Volume
VOLUME 0 177 22.21 The solid is a wedge, cut from a perfectly round tree of radius r by two planes, one perpendicular to the axis of the
tree and the other intersecting the first plane at an angle of 30° along a diameter. (See Fig. 22-13.)
Fig. 22-14 Fig. 22-15
22.24 Let 91 be the region between y = x3,
region 9? about v = — 1. x = \, and y = 0. Find the volume of the solid obtained by rotating The same volume is obtained by rotating about the *-axis (y = 0) the region obtained by raising 3$ one unit, that is, the region bounded by y = x3 + l, y = l, and x. = 1 (see Fig. 22-16). By the circular ring formula, this is
22.23 The tetrahedron formed by three mutually perpendicular edges of lengths a,b,c.
Let the origin be the intersection of the edges, and let the jc-axis lie along the edge of length c (Fig. 22-15). A typical cross section is a right triangle with legs of lengths d and e, parallel respectively to the edges of lengths a and b. By similar triangles,
By the cross-section formula
and So, the area
22.22 A square pyramid with a height of h units and a base of side r units.
Locate the x-axis perpendicular to the base, with the origin at the center of the base (Fig. 22-14). By similar right triangles,
formula,
and So, and, by the cross-section
Let the x-axis be the intersection of the two planes, with the origin on the tree's axis. Then a typical cross
section is a right triangle with base and height So, the area A
is By symmetry, we can compute the volume for x > 0 and
then double the result. The cross-section formula yields the volume
178
Fig. 22-16
22.25 Let 91 be the region in the first quadrant between the curves y = x2 and y = 1x. Find the volume of the solid obtained by rotating 3? about the jc-axis.
By simultaneously solving y = x2 and y = 2x, we see that the curves intersect at (0,0) and (2,4). The line y = 2x is the upper curve. So, the circular ring formula yields V= -n Jo[(2*)2 - (x~)~] dx = ir $„ (4x2 — x4) dx = ir(tx3 - ±x5) ]2 = TT(¥ - f ) = 6477/15.
22.26 Same as Problem 22.25, but the rotation is around the y-axis.
Here let us use the difference of cylindrical shells: V—2-n J02 x(2x — x2) dx =2ir Jj (2x~ — x*) dx = 27r(f.r'-^4)]^ = 27r(¥-4) = 87r/3.
22.27 Let 3? be the region above y = (x - I)2 and below y = x + 1 (see Fig. 22-17). Find the volume of the solid obtained by rotating SI about the ;t-axis.
Fig. 22-17
22.28 Same as Problem 22.27, but the rotation is around the line y = -I.
22.29 Find the volume of the solid generated when the region bounded by y2 = 4* and y = 2x - 4 is revolved about the y-axis.
CHAPTER 22
Raise the region one unit and rotate around the x-axis. The bounding curves are now y = x 4- 2 and y = (x - I)2 + 1. By the circular ring formula, V= TT J3 {(x + 2)' - [(x - I)2 + I]2} dx = TT /3 {(x + 2)2
-((
X-iy + 2(x-iy + i]}dx = 7rO(* + 2)
3-u*-i)
5-f(*-i)
3-*))o = ^[(
if
5-¥-¥-3)-n + 5 +
3)1 = 1177T/5.
By setting x + 1 = (x — I)2 and solving, we obtain the intersection points (0,1) and (3,4). The circular ring formula yields V= TT |03 {(x + I)2 - [(x - I)2]2} dx = ir J03 [(x + if - (x - I)4] dx = IT( \(x + I)3 $(x -I)5) ]2 = » K ¥ - ¥ ) - ( i + i)] = 72^/5.
VOLUME D 179
Fig. 22-18 Fig. 22-19
22.30 Let 9? be the region bounded by y = x3, x = l, x = 2, and y - x - \ . Find the volume of the solid generated when SI is revolved about the x-axis.
y = x3 lies above y = x-l for l<;t<2 (see Fig. 22-19). So, we can use the circular ring formula:
37477/21.
Fig. 22-20
Solving y = 2* - 4 and y2 = 4x simultaneously, we obtain y2 — 2y — 8 = 0, y = 4 or y = —2.
