hw-15-sec-16.2-solns - Sec l6. 2 17. If we divide R into mn...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Sec l6. 2 17. If we divide R into mn subrectangles, ffR k dA z E 2 f (mg), yz‘j) AA for any choice of sample points (sz , yfj). . . 1 1:13: 468 CI CHAPTER16 MULTIPLE INTEGRALS ET CHAPTER 15 But f (z;- , 34173) = k always and Z 2 AA 2 area of R = (b — a) (d — c). Thus, no matter how we choose the sample i=1j=1 points, i i f(m:j,y;j) AAzk z i AA=k(b—a)(d—c)andso iz‘ljzl i=1j=1 fkadAz lim in; zf(x:j,yzj)AA= lim kg; iAAz lim k(b—a)(d—c)=k(b—-a)(d—c). m,n—>oo,L-=1 i=1 mgr—>00 izl j=1 m,'n,—>oc> 18. Because sin 7m is an increasing function for 0 g m g i, we have sin0 3 sin 71m 3 sin g => 0 _<. sin 7w _<_ Similarly, cos Try is a decreasing function for i S y S 5, so 0 = cos g 3 cos 7ry 3 cos g : Thus on R, 0 3 sin 7m cos 7ry S g - g = Property (9) gives ffR OdA S ffR sinvrw cos 7ry dA g ffR % dA, so by Exercise 17 we haveO S ffRsinn'a: coswydA S fi— (fi — 0) (l - l) = —1—. 2 4 32 16.2 Iterated Integrals ET 15.2 ‘ 3 1:5 _ @[512x2y3 dzr 2 [12 1% 343] : 4x3y3]:;: = 4(5)3 y“ —« 4(0)3 y3 2 5003/3, x=0 y 4 y=0 . 4 11:1 fol 12.132313 dy = [12x2 i] = 3302314] :1 : 3x2(1)4 — 3x2(0)4 : 3952 y=0 2 93:5 2. f05(y +xey)dm = [my + $399] 2 (53} + 2—256y) — (0+ 0) 2 5y + 27569, I:0 2 y=1 fg(y+$ey)dy:[%—+mey] =(%+mel)v(0+meo)=%+em—m y=0 @ff f01(1+4my)dmdy : ff [m+2m2y]::; dy = L30 +2y)dy= [W112]? = (3+9) — (1+1) 2 10 4. f01f12(4$3 — 9552112) dy dx = fol [4m3y — 3m2y3l1y1: d1” 2 fol [(8933 — 24332) — (4:53 _ 3932)]‘173 : f01(4a:3 — 21x2) da: 2 [11:4 — 7:733]:J = (1 — 7) “ (0 — 0) = ‘6 m2 2 7r . 7r - I 5. f02 f0 /2ms1nydyda:=f02mdmf0 /2s1nydy [asmExampleS] = [—cosy] =(2—0)(0+1)=2 0 o 6. [7:762 I: cosydmdy: f: dzc [7:762 cosydy [by Equation 5] 0 f5 £093 + wgdx dy = [xlillsinyliéi = [5 —<e1>1<sin% —sin%> : 6<1 — 9 = 3 2 2 9 36:1 / dy [substitute u = 2:5 + y => d1: = % du] o 1 2 9 9 1 (2+y)10 ylo 2 _ 2 _ d :_ _____ 18 0 [( +24) (0+y)l y 18[ 10 10 z:0 H 0 fl 1 [(410 210) (210 010)] k 1104112015 28 2 261,332 SECTION 16.2 |TERATED INTEGRALS ET SECTION 15.2 469 1 2 x 1 2 / / me dy dm : / acez dzc / l dy [as in Example 5] 2 [£1361 — 6313 [In |y| [by integrating by parts] 0 1 y 0 1 y : [(e—e) ~ (0— 1)](1n2—0) =1n2 4 2 m y) 4 1 1 2 