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Unformatted text preview: MATH 138 Assignment 3 Solutions Due by: 11 a.m., Friday, January 29. 1. Rotate the region bounded by y = sec at, y = 0, x = 0 and x = 7r/4 about
the x—axis and ﬁnd the volume of the solid that is generated. Solution: Referring to the ﬁgure, we take vertical slices of the region into
disk—shaped elements. The circular face of the disk at xi has radius sec(a:i)
and depth Ami, so has volume 7r[sec(:ri)]2A:ri. Summing a ﬁnite number of
these volumes gives an approximation to the volume of the region. In the
limit, that sum becomes the integral: 77/4
/ 7r secg(a:) dx.
0 Recalling that tan(x) is an antiderivative of secz(:r), we evaluate: 7T/4
/ 7rsec2(:r) dx = 7r[tan(:r)]g/4
0 7r[tan(7r/4) — tan(0)]
7T[1 ~ 0] 77'. II The volume of the region is 7r cubed units. $2 2. The parabolas y = 3:2 and y = —2— + 2, along with the y—axis, enclose a bounded region in the ﬁrst quadrant. When this region is revolved about the
yaxis, a parabolic lens is formed. Find the volume of the lens. I '2
7‘)? ﬁ+3w 7?" i 2 Solution: Referring to the ﬁgure, we construct volume elements as cylin—
drical shells at radius x with height (2 + x2 / 2) — m2. Each cylinder has
volume 27rx( (2 + x2 / 2) — x2)Ax. Integrating over these elements gives 2
Volume = / 27rx((2 + x2/2) —~ :32) dx
0 27r/02x(2 — x2/2)d:r 2
27r/ 2x — $3/2dm
0
27r[x2 — CHI/4]?)
= 27r[4 — 16/813
477'. The volume of the region is 47r units cubed. 3. Water sits in a bowl formed by revolving the region above y = x4 and to
the right of the yaxis, about the y—axis. If h is the depth of the water, what is the volume of the water. q x1 \/:\x b) / V 4“»;
if“ / Ur. , Wt}? Solution: Referring to the ﬁgure, we see that a horizontal slice at height h
yields a disk of radius y1/4. Given the height h of the water, the volume is then
h h
/7T(y1/4)2dy = W/ 2/1/2619
0 O h
3/2 0
27r FM” — 0]. The volume is then 331i?“ 2 units cubed. Alternative Solution: Alternatively, we can construct cylindrical shells as
volume elements. The shell at radius x has height h — m4 so the volume is given by h1/4 h1/4
/ 27rm(h~:r4)dx = 27r/ mh~335dm
0 0
m2 x6 hm
= 27Tlh37l0
2 h3/2 h3/2
— wig«Ti
27f
= __h3/2
3 7 conﬁrming our previous calculation. 4. Here is another “water in the bowl problem”. Water sits in a hemispherical
bowl of radius 30 cm, to a depth of 20 cm. Find the volume of water in the
bowl. Solution: Referring to the ﬁgure, we slice horizontally into disks with ra dius 302 — (y —~ 30)2 and ﬁnd the volume as /0 mm ~ (:1 — 30)2>2dy 20
n/ 302 ~ (y — 30)2dy
O 20
7r/ 60y — 1/2 dy
O
7T[30y2 — 1/3/9430
7r[30(20)2 ~ (20)3/3]
280mm
3 II II 28000w 3 units cubed. The volume of water is 5. The region inside the circle 1/2 + (:3 —~ 2)2 = 1 is revolved about the y—axis
to form a donut. Use the method of shells to ﬁnd the volume of the donut. Solution: Referring to the ﬁgure, we address the top half of the region and construct volume elements as cylindiical shells at radius ac with height
‘/ 1 —~ (m —— 2)? Integrating over these elements gives the volume of the
whole region (twice the upper half) as 3 .
2/ 27rxx/1—(33—2)2dx.
1 Substituting u = m — 2 gives du = dx while the limits 1‘ = 1 and x = 3 become 11 = —1 and u = 1, respectively. Thus we can write the volume as
1
47r/ (u+2)x/1 ~u2du =
—1
1 1
47r uV1~u2du+2/ \/1 —u2du>
—1 —1 The ﬁrst of these integrals evaluates to zero, since the integrand is odd and
the domain of integration is symmetric about the origin. The second evalu—
ates to 7r / 2, since it is the area of a semicircle of radius 1. We conclude that
the volume is 47T2 units cubed. 6. Decide if the following improper integrals of type 1 converge, and if they
do, ﬁnd their value. (20/0 m“ “” mummy“ <°’/0 e2x+3d$ Solution: a) Using partial fraction decompositon, we ﬁnd 1 1 1
($+1)(x+2) x—l—l 5642. We calculate /00 1 d l‘ b 1 d
—————— x 1m ——~————— :r
0 ($+1)($+2) (MOO 0 ($+1)($+2) b 1 1
b—voo 0 x+1—x+2
lim [1n Ix + 1 — ln LE + 2]]3
b—«roo dm II
E . [ x+1]b
2 11m ln
b—voo $+2 O
. b+1 1
— 12131010111 b+2‘—1n = ln2, which indicates that the improper integral is convergent.
