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Unformatted text preview: 00:1; atﬂs Version UG4/AAABA — DGQlZc —— Gilbert — (78759) 1 This printout should have 3 questions.
Multiple—choice questions may continue on
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before answering. 
:l¥l unu
nan 001 10.0 points
Evaluate the triple integral Jeff/mz2w over the points (a: y,z in the rectangular box
B m[0,2]><[2,3]><[0,}1].
LI :2 2 correct
2. I = l
5
3. 2 ~
I 2
4. I x: 3
3
5.1 m —2—
Explanation: Since B consists of all points (as, y, 2) such
that Osm<2, 25.1153, I can be written as a repeated integral 1 e [/61 (EU: 3mz2dx)dy)dz. NOW 2 2
/ 333320513 = [—33x222] = 622,
0 2 0 White
3 3
/ 622dy r: [6y22] m 622.
2 In this case, Consequently, 002 10.0 points
The solid E shown in 1? is bounded by the graphs of 2 Elwin; $+y:1) and the coordinate planes :3, y, z m 0. Write
the tripie integral ﬂ/[V/Eﬂmywdl/ as a repeated integral, integrating ﬁrst with
respect to 2, then y, and ﬁnally :r. 1. 111(/1:(/01—$2 f(3:, y, z)dz)dy)dm Version'OOZL/AAABA — DGQIZC  Gilbert ~ (78759) 2 2. f; (/10 (Al—~33? f(9:, y, z) dz)dy)dm 6' f m ”(816 ml) ma: Explanation:
In Cartesian coordinates the region of inte 0 (3 . 0
{($111) : OSyS 16_.$2a 033334}? 1 1—55 1mm2
4' [0 (/0 (A “33’ y, z) dz) dy) dm which is the shaded region in
correct _ Explanation: Because E is bounded by the coordinate
planes :3, y, z m 0, the graph of the solid
shows that E consists of all points (as, y, z) in
3~space satisfying both the inequalities 3:,y,z Z 0, as well as 0. g 2: g 1‘332, 0 S y S l——:c, O S a: S 1' The graphic shows that in polar coordinates
the region of integration 18 Consequently, as a repeated integral, {(né’) = osrs4,osesw/2}. 1 lwm 1—332
I 2 A (/0 (/0 flx’ l ’ Z) dz) ally) dx Thus in polar coordinates 4 7r/2 2
..mwmm____._.__m_m I = f j 26?" Tdﬁdr
keywords: choker G 0
4 2 l 16
003 10.0 points x a/ “re?" d?" m gar/l e“ do,
~ 0 0 Evaluate the iterated integral
setting a m r2. Consequently, 4 v‘lﬁwm2 2 2
I = f / 2e”; +3 dyda: 1
o .0 I = §W(826_1)
by converting to polar coordinates.
1. I m é—Mel6 w 1) correct 2. I m gaﬁflml) ...
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 Fall '09
 Gilbert

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