11:2 4 3 1 2 3 4 9. —+— dydm2/ [m1ny+—~—y] dmz/ $1n2+— dx: —:cln2+—ln[w| /1/1 (3! m 1 ll 55 2 y 1 1 293 [2 2 h :81n2+%ln4—%1n2= 1—251112+31n41/2 : 2711112 1 10. f01f03 er+39 d3: dy : fol]: 618311 dz dy I 6:: dz]: 63y dy : [5631/10 2 (63 — 60)- % (e3 — (30) : §(e3 — 1)20r—;-(66 — 263 + 1) 1LflifiW-vfmwv:13gm—er3dv=éfi1u_m6—m_vmd. =wfluflfflflw=fl%u~w—%T =‘4—12[(0+1)~(1+0)]=o 11:1 12. folfol myx/zcz + y2 dydm = I; + y2)3/2] d1: :5 01 :c[($2 + 1)?”2 — $3] (19: = %f01[;r($2 + 1)?”2 — x4]dx y=0 1 =%HK+W”—fii=fiPW—L4+q=%@fi—U 13. fog]; Tsin2 0d6 d7“ = fog rdr f0" sin2 9d0 [as in Example 5] = I: TdT’ f0” %(1 —- cos 26) d0 2 E73]:- %[0 — %sin29]: = (2 — 0) - * %sin27r) — (0 v %sin0)] =2-%1(vr—o>—(0~o)1=vr s—l 14- folfolx/SHLtdsdt =f01[§(s+t)3/2] dt_ 5:0 1 I who % 01[(1 +0372 — t3/2]dt = §[§(1 +t)5/2 — §t5/2]0 =Ef—fi=%-3 16. ffR cos(m+ 2y) dA 2 f0" [OW/2 cos(ac + 2y) dy dx = f0" sin(:c + 234)]:3/2 d2: = % f0"(sin(m + 71') ~ sinzv) dm [—cos(:c+7r) +cosac]7r = é[~cos?7r+c0s7r —— (—cos7r +Cos0)] o NIH =§(—1—1‘(1+1))=—2 1+$2 /1/1 1+w2 /1 2 /1 1 |: 1 3]1 _1 1 18. dA = d dx: 1+m d2: d = :c+——a: tan f/Rl'1'1/2 0 0 1+yz y 0( ) 0 1+yz y 3 0[ do SQC MHZ 470 D CHAPTER 16 MULTIPLE INTEGRALS ET CHAPTER 15 19. for/6 J/smsin(:c+y)dydm __ 1r/6 — 0 [—2c0s<$+y>1::3/3 den = fo"/61mcosm—zcos<m+ a] dz = :1: [sinar — sin (w + 3/6 — far/6 [sin x — sin (a: + d9: [by integrating by parts separately for each term] =%[%—11—[—cosm+cos(m+">13/6=—e—Pew—(4%)] =fi;—1—% 1 1 1 1 :L' I :1 20. dA=// ddLBZ/ 1n1+zv y dmz/ ln1+m —ln1dcc //1;1+$y 0 0 1+3”?! y 0 [ ( y)]y:0 o [ ( ) i = [01an + at) dam = [(1 + ac) ln(1 + w) — at]; [by integrating by parts] WI :(21n2—1)—(ln1-0):2ln2—1 32:1 2 @ffrz $96221, dA: f02 fol 933183521] d3: dy : fr? [égrzy] dy = f02(ey ” 1) dy : %[ey _ yio 2%[(e2—2)—(1—0)]= (e2 —3) bah-a Nil" 1 2 x $22 22. / / $2 + W dm dy 2 L,1 [g 111(272 “flux:1 dy 2 H01 [1n(4 + f) _1n(1 + y2)] dy 0 1 22; 4+7;2 To evaluate the first term, we integrate by parts with u = ln(4 + 3/2) :> du = dy and dvzdy : v=y.Then 2y2 2 8 142d214+2#/ d:l4 —/2— d /n(+y)y yn( y) 4+y2 y yn( +21) 4+y2 y = yln(4 + y2) — 2y + 8: gum—10%) 2 yln(4 + 3,12) — 2y +4tan_1(%) Similarly, f1n(1 + yZ) dy : yln(1 + 3/2) — 2y + 2tan-1 3/. Thus, 12 x _ 1 ddzl/14+2—11+2d y/Oflmzflfimy 20[n( y) n( y)]y 1 1ylo = % [1111191 +. 