Thus, the curves intersect at (4,4) and (1, -2), as shown in Fig. 22-18. We integrate along the y-axis, using the circular ring formula:
22.31 Same as Problem 22.30, but revolving about the y-axis.
We use the difference of cylindrical shells: V = 2ir J2 x[x3 - (x - 1)] dx = 2ir J\ (x4 - x +x)dx =
2TT(\X5 - \X* + kx2) ]2 = 277[(f - 1 + 2) - (I - I + I)] = 1617T/15.
22.32 Let 5? be the region bounded by the curves y = x2 - 4x + 6 and y = x + 2. Find the volume of the solid generated when 3ft is rotated about the ^-axis.
Solving y = x2-4x + 6 and y = x + 2, we obtain x2-5x + 4 = Q, x = 4 or x=l. So, the curves meet at (4,6) and (1,3). Note that y = x2 - 4x + 6 = (x -2)2 + 2. Hence, the latter curve is a parabola with vertex (2, 2) (see Fig. 22-20). We use the circular ring formula: V= TT J7 {(x + 2)2 - [(x - 2)2 + 2]2}dx= 7rJ14[(^ + 2 )2- ( ^ - 2 )4- 4 ( x - 2 )2- 4 ] r f A : = 7r(K^ + 2 )3~ H ^ - 2 )5- ! ^ - 2 )3- 4 ^ ) ] : =
7r[(72-f - 7r[(72-f - 16) - (9 + L + | - 4)1 = 1627T/5.
V = TTtf((XX-l)2]dX = TTtf(x6-(X-l)2]dx = irO^-H*-!)3)]? = ^[(¥-~i)-0-0)] =
CHAPTER 22
Fig. 22-21 22.35 Same as Problem 22.34, but the region is revolved about the y-axis.
Fig. 22-22 180
22.33 Same as Problem 22.32, but with the rotation around the y-axis.
22.34 Let 9? be the region bounded by y = 12- x3 and y = 12 — 4x. Find the volume generated by revolving 91 about the jt-axis.
22.36 Let &i be the region bounded by y = 9 - x2 and y = 2x + 6 (Fig. 22-22). Find, using the circular ring formula, the volume generated when 2fl is revolved about the x-axis.
Solving 9 - x2 =2*+ 6, we get x2 + 2*-3 = 0, (.v + 3)(* - 1) = 0, x = -3 or x = l. Thus, the curves meet at (1,8) and (3,0). Then V= 77 J13 [(9 x2)2 (2x + 6)2] dx = 77 J13 (81 18*2 + x* 4x2 24;t 36) (it = 77 J13 (45 24x 22x2 + *4) rf* = 77(45* 12x2 f x3 + ^5) ]!_, = 7r[(45 12 f + i) -(-135- 108 + 198- 1?)] = 179277/15.
By symmetry, we need only double the volume generated by the piece in the first quadrant. We use the difference of cylindrical shells: V= 2-277 J2 x[(l2 - jc3) - (12- 4x)] dx=4Tt J2 x(4x - x3) dx = 477 Jj (4.V2 - .V4) dx = 477(|X3 - i.V5) ]' = 47T(f - f ) = 25677/15.
Solving 12 - x3 = 12 - 4x, we get x = 0 and x = ±2. So, the curves intersect at (0,12), (2, 4), and (-2,20). The region consists of two pieces as shown in Fig. 22-21. By the circular ring formula, the piece in the first quadrant generates volumeV= TT J02[(12-x3)2 -(12 -4x)2] dx = TT J02 [(144 -24jc3 + x6) - (144 -96* + 16*2)] dx = 77 J2 (x6 -24x3 - 16*2 + 96*) dx = ir( $x7 - 6x4 - ^x3 + 48x2) ]2 = *r( ^ - 96 - ^ + 192) = 150477 /21. Similarly, the piece in the second quadrant generates volume 252877/21, for a total of 150477/21 + 252877/21 = 19277.
We use the difference of cylindrical shells: V=277 J,4 x[(x + 2) (x2 4x + 6)] dx = 2ir J,4 x(5x x2 -4) dx = 277 j\4 (5*2 - x3 - 4x) dx = 2ir( §*3 - \x* - 2x2) ]* = 27r{[f (6-4) - 64 - 32) - (f - \ - 2)] = 4577/2.
22.38 Let SK. be the region bounded by y = x3 + x, y = 0, and A: = 1 (Fig. 22-23). Find the volume of the solid obtained by rotating &i about the y-axis.
