Assignment 18-solutions

Assignment 18-solutions - nanni(arn437 – Assignment 18...

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Unformatted text preview: nanni (arn437) – Assignment 18 – guntel – (54940) This print-out should have 4 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 and above the rectangle A= (x, y ) : 1 ≤ x ≤ 2 , 0 ≤ y ≤ 2 . 1. volume = 62 cu.units correct 10.0 points 2. volume = 69 cu.units Find the value of the integral 3. volume = 61 cu.units 2 I= 1 f (x, y ) dx 0 4. volume = 67 cu.units when 5. volume = 60 cu.units f (x, y ) = 4x + x2 y . Explanation: The volume is given by the double integral 8 1. I = 8 − y 2 3 2 (2 + 9x2 + 8y ) dxdy V= 2. I = 8 + y 0 3. I = 4y + 4y 2 2 V= 2 2 9x2 dxdy 2 dxdy + 0 1 0 2 8 5. I = 8 − y 3 1 2 + 8y dxdy = 4 + 42 + 16. 0 1 Consequently, 6. I = 4y + y 2 volume = 62 cu.units . Explanation: Since x varies but y does not, 2 4 x + x2 y d x = 0 1 of f (x, y ) over the rectangular region A. Integrating each term separately, we see that 2 8 4. I = 8 + y correct 3 I= 2 1 2 x2 + x3 y 3 003 2 0 . 10.0 points Evaluate the integral Consequently, xexy dxdy I= A 8 I = 8+ y . 3 over the rectangle A = { (x, y ) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 3 }. keywords: 1. I = 13 e − 4 correct 3 Find the volume of the solid under the graph of 2. I = 13 e −4 6 f (x, y ) = 2 + 9x2 + 8y 3. I = 13 e −3 3 002 10.0 points nanni (arn437) – Assignment 18 – guntel – (54940) 2 13 8 4. I = 13 e −3 6 2. I = ln 5. I = 13 e −2 6 3. I = 13 1 ln 2 4 6. I = 13 e −2 3 4. I = 1 4 ln 2 13 5. I = ln xexy dxdy I= 13 correct 4 6. I = ln Explanation: Since the integral with respect to y in 4 13 A can be evaluated easily using substitution (or directly making the substitution in one’s head), while the integral with respect to x requires integration by parts, this suggests that we should represent the double integral as the repeated integral 1 I= xe 0 3 dy dx. 0 Now after integration the inner integral becomes 3 exy = e3x − 1 . 0 0 1 −1 3 (e 3xy 2 dy dx , 4 + x2 xy 3 4 + x2 1 −1 − 1) dx = 0 e3x 3 1 −x 0 , 0 2x dx = ln(4 + x2 ) 4 + x2 Consequently, and so I= 004 13 e −4 3 . I = ln 10.0 points Determine the value of the double integral I= A 3xy 2 dA 4 + x2 over the rectangle A= 1. I = (x, y ) : 0 ≤ x ≤ 3, −1 ≤ y ≤ 1 13 1 ln 2 8 = 2x . 4 + x2 But 1 I= −1 3xy 2 dy = 4 + x2 Thus 3x 1 I= integrating first with respect to y . Now after integration with respect to y with x fixed, we see that 3 xy Explanation: The double integral over the rectangle A can be represented as the iterated integral . 13 4 . 3 0 . ...
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This note was uploaded on 11/02/2011 for the course M 408N taught by Professor Gualdini during the Fall '10 term at University of Texas.

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