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Unformatted text preview: Vincent (jmv777) m HW12 w Gilbert — (57495) 1 This print—out should have 20 questions. is the volume of the solid below the graph of
Multiple—choice questions may continue on f having the rectangle
the next column or page m find all choices before answering. A = {my} ; 0 g 1; g 4, 0 g y g 4} 001 10‘0 pOiﬂtS for its base. Thus the solid is the wedge The graph of the function 2 : ftyay) m 7_$ is the plane shown in and so its voiuine is the area of trapezoidal
face multiplied by the thickness of the wedge.
Consequently, 'I = 80 cu. units ,
Determine the value of the double integral ' I m f / any date!
A H ) y keywords: over the region _
002 10f) points A m {(m,y):ogazgzi, egg/54}
Find the value of the integral in the say—plane by first identifying it as'the 2
volume of a solid below the graph of f. I 2 f f (a: y) day
0
1. I = 82611. units
when
2. I m 80 cu. units correct ﬂax, y) m 455: ~ 53323;.
3. I = 81 cu. units 40 _
1. I = 8+ 3:!) 4. I z: 79 cu. units 2.1 e 4. m5 2
5. I m 830:1. units 9 y Explanation: 3 I __ 8 40
The double integral  w "H "gay correct I m j [A meme 4. — 8+£§y2 L...‘
l Vincent (jaw???) — HWIZ —— Gilbert w (57495) 3 Consequently, ‘ III!!! 005 10.0 points Evaluate the double integral 1+3:2
In dd
[[41+3JZ 33y when
m {(ay) 0<w31, ogygl} 1 1. I 2 "2‘71"
1 2. I m 67f
E" 3. I 2 gr 4. I = ":7?
1 5. I n gt” correct Explanation: Since A = {(3:,y):0§23$1, Ogygl} is a rectangle with sides parallel to the coor
dinate axes, the double integral can be repre»
sented as the iterated integral 111+$2
Im dd.
fe<fe Maggy Now
1 2
1+3: 1 1 3 3
d a [ m . .
[0 1+3}? 3: 1 y? t+3m10
Thus Consequently, 1
Izm— 7T . 806 10.0 points Calculate the value of the double integral I = // 2msin(cc+y)dmdy
A . when A is the rectangle {(m,y):05mg%, Ogygg}. 1. I = iﬁﬁ+2(ﬂ“l) 2.1 a iaﬂ~2(\/§mi) 3. I = 2(\/§m1)+—:w(2—V§) 4.1 = Ell—ﬂx/é—Zb/E—ll) 5.1 x 2(\/§+1)m»:«1r(2+\/§)
1 205 — 1) ~ _7r(2 ~ ﬂ) correct . I
6 4 H Explanation:
By treating I as an iterated integrai, inte~
grating first with respect to as, we see that / 2m sin(x + y) 013: m —22: cosh: + y) + 2/ cos(:c +y)dac = 2( may eosm + y) + sin(.’1: + 11)) after integration by parts. Thus WM . 1 7r
/0 2m51n(:c+y) dzt = w~~2~7rc30e(~&w +31)
. 75" .
+2s1n(1 +y) — Zsmy. vincent (jaw???) w HWIB m Gilbert w (57495) since
cos2u 2 Ecoszu.m1.
