Quiz 12 - Version 004/AAABA — DGQch “1 Gilbert m(78759 1 This print—out shoulé have 3 questions Multiplewchoéce questions may continue on

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Unformatted text preview: Version 004/AAABA — DGQch “1 Gilbert m (78759) 1 This print—out shoulé have 3 questions. Multiplewchoéce questions may continue on the next column or page - find all choices before answering. 1 2 3 4 5 6 as an m 001 10.0 points Evaluate the triple integral Jeff/£33mng over the points (3:, y, z) in the rectangular box B m [0,2] X [2, 3] X [0, 1}. 1. I 2 2 correct 2.I r: 1 5 3.1 —- é 4.I = 3 3 5. I —— E Explanation: Since B consists of all points (at, y, 2) such that 03,231, was; 23;:53, I can be written as a repeated integra} I =/01(/:(/023mz2dm)dy)az. Now 2 2 / 311222053: 2 [gmgzz] = 622, 0 2 {3 While /: 622dy : [@232]: = 622. In this case, Consequently, 002 10.9 90133133 The solid E shown in x is bounded by the graphs of 2 z=l—-$, m+ym1, and the coordinate planes :9, y, z = 0. Write the tripie integral I: ff/Eflagyfldlf as a repeated integrai, integrating first with reSpect to 2, then 1 , and finally :3. 1. (fol—$2 f(a:,y,z)dz)dy)dm "x Version_004/AAABA m Doom: "- Gilbert — (78759) 2 ' 2. few/1‘0 (/01fo f(:1:, y3 z) dz)dy)dm ‘ 6' I : Mel's “1.). Mm Explanation: In Cartesian coordinates the region of into 1 1mm2 1wy . . 3° / (f ( ' f($;y,z)dz)dy)dm gratioms I O 0 {3 2 {Law:Ogygx/mficz’ogmgtl}? l 1—$ lmx 4' f0 (/0 (f0 We 9’ ’3) dz) ‘1de53 V 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 soiid shows that E consists of all points (as, y, z) in 3—space satisfying both the inequalities $?yflz 2 0} as weil as 0 S 2 S 1~$23 g gig 1mm: 0 S m S I. The graphic shows teat in poiar coordinates - the region of integration 18 Consequently, as a repeated integral, 1 l~m l—$2 ' I. I _:. f0 (f0 [G _- ft”, 9? 3) d3) d9) 0:53 Thus in polar coordinates ' __ : _ ' 4 fl/Q 2 - I = f f 231* macaw keywords: clicker ‘ 0 0 2 1 . 16 n/ re?" ctr m ~7r/ ends, 0 3 0 . setting a 2 r2. Consequently, {(r,9):0£r§4,059§n/2}. 003 10.0 points Evaluate the iterated integrai 4 V16me 2 2 I 2 f f 23‘” +9 dyda: 0 0 by converting to polar coordinates. E I :3 27r(616- 1) 1 . iflem —— 1) correct 2. I m %w(e’1~l) 1.1 EE 3. I = 27T(816m 1)‘ 4. I m 77(64— 1) 5. I =. 27r(e4 — 1) ...
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This note was uploaded on 03/24/2010 for the course M 57500 taught by Professor Gilbert during the Fall '09 term at University of Texas at Austin.

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Quiz 12 - Version 004/AAABA — DGQch “1 Gilbert m(78759 1 This print—out shoulé have 3 questions Multiplewchoéce questions may continue on

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