# Test 3 - Version 102 – Exam 3 – Gilbert – (59825) 1...

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Unformatted text preview: Version 102 – Exam 3 – Gilbert – (59825) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 10.0 points Suppose (1 , 1) is a critical point of a func- tion f having continuous second derivatives such that f xx (1 , 1) = − 3 , f xy (1 , 1) = 5 , f yy (1 , 1) = − 5 . Which of the following properties does f have at (1 , 1)? 1. a local minimum 2. a saddle point correct 3. a local maximum Explanation: Since (1 , 1) is a critical point, the Second Derivative test ensures that f will have (i) a local minimum at (1 , 1) if f xx (1 , 1) > , f yy (1 , 1) > , f xx (1 , 1) f yy (1 , 1) > ( f xy (1 , 1)) 2 , (ii) a local maximum at (1 , 1) if f xx (1 , 1) < , f yy (1 , 1) < , f xx (1 , 1) f yy (1 , 1) > ( f xy (1 , 1)) 2 , (iii) a saddle point at (1 , 1) if f xx (1 , 1) f yy (1 , 1) < ( f xy (1 , 1)) 2 . From the given values of the second deriva- tives of f at (1 , 1), it thus follows that f has a saddle point at (1 , 1). 002 10.0 points Which, if any, of the following are correct? A. If I = integraldisplay integraldisplay x 2 + y 2 ≤ 1 radicalbig 1 − x 2 − y 2 dxdy , then I = 4 π/ 3 . B. If I = integraldisplay integraldisplay x 2 + y 2 ≤ 1 f ( x, y ) dxdy , then I = integraldisplay 1 integraldisplay 2 π rf ( r cos θ, r sin θ ) dθdr. 1. B only correct 2. both of them 3. A only 4. neither of them Explanation: A. FALSE: the integral is the volume of the upper hemi-sphere of the sphere of radius 1 centered at the origin, hence has value 2 π/ 3, not 4 π/ 3. B. TRUE: when changing to polar coordi- nates, dxdy = rdθdr . 003 10.0 points Evaluate the triple integral I = integraldisplay integraldisplay integraldisplay E 4 xdV, Version 102 – Exam 3 – Gilbert – (59825) 2 where E is the set of all points ( x, y, z ) in 3-space such that ≤ y ≤ 4 , ≤ x ≤ radicalbig 16 − y 2 , ≤ z ≤ y . 1. I = 257 2 2. I = 255 2 3. I = 259 2 4. I = 129 5. I = 128 correct Explanation: As an iterated integral, I = integraldisplay 4 integraldisplay √ 16 − y 2 integraldisplay y 4 xdzdxdy = integraldisplay 4 integraldisplay √ 16 − y 2 4 xy dxdy = integraldisplay 4 2 y (16 − y 2 ) dy . Consequently, I = 128 . keywords: 004 10.0 points If E is the solid wedge that is cut from the cylinder x 2 + y 2 = 9 by the planes z = 0 and z = y , express the volume of E as an iterated triple integral. 1. volume = integraldisplay 3 integraldisplay √ 9 − x 2 integraldisplay y xdzdydx 2. volume = integraldisplay 3 − 3 integraldisplay √ 9 − x 2 integraldisplay y 1 dzdydx correct 3. volume = integraldisplay 3 − 3 integraldisplay x integraldisplay √ 9 − x 2 1 dzdydx 4. volume = integraldisplay 3 − 3 integraldisplay √ 9 − x 2 integraldisplay x 1 dzdydx 5. volume = integraldisplay 3 integraldisplay √ 9 − x 2 integraldisplay x xdzdydx 6. volume = integraldisplay 3 integraldisplay x integraldisplay √ 9 − x...
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## This homework help was uploaded on 03/19/2008 for the course M 408M taught by Professor Gilbert during the Fall '07 term at University of Texas at Austin.

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Test 3 - Version 102 – Exam 3 – Gilbert – (59825) 1...

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