VOLUME
175
We use the circular ring formula for the region in the first quadrant, and then double the resultV= * Jo2 l(5)2 - (1 + x2)2] <** = " Jo* [25 - (1 + 2xz + *4)] dx = 77 J (24 - 2*2 - *4) dx = 77(24* - tf - ^) ]*
0
2
= 7r(48 - ¥ -7) = 5447T/15. Doubling this, we obtain 108877/15.
22.11
Solve Problem 22.10 by means of the cylindrical shell formula.
We compute the volume in the first quadrant and then double it. V= 2ir /f xy dy = 2w Jf
^3,, "'~1' y = u + l< du = dy. Then V=2ir ft Vu(u + 1) du =2*- J
4
=>
0
4 («
3 '
2 + H "
2 ) du = 27r(| M
5 '
2 +
S" ")]o = 27r(" + T) = 544?7/15, which, doubled, yields the same answer as in Problem 22.10.
22.12
The region inside the circle x
2 + (y - b)
2 = a
2 (0
Refer to Fig. 22-8; we deal with the region in the first quadrant, and then double it. Use the cylindrical
shell formula:
integrand is an odd function (see Problem 20.48). The second integral is the area of a semicircle of radius a (see
Problem 20.71) and is therefore equal to \^a
2 . Hence, V= 1-nb • \-na~ = Tr
2 ba
2 , which, doubled, yields the
answer 2ir
2 ba
2 .
The first integral is 0, since the
Then
Fig. 22-8
Fig. 22-9
22.13
The region bounded by x
2 = 4y and y = {x; about the y-axis.
See Fig. 22-9. The curves A:" = 4_y and y = {x intersect at (0,0) and (2,1). We use the circular ring
formula: V= 77 /„' (4y - (2y)2] dy = 77 J0' (4y - 4y2) dy = 77(2>.2 - jy3) ],', = 77(2 - $) = 2w/3.
22.14
Solve Problem 22.13 by means of the cylindrical shell formula.
22.15
The region of Problem 22.13; about the *-axis.
Use the circular ring formula:
22.16
The region bounded by y = 4/x and y = (*-3)
2 ; about the x-axis. (See Fig. 22-10.)
From >> = 4/jc and .y = (.r-3)
2 , we get 4 = x
3 - 6x
2 + 9x, x
3 - 6x
2 + 9x: - 4 = 0. x = 1 is a root,
and, dividing x
3 - 6x
2 + 9x - 4 by x - 1, we obtain x
2 -5x + 4 = (* - l)(jr -4). with the additional
root x = 4. Hence, the intersection points are (1,4) and (4,1). Because x = 1 is a double root, the slopes
of the tangent lines at (1,4) are equal, and, therefore, the curves are tangent at (1, 4). The hyperbola xy =
4 is the upper curve. The circular ring formula yields V= 77 tf {(4/x)
2 - [(x - 3)
2 ]
2 } dx =TT J
4 [I6x~
2 -
(x - 3)
4 ] dx = Tr(-\6x~
l - i(* - 3)
5 ) ]J = 77[(-4 - *) - (-16 + f)] = 2777/5 .
ydy. Let u=y-b, y = u + b, du = dy.
(u + b)du = 2ir
u du + b
V=2ir
ydy.
175
We use the circular ring formula for the region in the first quadrant, and then double the resultV= * Jo2 l(5)2 - (1 + x2)2] <** = " Jo* [25 - (1 + 2xz + *4)] dx = 77 J (24 - 2*2 - *4) dx = 77(24* - tf - ^) ]*
0
2
= 7r(48 - ¥ -7) = 5447T/15. Doubling this, we obtain 108877/15.
22.11
Solve Problem 22.10 by means of the cylindrical shell formula.
We compute the volume in the first quadrant and then double it. V= 2ir /f xy dy = 2w Jf
^3,, "'~1' y = u + l< du = dy. Then V=2ir ft Vu(u + 1) du =2*- J
4
=>
0
4 («
3 '
2 + H "
2 ) du = 27r(| M
5 '
2 +
S" ")]o = 27r(" + T) = 544?7/15, which, doubled, yields the same answer as in Problem 22.10.
22.12
The region inside the circle x
2 + (y - b)
2 = a
2 (0
Refer to Fig. 22-8; we deal with the region in the first quadrant, and then double it. Use the cylindrical
shell formula:
integrand is an odd function (see Problem 20.48). The second integral is the area of a semicircle of radius a (see
Problem 20.71) and is therefore equal to \^a
2 . Hence, V= 1-nb • \-na~ = Tr
2 ba
2 , which, doubled, yields the
answer 2ir
2 ba
2 .
The first integral is 0, since the
Then
Fig. 22-8
Fig. 22-9
22.13
The region bounded by x
2 = 4y and y = {x; about the y-axis.
See Fig. 22-9. The curves A:" = 4_y and y = {x intersect at (0,0) and (2,1). We use the circular ring
formula: V= 77 /„' (4y - (2y)2] dy = 77 J0' (4y - 4y2) dy = 77(2>.2 - jy3) ],', = 77(2 - $) = 2w/3.
22.14
Solve Problem 22.13 by means of the cylindrical shell formula.
22.15
The region of Problem 22.13; about the *-axis.
Use the circular ring formula:
22.16
The region bounded by y = 4/x and y = (*-3)
2 ; about the x-axis. (See Fig. 22-10.)
From >> = 4/jc and .y = (.r-3)
2 , we get 4 = x
3 - 6x
2 + 9x, x
3 - 6x
2 + 9x: - 4 = 0. x = 1 is a root,
and, dividing x
3 - 6x
2 + 9x - 4 by x - 1, we obtain x
2 -5x + 4 = (* - l)(jr -4). with the additional
root x = 4. Hence, the intersection points are (1,4) and (4,1). Because x = 1 is a double root, the slopes
of the tangent lines at (1,4) are equal, and, therefore, the curves are tangent at (1, 4). The hyperbola xy =
4 is the upper curve. The circular ring formula yields V= 77 tf {(4/x)
2 - [(x - 3)
2 ]
2 } dx =TT J
4 [I6x~
2 -
(x - 3)
4 ] dx = Tr(-\6x~
l - i(* - 3)
5 ) ]J = 77[(-4 - *) - (-16 + f)] = 2777/5 .
ydy. Let u=y-b, y = u + b, du = dy.
(u + b)du = 2ir
u du + b
V=2ir
ydy.