Fig. 22-23
We use the cylindrical shell formula: V=2v JJ *(*3 + x) dx = 2ir $„ (x4 + x2) dx = 2ir(^5 + i*3) ]J =
2ir(\ + £ ) = 16-77/15.
22.42 Find the volume of the solid of rotation generated when the curve y = tan x, from x = 0 to x = IT 14, is rotated about the jc-axis.
22.43 Find the volume of the solid generated by revolving about the x-axis the region bounded by y = sec x, y = 0, x = 0, and x = ir/4.
See Fig. 22-24. By the disk formula, V= TT Jo"4 sec2 Jt dx = w(tan x) ]„'" = ir(l - 0) = IT.
The disk formula gives
22.37 Solve Problem 22.36 by the cylindrical shell formula.
VOLUME 181
We must integrate along the y-axis, and it is necessary to break the region into two pieces by the line y = 8.
Then we obtain
We have to find Let
Then
Hence,
22.39 Same as Problem 22.38, but rotating about the x-axis.
We use the disk formula:
22.40 Same as Problem 22.38, but rotating about the line x = -t2.
22.41 Find the volume of the solid generated when one arch of the curve y = sin x, from x = 0 to x = IT, is rotated about the jc-axis.
The disk formula yields
The same volume can be obtained by moving the region two units to the right and rotating about the _y-axis.
The new curve is y = (x - 2)3 + x - 2, and the interval of integration is 2 < x< 3. By the cylindrical shell formula, V= 2n J23 x[(x -2f + x-2}dx = 2Tr J23 x(A:3 - 6x2 + 12x - 8 + x - 2) dx = 2ir J23 (^4 - 6:c3 + 13A:2 - 10^) dx = 277-G*5 - Ix4 + f x3 - 5x2) ]2 = 2ir[( ^ - ^ + 117 - 45) - (f - 24 + ^ - 20)J = 617T/15.
182
Fig. 22-24
(b) Interchange a and b in (a):
Fig. 22-25 Fig. 22-26
22.49 Find the volume of the solid obtained by rotating about the Jt-axis the region in the first quadrant under the line segment from (0, r j to (h, r2), where 0 < r , < r2 and 0<h. (See Fig. 22-26. Note that this is the volume of a frustum of a cone with height h and radii rl and r2 of the bases.)
The equation of the line is By the disk formula,
(a) We double the value obtained from the disk formula applied to the part of the region in the first quadrant (see Fig. 22-25):
22.48 Find the volume of the ellipsoid obtained when the ellipse about the y-axis.
is rotated (a) about the *-axis, (b) 22.47 Same as Problem 22.46, but with the rotation around the y-axis.
Use the cylindrical shell formula:
The disk formula applies:
Note that this approaches TT as
22.46 Find the volume of the solid generated when the region in the first quadrant under the hyperbola xy = 1, between x = 1 and x = b>l, is rotated about the *-axis.
22.45 Same as Problem 22.44, but the rotation is about the y-axis.
Use the cylindrical shell formula:
Use the disk formula:
22.44 Calculate the volume of the solid paraboloid generated when the region in the first quadrant under the parabola x = y". between x = 0 and x = b, is rotated about the *-axis.
CHAPTER 22
VOLUME 183 22.50 A solid has a circular base of radius r. Find the volume of the solid if every planar section perpendicular to a fixed
diameter is a semicircle.
Fig. 22-27 22.51 Same as Problem 22.50, but the planar cross section is a square.
The area A(x) of the dashed square in Fig. 22-27 is 4(r2 - x2). Hence the volume will be S/TT times that found in Problem 22.50, or 16r3/3.
22.52 Same as Problem 22.50, except that the planar cross section is an isosceles right triangle with its hypotenuse on the base.
22.53 Find the volume of a solid whose base is the region in the first quadrant bounded by the line 4x + 5y = 20 and the coordinate axes, if every planar section perpendicular to the x-axis is a semicircle (Fig. 22-28).
The radius of the semicircle is |(5-x), and its area ^W 's> therefore, ^Tr(5-x)2. By the
cross-section formula, V= %tr J (5 - x)2 dx = £TT(- $)(5 - x)1 }50 = - ^| (5 - x)3 ]« = - jj (0 - 125) = 2077/3 .
