Exam 3 - Version 086 – Exam 3 – Gilbert – (59825) 1...

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Unformatted text preview: Version 086 – 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) = − 5 , f xy (1 , 1) = 4 , f yy (1 , 1) = 2 . Which of the following properties does f have at (1 , 1)? 1. a local maximum 2. a local minimum 3. a saddle point correct 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 = 2 π/ 3 . B. If I = integraldisplay integraldisplay x 2 + y 2 ≤ 1 f ( x, y ) dxdy , then I = integraldisplay 1 integraldisplay 2 π f ( r cos θ, r sin θ ) dθdr . 1. neither of them 2. A only correct 3. both of them 4. B only Explanation: A. TRUE: 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. B. FALSE: when changing to polar coordi- nates, dxdy = rdθdr . 003 10.0 points Evaluate the triple integral I = integraldisplay integraldisplay integraldisplay E 2 xdV, where E is the set of all points ( x, y, z ) in 3-space such that ≤ y ≤ 2 , ≤ x ≤ radicalbig 4 − y 2 , ≤ z ≤ y . Version 086 – Exam 3 – Gilbert – (59825) 2 1. I = 3 2. I = 4 correct 3. I = 5 4. I = 7 2 5. I = 9 2 Explanation: As an iterated integral, I = integraldisplay 2 integraldisplay √ 4 − y 2 integraldisplay y 2 xdzdxdy = integraldisplay 2 integraldisplay √ 4 − y 2 2 xy dxdy = integraldisplay 2 y (4 − y 2 ) dy . Consequently, I = 4 . 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 x integraldisplay √ 9 − x 2 1 dzdydx 3. volume = integraldisplay 3 − 3 integraldisplay √ 9 − x 2 integraldisplay x 1 dzdydx 4. volume = integraldisplay 3 integraldisplay x integraldisplay √ 9 − x 2 xdzdydx 5. volume = integraldisplay 3 − 3 integraldisplay √ 9 − x 2 integraldisplay y 1 dzdydx correct 6. volume = integraldisplay 3 integraldisplay √ 9 − x 2 integraldisplay x xdzdydx Explanation: Since E is the set of all points (...
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This note was uploaded on 05/08/2008 for the course M 408 M taught by Professor Gilbert during the Fall '07 term at University of Texas at Austin.

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Exam 3 - Version 086 – Exam 3 – Gilbert – (59825) 1...

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