b) Here we will use the substitution u = ln x that gives du = idm while the limits are modiﬁed accordingly. We ﬁnd /00 1 d 1' b 1 d
1
e x(lnx)2 a: (1—320 8 :r(lnsc)2 x “00 ln(e) U _1 ln(b)
lim ll b—«roo ’U, 1
_ 1, ——1 —1
— 1:330 b 1
= 1. This improper integral is convergent.
c) In this case we substitute u = 63”, so (111 = emdm and the limits are
modiﬁed accordingly. We ﬁnd 00 a: b a:
e . e
/ dcc = 11111 dx
0 e2w+3 (Hoe O e2w+3
5b 1
= 111—1320 1 112 + 3 du
1 6" 1
= — lim
3 (we 1 (uh/EV +1
1 6W5 1
= — lim 2 ﬁdv (v = 3 b—+oo l/ﬁ "U + 1
1 . _1 eb/\/§
= .gblirpo [tan (12)]1Ng
_ i  —1 b —1 )
— ﬂ (blirgiotan (e tan 
male
. /—\
(01>? l
03>}
v This improper integral is convergent. 7. Decide if the following improper integrals of type 2 converge, and if they
do, ﬁnd their value. 2 1 7T/2
a) / dx (b) / tanmdm
1 $"1 0 Solution: a) Since the integrand is not deﬁned at :1: = 1, we set 2 2
1 1
/ dm = lim dx
1 x——1 a—.1+a:1:—~1 and calculate 2
1
lim dx = lim [Inky —~ a—v1+ a (I? "1 (1—41+
= lim [lnl—lnla—lﬁ
a—vl‘l”
= 00. This improper integral is divergent. b) To integrate tan(x) == sin(x)/ cos(x) we substitute u = cos(:r) so that
du = — s1n(x)d:r and the limits m = 0 and :3 == 7r/2 become 11 = 1 and
u = 0 respectively. Then, since tan(:r) is not deﬁned at 7r / 2, we have 77/2 b
/ tanxdm = lim tanxdm
O b—«nr/Z” O
b
1
= — 11m «du
b—>O+ 1 U
1
1
2 11m wdu
b—*O+ b U
= 1' 1 1
3311111111,
= lim [1n1—~lnb]
b—>O+
== 00. This improper integral is divergent. 8. (a) If p is any real number and E is a small positive number, make a sub
stitution to prove that l (b) The integral / mp dx is improper of type 2 when p < 0. Show that
0 this improper integral converges if and only if the type 1 improper ' l 00 1 dt
1ntegra l W COHVCI‘gCS. (c) Using the above and the information of example 4 on page 511 of 1
Stewart, ﬁnd all p for which / mp dx converges.
0 Solution: a) We make the substitution :2 == 1/t which yields dx = {aldt
while the limits at = 1 and :r = 6 become t = 1 and t = 1/6. Then 1, 1 1 a
/:rpd:r / —— dt
6 1/6 tp t ll
' <
m
a.
3%
N)
R.
b,— as required. b) These impropcr integrals are equivalent to the integrals in part (a) in the
limit as 6 tends to zero. By part (a), these limits are equal. c) From page 511 of Stewart, we have that 00
1
/1 “t?” dt converges when p + 2 > 1, i.e. when p > ——1. From part (b), we conclude that 1
/ 56" d2:
0 converges under precisely the same condition: p > ~1. . For a constant k > 0 the region under y = 1 /:r’° and above the :naxis, from
x = 1 to 00, is revolved about the xaxis. This forms an object in the shape
of an inﬁnitely long horn. Find all k > 0 for which the volume created is
ﬁnite, and then ﬁnd those ﬁnite volumes in terms of k. Solution: Referring to the ﬁgure we take vertical slices of radius x‘k so
ﬁnd the total volume as 00 b
/ Mac—k)2 dm = lim Wilt—2k dx
1 b—+Oo 1
b SUI—2k
lim 7r [ ] provided k 7é 1/2 b—voo 1 — 2k 1
b1—2k
= l ~ 1
‘birgoﬂ [1 — 219 J
_ oo ifk < 1/2
“ 2kg ifk > 1/2
In the case k = 1/2 we have
b
lim 7w“1 dm = lim [1n b—voo 1 b—«roo
= blim [1n lbl — 1111]?
= 00. We conclude that the volume is ﬁnite for k > 1/2, in which case it has volume 21:11 units cubed. 10 ...
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This note was uploaded on 03/16/2011 for the course MATH 138 taught by Professor Anoymous during the Winter '07 term at Waterloo.
 Winter '07
 Anoymous
 Calculus

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