calc hw 6 - lawrence (cdl678) Homework 6 ODELL (56280) 1...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: lawrence (cdl678) Homework 6 ODELL (56280) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points The shaded region in lies inside the polar curve r = 3 cos and outside the polar curve r = cos . Which of the following integrals gives the area of this region? 1. I = 1 2 integraldisplay / 2 2 cos d 2. I = 1 2 integraldisplay / 2 / 2 8 cos 2 d correct 3. I = 1 2 integraldisplay / 2 8 cos 2 d 4. I = integraldisplay / 2 / 2 2 cos d 5. I = 1 2 integraldisplay / 2 / 2 2 cos d 6. I = integraldisplay / 2 / 2 8 cos 2 d Explanation: As the graphs show, the polar curves inter- sect when 3 cos = cos , i.e. at = / 2. Thus the area of the shaded region is given by I = 1 2 integraldisplay / 2 / 2 braceleftBig (3 cos ) 2 (cos ) 2 bracerightBig d . Consequently, I = 1 2 integraldisplay / 2 / 2 8 cos 2 d . keywords: area, polar cooordinates, definite integral, circle, 002 10.0 points Find the area of the region bounded by the polar curve r = 4 ln + 3 as well as the rays = 1 and = e . 1. area = 1 2 (3 e + 1) correct 2. area = 1 4 (3 e + 2) 3. area = 3 e + 2 4. area = 3 e + 1 5. area = 1 4 (3 e + 1) 6. area = 1 2 (3 e + 2) Explanation: The area of the region bounded by the graph of the polar function r = f ( ) as well as the rays = , 1 is given by the integral A = 1 2 integraldisplay 1 f ( ) 2 d . When f ( ) = 4 ln + 3 , = 1 , 1 = e, lawrence (cdl678) Homework 6 ODELL (56280) 2 therefore, the area of the enclosed region is thus given by the integral A = 1 2 integraldisplay e 1 ( 4 ln + 3) 2 d = 1 2 integraldisplay e 1 (4 ln + 3) d . To evaluate this last integral we use Integra- tion by Parts, for then A = 1 2 bracketleftBig 4 ln + 3 bracketrightBig e 1 1 2 integraldisplay e 1 4 d = 1 2 bracketleftBig 4 ln 1 bracketrightBig e 1 . Consequently, area = A = 1 2 (3 e + 1) . 003 10.0 points Find the area of the region enclosed by the graph of the polar function r = 2 sin . 1. area = 7 2 2. area = 17 2 3. area = 5 2 4. area = 9 2 correct 5. area = 3 Explanation: The area of a region bounded by the graph of the polar function r = f ( ) and the rays = , 1 is given by the integral A = 1 2 integraldisplay 1 f ( ) 2 d . On the other hand, the graph of r = 2 sin is the cardioid similar to the one shown in so in this case we can take = 0 and 1 = 2 . Thus the area of the region enclosed by the graph is given by the integral A = 1 2 integraldisplay 2 (2 sin ) 2 d ....
View Full Document

This note was uploaded on 04/09/2010 for the course M 56280 taught by Professor Odell during the Spring '10 term at University of Texas at Austin.

Page1 / 12

calc hw 6 - lawrence (cdl678) Homework 6 ODELL (56280) 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online