Quiz 12 a - 00:1 atfls Version UG4/AAABA — DGQlZc ——...

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Unformatted text preview: 00:1; atfls Version UG4/AAABA — DGQlZc —— Gilbert — (78759) 1 This print-out should have 3 questions. Multiple—choice questions may continue on the next column or page — find all choices 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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