Consequentiy,
1 1 77/6
I = 2/ y[u+wsin2u] dy
0 2 0
1 i
z 2 hf: [W2] ,
6 4 2 e 009 10.0 points Evaluate the iterated integral I m f:{fyﬂ 9ydm}dy. . 6
1. I m —
5
2. I r: 3
9
3. I = —
5
12
4. I 2 m»
5
3
5. I m ~5 correct
Explanation: AS an interated integral, integrating ﬁrst
with respect to x, we see that .I m £1[9xy]:§dy 3
m 9/ (213/2~y2)dy
0 Consequently, [l CHICO 010 10.0 points Find the value of the integral I m f/A{2(cc—2)—3y}dmdy when A is the region {(w.y):osys:cw2. 29:4}. 7
1. I = w 3
2. I = 2 8
3. I = _. 3 5
4. I m — 3
5. I m georrect
Expianation: The integral can be written as the repeated
integrai I m fmmmaamamagywy} day. Now . x—Z
[0 {2&3 ~ 2) m em 3 2 = Plaid)?” “2’9 if? Consequently I = [24;(x2)2dm = [éedﬁ]. and so
4
I m —
3 011 10.0 points Evaluate the double integral 89
I:
f/D m2+1dmdy Vincent {jmv777) — HVV 12 —— Gilbert * (57495) . 7 Thus ' Explanation:
7 The volume of the solid under the graph of
I = f {in 7 m In 3;} d3; f is given by the double integral
. i '
7 =
:ﬁln’?——[azln:cwm]l. V f/Af(az,y)dmdz,
Consequently, which in turn can be written as the repeated
integral V 2 6—1137. 013 10.0 points f: (fay/m 82129013,!) disc. N OW the inner integral is equal to m ﬂing) x 893g [4ey2]0 z: 4m(4 w 3:9). rI‘he graph of over the bounded region A in the ﬁrst qued— Thus rant enclosed by
2 W Vm4/02m(4—x2)am=[menaﬁﬂd and the m, y~axes is the surface y Consequently, Volume = 16 cu. units. 014 10.0 points Evaluate the double integral Im /L(3m+2y)dxdy when A is the region enclosed by the graphs Find the volume of the solid under this graph Of :2: = 1, m m y m 1’ y m 1'
over the region A. 1. Volume 2 32 cu. units , 1. I = Scarred; 2. Volume m80‘u. units 2. I = g, 3. Volume 221604. units correct 3. I m 3 4. Volume 3 4cu. units 4_ I = 2 . {xvVolume m g on units 5. I = Z 3 3 vincent (jinx/777) w HW12 w Gilbert M (57495) 9 _ Thus 7?
[2/ (27rm233msinsc)dm
U 2 71'
= [21m —— a: + cosmL).
' Consequently, I I=7r2w . 016 10.0 noints Find the volume of the solid in the ﬁrst
octant bounded by the cylinders $2+y2=16, y2+z2 16. ll Hint: in the first octant the cylinders are
shown in 128
1. volume = Mg— cu. units correct 2. volume at 44 cu. units 124 3. volume m —3— cu. units
116 4. volume : —— cu. units 3 5. volume = 40 cu. units Explanation:
As the ﬁgure shows, the solid in the first
octant bounded by the cylinders x2+y2 m :6, y2+32 16 is the soiid below the graph of
z = 16 —— yg
above that part of the circte
2:2 + 3,12 = 16
lying in the first quadrant of the xy{plane Thus the volume of the solid is given by the
double integra} V m // x/16my2dzcdy
A Where A is the region in the ﬁrst quadrant of
the w—y plane bounded by the quarter—circle {(x,y):0_<_'xg x/l6—y2,{)§yg4}, and so V can be represented as the iterated
integral V w [)4{/0W Wdac}dy. In this case,
4 16— 2
V2/[93x/16my3L) ydy
0
4
=/ (16~92)dy
G
Consequently, "42 — cu. units .
0 3 1 3]4 128 [mi/‘39 017 10.0 points Vincent (jmv???) w HVV12  Gilbert — (57495) 11 4. I m /:( 25 f(.’13, y)dm)dy ‘x/g
5.1 = /.25(/Gﬂ f(.r,y)d:c)dy o
n w e. I z] (/ f{m,y)d$)dy
0 5 Explanation: The region of integration is the set of all
points hayhﬁﬁySmKOSmg5} in the plane bounded by the ar—axis and the
graphs of This is the shaded region in (not drawn to scale). Integration is taken ﬁrst
with respect to y for ﬁxed :3 along the dashed
vertical line. To change the order of integration, first ﬁx
y in the interval [0, 25] and let CC vary along
the solid horizontal line. Then m varies from
V5 to 5. Hence, after changing the order of
integration, keywords: double integral, reverse order inte—
gration, parabola, 019 10.0 points Evaluate the double integral I = ff 43/8932 dmdy
A when A is the region in the ﬁrst quadrant
bounded by the graphs of N.
5...!
H
CD 9"
>...(
ll (e m» 1) correct
4. I = 2(em l) 5.1 m 4(e+r) Explanation:
2 Since the function ﬁx) = em cannot be
integrated directly, we have to represent I
as a repeated integral, integrating ﬁrst with
respect to y, where the region of integration
is similar to the shaded region in Thus as a repeated integral fol/f 1/2 4318552 rig) d1: and so 1
Im
fog+ Consequently, 2y
y2 dy Vincent ﬁrm/T777) — HVV 12 w Gilbért ~ (M495) 1 = [1n(9+y2)]0. l3 ...
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 Fall '09
 Gilbert

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