05Fig. 22-28 Fig. 22-29
22.54 The base of a solid is the circle x2 + y2 = 16*, and every planar section perpendicular to the jr-axis is a rectangle whose height is twice the distance of the plane of the section from the origin. Find the volume of the solid.
Refer to Fig. 22-29. By completing the square, we see that the equation of the circle is (x - 8)2 + yz = 64.
So, the center is (8,0) and the radius is 8. The height of each rectangle is 2x, and its base is 2y = 2V64 - (A- - 8)2. So, the area A(x) is 4x^/64 - (x - 8)2. By the cross-section formula, V= 4 J016 jc[64 -(;t-8)2]"2 dx. Let M = jc-8, x = u + 8, du = dx. Then V=4 Jfg (« + 8)(64- ir)"2 d« = 4 J!8 u(64 -H2)"2 dw + 32 J!8 (64 - u2)"2 du. The integral in the first summand is 0, since its integrand is an odd function.
The integrand in the second summand is the area, 3277, of a semicircle of radius 8 (by Problem 20.71). Hence, V= 32(327r) = 102477.
22.55 The section of a certain solid cut by any plane perpendicular to the jc-axis is a square with the ends of a diagonal lying on the parabolas y2=9x and x2 =9y (see Fig. 22-30). Find its volume.
The parabolas intersect at (0,0) and (9, 9). The diagonal d of the square is 3*"2 - §jt2 = $(27;c"2 - x2).
Then the area of the square is A(x) = \d2 = ^ [(27)2x - 54x5'2 + x4], and the cross-section formula yields the volume K= jfe /09 [(27)2* - 54jc5'2 + x4] dx = Tfe((27)2 • {x2 - ^x"2 + i*5) ft = |(81)2.
Let the *-axis be the fixed diameter, with the center of the circle as the origin. Then the radius of the semicircle at abscissa x is V/-2 - x2 (see Fig. 22-27), and, therefore, its area A(x) is \TT(T* - x1). The cross-section formula yields V= \'_r A(x) dx = ±TT fr_r (r2 - x2) dx = ±ir(r2x - $x3) ]r_r = ±ir[(r3 - ^r3) -(-r*+$r3)}=t7rr\
The area A(x) of the dashed triangle in Fig. 22-27 (which is inscribed in the semicircle) is j(2Vr2 - A-2)(Vr - x2) = r2 - x2. Hence the volume will be one-fourth that of Problem 22.51, or 4/-3/3.
184 0 CHAPTER 22
22.56 Find the volume of the solid of revolution obtained by rotating about the jc-axis the region in the first quadrant bounded by the curve x2'3 + y2'3 = a2'3 and the coordinate axes.
22.57 The base of a certain solid is an equilateral triangle of side b, with one vertex at the origin and an altitude along the positive j;-axis. Each plane perpendicular to the .it-axis intersects the solid in a square with one side in the base of the solid. Find the volume.
See Fig. 22-31. The altitude h = b cos 30° = \bV3. For each x, y = x tan 30° = (1A/3)*. Hence, the side of the square is 2y = (2/V3)x and its area is A= \x2. Hence, the cross-section formula yields
Fig. 22-31 Fig. 22-32
22.58 What volume is obtained when the area bounded by the line y = x and the parabola y = x2 is rotated about the bounding line?
Refer to Fig. 22-32. The required volume is given by the disk formula as V = TT J0 2 r2 ds; so our strategy will be to find r2 and s as functions of x, and then to change the integration variable from s to x. Now, by the Pythagorean theorem,
and, by the distance formula,
Eliminating r2 between (1) and (2), we obtain
(I)
(2)
so
and then, from (1), r2 = \x2 - x3 + j*4. Carrying out the change of variable, we have with x ranging from 0 to 1. Hence
Fig. 22-30
By the disk formula, V= IT ft / dx = TT $°0 (a213 - x2'3)3 dx = IT J° (a2 ~ 3a*'3x2'3 + 3«2'V'3 - x2) dx = TT(a2x - 3fl4'3 • lx513 + 3a2'3 • lx"3 - \x3) }"0 = Tr(a3 - \a3+ |«3 - |a3) = 167r«3/105.
i- + s2 = h2 = x2 + x4