLUME
EXAMPLES
(a) Assume that half of a salami of length h is such that a cross section perpendicular to the axis of the salami at a distance x from the end O is a circle of radius kx (see Fig. 33-8). Thus, A(x) = ( kx)2 = kx and the cross-section formula gives V=
kx dx = k
x dx = k
x2 2
kh2 2
Note that for this solid of revolution the disk formula would give the same expression for V . (b) Assume that a solid has a base which is a circle of radius r. Assume that there is a diameter D such that all plane sections of the solid perpendicular to diameter D are squares (see Fig. 33-9). Find the volume. Let the origin be the center of the circle and let the x-axis be the special diameter D. For a given value of x, with r x r, the side s(x) of the square cross section is obtained by applying the Pythagorean theorem to the right triangle with sides x, s/2, and r (see Fig. 33-9), x2 + s 2 = r2 2 s2 = r2 x2 + 4 s2 = 4 r 2 x 2 = A(x)
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