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Unformatted text preview: Miller, Kierste – Homework 9 – Due: Mar 27 2007, 3:00 am – Inst: Gary Berg 1 This printout should have 15 questions. Multiplechoice 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 6 dxdy with A = n ( x, y ) : 6 ≤ x ≤ 8 , 2 ≤ y ≤ 5 o by first identifying it as the volume of a solid. 1. I = 36 correct 2. I = 38 3. I = 40 4. I = 32 5. I = 34 Explanation: The value of I is the volume of the solid below the graph of z = f ( x, y ) = 6 and above the region A = n ( x, y ) : 6 ≤ x ≤ 8 , 2 ≤ y ≤ 5 o . Since A is a rectangle, this solid is a box with base A and height 6. Its volume, therefore, is given by length × width × height = (8 6) × (5 2) × 6 . Consequently, I = 36 . 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 3 1 ( x + y ) 2 dx o dy . 1. I = ln 6 5 2. I = 2 ln 2 3. I = 2 ln 6 5 4. I = 1 2 ln 6 5 5. I = ln 2 correct 6. I = 1 2 ln 2 Explanation: Integrating the inner integral with respect to x keeping y fixed, we see that Z 3 1 ( x + y ) 2 dx = h 1 x + y i 3 = n 1 y 1 3 + y o . In this case I = Z 3 1 n 1 y 1 3 + y o dy = h ln y ln(3 + y ) i 3 1 . Consequently, I = ln ‡ (3)(1 + 3) (3 + 3) · = ln 2 . keywords: iterated integral, rational function, log integral 003 (part 1 of 1) 10 points Miller, Kierste – Homework 9 – Due: Mar 27 2007, 3:00 am – Inst: Gary Berg 2 Evaluate the iterated integral I = Z ln 4 ˆ Z ln 5 e 2 x y dx ! dy . 1. I = 7 2. I = 8 3. I = 5 4. I = 6 5. I = 9 correct Explanation: Integrating with respect to x with y fixed, we see that Z ln 5 e 2 x y dx = 1 2 h e 2 x y i ln 5 = 1 2 ‡ e 2 ln 5 y e y · = ‡ 5 2 1 2 · e y . Thus I = 12 Z ln 4 e y dy = 12 h e y i ln 4 = 12 ‡ e ln 4 1 · . Consequently, I = 12 ‡ 1 4 1 · = 9 . 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 ≤ 2 , 3 ≤ y ≤ 3 o , integrating first with respect to y . 1. I = 27 ln ‡ 9 13 · 2. I = 27 2 ln ‡ 13 9 · 3. I = 27 2 ln ‡ 9 13 · 4. I = 27 2 ln ‡ 13 18 · 5. I = 27 ln ‡ 13 9 · correct 6. I = 27 ln ‡ 13 18 · Explanation: The double integral over the rectangle A can be represented as the iterated integral I = Z 2 µZ 3 3 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 3 3 3 xy 2 9 + x 2 dy...
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This note was uploaded on 05/08/2008 for the course M 408 L taught by Professor Cepparo during the Spring '08 term at University of Texas.
 Spring '08
 Cepparo

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