This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Cheung, Anthony – Homework 9 – Due: Nov 1 2006, 3:00 am – Inst: David Benzvi 1 This printout should have 20 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 3 2 µZ 1 7 x 6 y dx ¶ dy. Correct answer: 2 . 5 . Explanation: The double integral I can be written as the repeated integral I = Z 3 2 µZ 1 7 x 6 y dx ¶ dy, integrating first with respect to x . Now after use of the Fundamental theorem of integral calculus, the inner integral becomes h 7 x 7 y 7 i 1 = y. Thus I = Z 3 2 y dy = h‡ y 2 2 ·i 3 2 . Consequently, I = 5 2 . keywords: 002 (part 1 of 1) 10 points Evaluate the integral I = Z 2 1 Z 4 1 ( x 2 2 xy ) dydx. 1. I = 25 2 2. I = 21 2 3. I = 27 2 correct 4. I = 13 2 Explanation: The integral can be written in iterated form I = Z 2 1 ‰Z 4 1 ( x 2 2 xy ) dy ¾ dx. Now Z 4 1 ( x 2 2 xy ) dy = h x 2 y xy 2 i 4 1 = (4 x 2 16 x ) ( x 2 x ) = 3 x 2 15 x. But then I = Z 2 1 (3 x 2 15 x ) dx = h x 3 15 x 2 2 i 2 1 . Consequently, I = 27 2 . keywords: definite integral, iterated integral, polynomial function, 003 (part 1 of 1) 10 points Evaluate the iterated integral I = Z 5 1 n Z 5 1 ‡ x y + y x · dy o dx. 1. I = 5ln12 2. I = 24ln12 3. I = 24ln5 correct 4. I = 5ln24 5. I = 12ln5 Cheung, Anthony – Homework 9 – Due: Nov 1 2006, 3:00 am – Inst: David Benzvi 2 6. I = 12ln24 Explanation: Integrating with respect to y keeping x fixed, we see that Z 5 1 µ x y + y x ¶ dy = • x ln y + y 2 2 x ‚ 5 1 = (ln5) x + 12 µ 1 x ¶ . Thus I = Z 5 1 • (ln5) x + 12 µ 1 x ¶‚ dx = •µ x 2 2 ¶ ln5 + 12ln x ‚ 5 1 . Consequently, I = 24ln5 . keywords: 004 (part 1 of 1) 10 points Evaluate the double integral I = Z 3 2 Z 2 e x y dxdy . 1. I = e 3 e 2 e 1 + 1 correct 2. I = e 3 e 2 + e 1 + 1 3. I = e 3 e 2 e 1 1 4. I = e 3 + e 2 e 1 + 1 Explanation: After integration with respect to x , I = Z 3 2 £ e x y / 2 dy = Z 3 2 ( e 2 y e y ) dy . But then, after integrating next with respect to y we see that I = £ e 2 y + e y / 3 2 = e 1 + e 3 ( 1 + e 2 ) . Consequently, I = e 3 e 2 e 1 + 1 . keywords: 005 (part 1 of 1) 10 points Determine the value of the double integral I = Z Z A 3 xy 2 16 + x 2 dA over the rectangle A = n ( x,y ) : 0 ≤ x ≤ 3 , 3 ≤ y ≤ 3 o , integrating first with respect to y . 1. I = 27 2 ln ‡ 25 16 · 2. I = 27 2 ln ‡ 25 32 · 3. I = 27ln ‡ 25 32 · 4. I = 27ln ‡ 16 25 · 5. I = 27ln ‡ 25 16 · correct 6. I = 27 2 ln ‡ 16 25 · Explanation: The double integral over the rectangle A can be represented as the iterated integral I = Z 3 µZ 3 3 3 xy 2 16 + x 2 dy ¶ dx, Cheung, Anthony – Homework 9 – Due: Nov 1 2006, 3:00 am – Inst: David Benzvi 3 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 16 + x 2 dy = h xy 3 16 +...
View
Full
Document
This note was uploaded on 01/27/2010 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas.
 Spring '08
 RAdin
 Calculus

Click to edit the document details