Polar integrals

Test Preparation AP Test Preparation
1

<p>In Peterson's Calc. chapter 9, ex. 6, problem 5, how did they figure out which intervals to use? Why did they do what they did?</p>

2

<p>Maybe someone can help if you type the problem.</p>

3

<p>The problem is: Find the area bounded by r=3-2sin(x) and r=4cos(x).</p>

4

<p>If it's a calculator problem, just solve 3-2sin(x) = 4cos(x) to find where they intersect, at x = -0.371 and x = 1.299. So that's the interval... isn't it just</p>

<p>(1/2)int((4cos(x) - (3-2sin(x)))^2,x,-0.371,1.299)</p>

<p>That's part of it. To get the part that's in the 4th quadrant, it's (1/2)int((cos(x))^2,x,3pi/2,2pi+1.299)</p>

5

<p>Thanks. Actually,they did it like this: 1/2{int[(3-2sin(x))^2 from 0 to 1.2991]+int[(4cos(x))^2 from 1.2991 to 5.9114-pi]+int[(3-2sin(x))^2 from 5.9114 to 2*pi]}. Why did they do it that way? I can scan the diagram if it would help.</p>

6

<p>Oh, that way works, too. This is what they did:</p>

<p><a href="http://img.photobucket.com/albums/v300/silent_rain/polarintegrals.jpg%5B/url%5D"&gt;http://img.photobucket.com/albums/v300/silent_rain/polarintegrals.jpg&lt;/a&gt;&lt;/p>

7

<p>One thing you got wrong there:</p>

<p>Calculus isn't fun. :P
well at least not polar integrals.</p>

8

<p>I have to say I disagree :D</p>

<p>The first week of class, we had to print out this worksheet that said across the top in inch-tall type "I <3 CALCULUS", and I had my doubts at first, but now I'm a convert ^_^</p>