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Unformatted text preview: Ma 231
Exam 111
Version B
Solutions (1) (16 points) LetR be the region inside x2 + y2 = 25 and below the line x = 2 y — 5 . (a) Sketch the region (label any intersection points of the curves): Solution: Intersection: x2 + y2 = 25 and x = —2 y — 5
(~2y—5)2 +y2 =25, 5y2 +20y=0:>y=0(x=5), y=—4(x=3)
Intersection points: (5,0) and (3, —4) (3.4) (b) Write I I f (x, y) (M as an iterated integral integrating in y ﬁrst:
R Solution: [if 3 f (x, y) dyde (c) Write I I f (x, y) (M as a sum of iterated integrals integrating in x ﬁrst:
R 4 25—1" 0 —2_\‘—5 _4 Solution: I f(x,y)dx]dy+f: I f(x,y)dx]afv _ “€_‘~' dx (Do n_0t do directly) Solution: Switch to polar; ' 1 7r 2 r
{$17.73de dyzghrﬁldng Ht” [(2 l)
={[—"—'—;—] 5 111(5) _7rln(5)
2 d6— 7 0 0 I” l l l 1n(r3+l)
at. 2— _ =_ ‘=— ‘
(). Ir2+ldr 2[Nam ZIDMH 2 H
\__"—_4
Sub1u=rz+1 rdrzldu
2 1. 4
[ J L — l dx (Do my do directly)
‘5. y 16
(3) (16 points) Evaluate: J 0 Solution: Switch order: (4) (16 points) Determine the surface area of the portion of the graph 2 = 2x + y2 above
the region in the x, yplane between the lines x = y and x 2 3y with 0 s y s 1. Solution: : = 2x + y2 = g(x, y), (15' = ‘fg: + +1 dxdy = 44y2 + 5 dxdy
Surface Area = HER/4.1!: + 5 dy]dx = 2y 4y2 + 5¢1f1":‘é'(4y2 + 5)3 2 (I l 0 _27—5J§
— 6 , 1, 12 . , 4v3+532 ,
(*): Illa/4y +5dy=ZIu' 'du =Z(§113 'j+( =£'—6—)—~+( —‘,——J
Sub: u=4y2+5, 1
——du=2 dy
4 y (5) (10 points) Write f(x, y, z) dV as an iterated integral where E is the region in the
E ﬁrst octant below the plane 2x + y + z = 8. (Do _n_ot evaluate) Solution: Use : = constant slices 6—2
2 Slice range: 2 = 0 (given) to z = 6 (slice collapse) 6—y—z
6—2 mﬂwwq I Mama dy dz 0 Note: other 5 ways are equally easy (6) (10 points) Write I I I f (x, y, z) dV as an iterated integral where E is the region
E bounded by x2 + y2 = 9 and y2 + 22 =16. (Do n_ot evaluate) Solution: Easiest is y = constant slices Slice collapses when: y = i3 Jﬂf(x’y’:) dV :1 I [ f(x,Jv’,') dx:Id: dy
E ‘3 Note: slicing in x is harder; slicing in z is much harder (7) (16 points) Determine the volume of the solid E which occupies the region in the ﬁrst
octant inside the sphere x2 + y2 + z2 = 9 and below the cone 2 = ‘le + y: . Solution: Use spherical coordinates, Inside the sphere x2 +y2 +22 2 92> O s p s 3 In ﬁrst octant and below the cone 2 = ‘le +y2 :> 7r/4 3 ¢ 3 7r/ 2
First octant 2;» 0 $6 3 7r/2 Volume = dV = If [if p2 sin(¢) dp]d¢]d0 =
E 0 4 0 )r/ j d¢]d6 = ﬂjz[ﬂjz9sin(¢)d¢]d0 = "glaew t, H [b 3 _ ’7 __ o
13 79£d6=l87r 9N: "I(—9cos(¢)): d6 = If] 4 ...
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 Spring '08
 Toland
 ﬁrst octant, iterated integral integrating, constant slices

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