15_7.pdf - FromEquations l,m=rcos0=4cosg =4%=2...

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Unformatted text preview: FromEquations l,m=rcos0=4cosg =4-%=2, y=rsino=4s‘ g: A? =2\/§,z=—2,sothepointis (2, 2J5, —2) inrectangular coordinates. a: = 2cm(—§) = 0, y = 2sin(—§) = -2, and z = 1, so the point is (0, —2, l) in rectangular coordinates. so the point is (—1, l, 2) in rectangular coordinates. x: lcosl =cosl,y= lsinl =sinl,andz= l, sothepointis (cos l,sin l, 1) z (0.54, 0-84,1) inrecmngular coordinates. 4. (3)1'2 = (—~/§)’ + (JEY =4sor=2; tan0= 3% = —1 andthepoint (—J§,¢§) isinthesecondquadrantofthe try-plane, so 6 = 3—: + 2n7r; z = 1. Thus, one set ofcylindrical coordinates is (2, 37", 1). (b) 1'2 =22+22 =850r=\/§=2\/§; tan0= % = landthepoint(2,2)isinthefirstquadmntoffllezy-plane,so = § + 21m; 2 = 2. Thus, one set of cylindrical coordinates is (2J5, f, 2). . Sincer = 2, the distancefromanypointtothe z-axisis 2. Becausefiandzmayvary, the sm'filceisacircularcylinderwith radius2andaxisthez-axis. (See Figure 4-) Also, 2:2 +312 = r2 = 4,whichwerecognizeasanequationofthis cylinder. . Since 0 = % but 1' and 2 may vary, the surface is a vertical plane including the z—axis and intersecting the zy-plane in the line y = 7152. (Here we are assuming that r can be negative; ifwe restrict r 2 0, then we get a half-plane.) . Sincerz+z2 =4and1'2 =:z:2+y2,wehave:z:2 +1112+z2 =4,aspherecente1edattheoriginwithradius2. 2 =1'2 c) z =32+y2,acirc1flarparaboloidopeningupwardwithvertexfl1eorigin, andz = 8—r2 a z = 8— (:1:2 +112), acircnflarparaboloidopening downwardwith vertex (0, 0, 8). The paraboloids intersect when 'r2 = 8 - 1'2 c) r2 = 4. Thus 1'2 5 2 S 8 — 7'2 describes the solid above the paraboloid z = :52 + y2 and below the parabolaidz=8—a:2 —y2 for22+y2 S 4. 17. Incylindricalcoordinates,Eisgivenby {(r,0,z) | 0 S 0 S 21r,0 S r S 4, —5 5 2 S 4}. So fffz W2 +112” = [02' f: ff; WNW“: 12"“ 104’”? [3st = [013" [as]: [2115 = mus—34x9) = 384w 21. Incyhndticalcoordinates,Eisbmmdedbythecyhnderr= l,theplanez = 0,3ndtheoonez =2r. So E={(r,0,z)|0$0$27r,05r$1,05252r}and z’dv= 2" ‘ 02'2rzoos 0rdzdrd0= 0’" r3 003 oz “2’ drd0= 2" 12r4c0320drd0 fffE f0 0 01 1:0 0 0 =fo* [§:r cos 0:;do=§ 02" cos’odo=§ :*l(1+oos2o)d9=;[o+-sin2o]§' =? 27. The paraboloidz = 412 +4312 intersects the planez = awhena = 41:2 + 4y2 on:2 +312 = %a. So,incylindrical coordinates,E= {(r,0,z)|0§r§ lfi,og 93211341'2 gzga}.Thus 21l- f/2 2r f/2 m: / /: / Krdzdrdfi: K/ / (ar—4r3 )drdO 4!-2 Since the region is homogeneous and symmetric, Mw— — Mg.z = 0 and 2K J-[2 21r \/-/2 M,,=/ / / Krzdzdrdfi: K] ./o (-a2 r—8r5)drd0 4,2 _ 2r =K02 [la2r2—-rs]:;o fifldfl: K/ 2—14a3 Huce (5, 5,3): (0,0, 2a). 29. The region ofintegration is the region above the cone 2 = z’ + y’, or z = r, and below the plane 2 = 2. Also, we have —2 S y S 2with —‘/4—y2 S a: S Mil—y2 whichdescribesacircleofmdius2inthezy-planecolte1edat (0,0). Thus, /_: /‘/::;v:2 szdzdzdy= /2"/2/2 (roosfl)zrdzdrd0= /2fi/2/2 r2 (coso)zdzdrd0 _ :1'202cr2(060)[;2]z_2drd0—2f r2(c089)(4—r2)drdfl Z—f "cosfldo f: (4r2 —r4) dr:— 2 [sin0]2’r gr 3——r5]o= 30. The region of integration is the region above the plane 2 = 0 and below the paraboloid z = 9 — 22 — 312. Also, we have —3 S a: S 3with0 S y S vg— 22 whichdesaibestheupperhalfofacircleofradius3inthezy—plane centeredat (0,0). ‘15, 3 «9—9 9—=’—u’ 3 9—r’ 1r 3 9—r’ // / \/x2+y2dzdydx=/w// Vr’rdzdrdasz/ rzdzdrda _3 o o o o o o o o =ff03r2(9—r2)drdo=fi,“ dflj:(S}r2— 4)dr =[010 [3" ‘ _r5]o_ — 7' (81 — 2—33)_ — Tn“ ...
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