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M408L hwk9 solutions - Pham Quoc Homework 9 Due 3:00 am...

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Pham, Quoc – Homework 9 – Due: Mar 27 2007, 3:00 am – Inst: Eric Katerman 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Evaluate the double integral I = Z Z A 2 dxdy with A = n ( x, y ) : 4 x 6 , 3 y 6 o by first identifying it as the volume of a solid. 1. I = 12 correct 2. I = 10 3. I = 8 4. I = 14 5. I = 6 Explanation: The value of I is the volume of the solid below the graph of z = f ( x, y ) = 2 and above the region A = n ( x, y ) : 4 x 6 , 3 y 6 o . Since A is a rectangle, this solid is a box with base A and height 2. Its volume, therefore, is given by length × width × height = (6 - 4) × (6 - 3) × 2 . Consequently, I = 12 . keywords: volume, double integral, rectangu- lar region, rectangular solid 002 (part 1 of 1) 10 points Evaluate the iterated integral I = Z 3 1 n Z 4 0 2 ( x + y ) 2 dx o dy . 1. I = ln 3 2 2. I = 2 ln 15 7 correct 3. I = ln 15 7 4. I = 1 2 ln 3 2 5. I = 2 ln 3 2 6. I = 1 2 ln 15 7 Explanation: Integrating the inner integral with respect to x keeping y fixed, we see that Z 4 0 2 ( x + y ) 2 dx = h - 2 x + y i 4 0 = 2 n 1 y - 1 4 + y o . In this case I = 2 Z 3 1 n 1 y - 1 4 + y o dy = 2 h ln y - ln(4 + y ) i 3 1 . Consequently, I = 2 ln (3)(1 + 4) (4 + 3) · = 2 ln 15 7 . keywords: iterated integral, rational function, log integral
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Pham, Quoc – Homework 9 – Due: Mar 27 2007, 3:00 am – Inst: Eric Katerman 2 003 (part 1 of 1) 10 points Evaluate the iterated integral I = Z ln 4 0 ˆ Z ln 3 0 e 2 x - y dx ! dy . 1. I = 6 2. I = 3 correct 3. I = 4 4. I = 2 5. I = 5 Explanation: Integrating with respect to x with y fixed, we see that Z ln 3 0 e 2 x - y dx = 1 2 h e 2 x - y i ln 3 0 = 1 2 e 2 ln 3 - y - e - y · = 3 2 - 1 2 · e - y . Thus I = 4 Z ln 4 0 e - y dy = - 4 h e - y i ln 4 0 = - 4 e - ln 4 - 1 · . Consequently, I = - 4 1 4 - 1 · = 3 . keywords: 004 (part 1 of 1) 10 points Determine the value of the double integral I = Z Z A 3 xy 2 9 + x 2 dA over the rectangle A = n ( x, y ) : 0 x 4 , - 2 y 2 o , integrating first with respect to y . 1. I = 8 ln 25 9 · correct 2. I = 8 ln 9 25 · 3. I = 8 ln 25 18 · 4. I = 4 ln 25 18 · 5. I = 4 ln 9 25 · 6. I = 4 ln 25 9 · Explanation: The double integral over the rectangle A can be represented as the iterated integral I = Z 4 0 Z 2 - 2 3 xy 2 9 + x 2 dy dx , integrating first with respect to y . Now after integration with respect to y with x fixed, we see that Z 2 - 2 3 xy 2 9 + x 2 dy = h xy 3 9 + x 2 i 2 - 2 = 16 x 9 + x 2 .
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