Unformatted text preview: {V} ’2 I," (a) e 244, w 91$: 37. The surface S is .given by z = f(m, y) = 6 — 3:1; — 2y which intersects the any—plane in the line 3a: + 2y = 6, so D is the triangular region given by {(a:,y) I 0 S a: 3 2,0 g y S 3 — gm}. By Formula 9, the surface area of S is 1 <=S> // 1+<3—>+(Z:> dA
=ffDWdA=¢ﬁffDdAzmA(D)=\/ﬁ(—;23):Bx/ﬁ. 38. z : f(a:, y) = 10 — 2a; — 5y and D is the disk (1:2 + y2 S 9, so by Formula 9, M) = MD V1 + (—2? + <~s>2 dA = «56m em = «Mm = «37w  32> = Wm 39 Z=f(w,y)= §($3/2+y3/2) andD={(w:1/)IOS$SLOSZJ31}. Thenfw =961/2,fy=yl/2 and =ffD 1+ (f) +( y)2dA=folfolx/1+w+ydydw =f01 i§(“+y+1)3/2iy=1dm §fol [(m+2)3/2— (w+1)3/2]dz y=0 1 =%[§<z+2>5/2—§<x+1>5”]=é<35/2125/2 25/2+1>=~3(35/2—2"2+1> 0
40. ru = (0,1, —5), r1, = (1,—2,1V), and r“ X r1, = (#9, —'5, —1). Then by Deﬁnition 6,
= ffp ru x r91 (M = fulf01(—9,—5,—1) dudv = x/107f01 du fol d1) = V107 41. z:f(m,y)=mywith0$a:2+y2Sl,sofm=y,fy=a; => =ffDmdA=f§WJW2+1TW9=f0" l%<+1>3”i:: d0
:O2n1(2f—1)d0=23£(2\/—— 1) 42. z = f(a:,y) = 1 + 337+ 2y2 withO S as g 2y,0 S y S 1. Thus, byFormula 9, 11(5) = Up 1 + 32 + (4y)2 dA = [J 029 1/10 + my dm dy = fol 2y J10 + my dy 1
= ﬁ . §(10 + 16y2)3/2] : 511(263/2 ~ 103/2) 0 ...
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 Fall '08
 Napier

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