212) e 21/ + Man—102‘)— 111110 + 92) + 274 ‘ 2m“— : %[(ln5+4tan‘1(%) ~ 1112 — 2tan_1 1) — 0] §[ln5 — 1n2+4tan—1(§) 7 2(3] = §1n3+2tan—1(§) — g Z 4 23. z=f(;v,y) =4—m—2y20for03wg landOSyg 1. Sothesolid is the region in the first octant which lies below the plane 2 : 4 — x — 2y and above [0,1] x [0,1]. SECTION 16.2 ITERATED INTEGRALS ET SECTION15.2 l: 471 Z22—at2—y2ZOforOSmgland0§y<1.Sothesolidisthe region in the first octant which lies below the circular paraboloid z 2 2 — 3:2 — y2 and above [0,1] x [0,1]. X 25. V 2 ffR(12 a 3m — 2y) dA 2 f: f01(12 — 3m — 2y) dx dy 2 f: [12x — €932 — 2$y]I r20 :4fo2 liv— fiws — éyszZS dy=4f02 (% - $312) dy=4l%y- 2-17y3l0 =4 2-3 = $79 @V 2 fil foflfl +6z siny) dydm 2 f_I1[y — 6T“ cosy]::;r dw 2 fi1(7r+er — 0+ez)d$ 2 fi1(7r+ 265cm 2 [7rac+2675]:1 2 27r+26 —% 29. Here we need the volume of the solid lying under the surface 2 2 $ sec2 y and above the rectangle R 2 [0, 2] X [0, 7r / 4] in the nay-plane. 7r/4 V 2 [02 0M4 itsec2 ydy div : fez 5” d5” Inf/4 secz ydy I léwzlrz) [tan ylo 2 (2 — 0)(tan% — tan0) 2 2(1 20) 2 2 30. The cylinder intersects the any—plane along the line 1v 2 4, so in the first octant, the solid lies below the surface 2 2 16 — m2 and above the rectangle R 2 [0, 4] X [0, 5] in the Lug-plane. 4 V 2 [05f;(16 -— 2:2) dm dy 2 f04(16 ~ 3:2)(11: fosdy 2 [16m — éatflo [y]:=(64~%—0)(5—o):§§_0 31. The solid lies below the surface 2 2 2 + $2 + (y — 2)2 and above the plane z 2 1 for —1 g x g 1, 0 S y S 4. The volume of the solid is the difference in volumes between the solid that lies under z 2 2 + m2 + (y ~ 2)2 over the rectangle R 2 [~1, 1] X [0, 4] and the solid that lies under 2 2 1 over R. V 2 [04f_l1[2 + $2 + (y — 2)2] dwdy — f04f:1(1) dmdy 2 I: [2x + §x3 + w(y — 2)2]::1_1 dy — f_11 dx f:dy = I: [<2 : §+ (y 2)2) < 2 % (y 2m] dy — [$111 [y]: = I: [% + 2<y — 2?] dy - [1 — (—1m4 — o] z [1741/ + §<y — 2F]: — <2)(4) —[(%6+1:) (0 an 8:? 8:634 ...
View Full Document

This note was uploaded on 05/02/2011 for the course MATH 408D taught by Professor Chu during the Spring '09 term at University of Texas at Austin.

Page1 / 4

hw-15-sec-16.2-solns - Sec l6. 2 17. If we divide R into mn...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online