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asst3soln

# asst3soln - MATH 138 Assignment 3 Solutions Due by 11 a.m...

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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 t-an(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 y-axis, 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 y-axis, 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 564-2. 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 x-axis. 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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asst3soln - MATH 138 Assignment 3 Solutions Due by 11 a.m...

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