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( f (x))2 dx = in .NET
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Code39 Creator In ObjectiveC Using Barcode generation for iPad Control to generate, create Code 3 of 9 image in iPad applications. Creating Bar Code In VB.NET Using Barcode creator for .NET Control to generate, create barcode image in VS .NET applications. Fig. 332 If we interchange the roles of x and y and revolve the area under the graph of x = g(y) about the yaxis, then the same reasoning leads to the disk formula V = Generating Code 128 Code Set B In Java Using Barcode maker for Eclipse BIRT Control to generate, create ANSI/AIM Code 128 image in Eclipse BIRT applications. Create Code39 In Java Using Barcode generation for Android Control to generate, create Code 39 Extended image in Android applications. (g(y))2 dy =
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Generating Barcode In None Using Barcode creator for Software Control to generate, create barcode image in Software applications. Barcode Recognizer In VS .NET Using Barcode Control SDK for ASP.NET Control to generate, create, read, scan barcode image in ASP.NET applications. EXAMPLE Applying the disk formula to Fig. 332(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 xaxis (see Fig. 333). 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. 333 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. 334). The curves obviously intersect in the points (0, 0) and (1, 1). The bowlshaped solid of revolution generated by revolving R about the xaxis 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. 334 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 xaxis, and between x = a and x = b (see Fig. 335). Now, however, revolve R about the yaxis. 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. 335 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 yaxis of the region R under the graph of f (see Fig. 336) 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. 336 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 xaxis at x = a and the plane perpendicular to the xaxis at x = b. For a x b, let the plane perpendicular to the xaxis at that value of x intersect the solid in a region of area A(x), as indicated in Fig. 337. Then the volume V of the solid is given by V= A(x) dx
crosssection formula
For a derivation, see Problem 33.6.
Fig. 337 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. 338). Thus, A(x) = ( kx)2 = kx and the crosssection 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. 339). Find the volume. Let the origin be the center of the circle and let the xaxis 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. 339), x2 + s 2 = r2 2 s2 = r2 x2 + 4 s2 = 4 r 2 x 2 = A(x)

