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Unformatted text preview: Ma 231
Exam III
Version A
Solutions (1) (16 points) Let R be the region inside x2 + y2 = 25 and below the line x 2 3y + 5 .
(a) Sketch the region (label any intersection points of the curves): Solution: Intersection: x2 + y2 = 25 and x = 3y + 5
(3y+5)2+y2 =25,10y2 +30y=0=>y=0(x=5), y=—3(x=—4)
Intersection points: (5,0) and (—4, —3) (4.3) (b) Write If f (x, y) dA as an iterated integral integrating in y first:
R Solution: J:  I ‘ f(x._1')dy]dx
4 “\[~—— (c) Write If f (x, y) dA as a sum of iterated integrals integrating in x ﬁrst:
R 0 ‘15—‘11: _1 Zia3
Solution: j j f(x,y)dx dv+j j f(x,y)dx]dfv
4 3.1% _< _JET: dy dx (Do M do directly) (2) (16 points) Evaluate: I ‘ _
0 _J.;_—‘" + y + Solution: Switch to polar: 2 4x3 1 n2 2 l'
I I "2—f—dy d": I [I,.z+ld"]d‘9 0 _,——4_x2x +y +1 _,,3 0 1; I (1110.2 +1)] d6 = ”I ln(5)d6 : ”111(5)
_,, 2 0 —Ir 5 2 ln(r2 +1) +(1
2 l6
(3) (16 points) Evaluate: I O 4 J l—cos(x) dx
5 x Jail) (Do £29.! do directly) Solution: Switch order: 4 = (8 —4sin(4) — cos(4)) — (0 —~ 0 — l) = 0 [x7 — xsin(x) — cos(x)]
9  4sin(4) — cos(4) (*) : Ixcos(x) dx = xsin(x) ~ Isin(x)dx : xsin(x) + cos(x) + C y__~___4
puns: u =x, du 2dr
dv=cos(x)dr, v=sin(x) (4) (16 points) Determine the surface area of the portion of the graph 2 z x2 + 2 y above
the region in the x, yplane between the lines y = x and y = 2x with 0 S x s 1 . Solution: 2 = x2 +2y=g(x,y), d5 = ,lgi +gf. +1 dxdy = \/4x2 +5 dxafv
l 21‘ l a. I 1
Surface Area = I“ \/4x1 +5 aft]dx = I x 4xI +5alx=i(4xI +5)‘ 12 l 0 x 0 _27—5J§ 12
. 2 _l 1: _123: w_(4x:+5)3I w
(an). Ix 4x +5dx—éJ'u du#§[§u J+C —~—l—2———+(
Sub1u=4x2 +5,
lduzxdx
s (5) (10 points) Write ”I f (x, y, z) dV as an iterated integral where E is the region in the
E ﬁrst octant below the plane x + 2 y + z : 6. (Do n_0t evaluate) Solution: Use .. = constant slices Slice range: 2 = 0 (given) to .1 = 6 (slice collapse) 6—2 6.2y—z Ill/(x’y’z)dV=i H I f(x,y,2)dx]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 x2 + 22 = 4. (Do 101 evaluate) Solution: Easiest is x = constant slices Slice collapses when: x = i2 Ill~f<xnuzld"zj[ I [ J f<x,,v,z)cb’]dz]dx Note: slicing in y 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 above the cone; 2 ,/x: + y: ‘ Solution: Use spherical coordinates, Inside the sphere x2 +y2 +22 :9 :> 0 s p s 3 Above the cone: = ‘le +_v2 :> 0 s (1) s 7r/4 First octant 2036’ s 7r/2 Volume = j j j 1 (IV = Trfﬁ p2 sin(¢) dp]d¢]d0 = 0 ”Iz[”f[_3:sin(¢)] d¢]d6= ”jzrjssmwwdda:
"18— 95%! =18n—9mfz’ ”IZ(—9cos(¢))‘: (102 J:———— 4 ...
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 Spring '08
 Toland

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