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Homework Help: Double integrals of polar equations help!

  1. Mar 20, 2010 #1
    Hey, everyone. I'm new to the forums and am hoping that someone can help me with a tricky problem from my multivariable calculus class. We're covering double integrals in polar coordinate systems, and there's a problem from my problem set for homework that I can't seem to get the grasp of.

    The question asks: Use a double integral to find the area of the region within both of the circles [tex]r=cos( \theta )[/tex] and [tex]r=sin( \theta )[/tex]

    I'm having a lot of difficulty figuring out where to go with this, though. It seems to me that the region is traced by [tex]r=sin( \theta )[/tex] as [tex]\theta[/tex] goes from [tex]0[/tex] to [tex] \pi /4[/tex] and by [tex]r=cos( \theta )[/tex] as [tex]\theta[/tex] goes from [tex]\pi /4[/tex] to [tex]\pi /2[/tex]. Other than that, I'm clueless as to how to solve this using a double integral. Can someone please help me?
     
  2. jcsd
  3. Mar 20, 2010 #2
    Do you know what the answer is for the cos integral?
     
  4. Mar 20, 2010 #3
    Do you just mean the antiderivative? Sure. The antiderivative of cos is sin.
     
  5. Mar 20, 2010 #4
    No the numerical answer. I have an answer but I am not 100% on it so before I make any suggestions I need to know if it is correct.
     
  6. Mar 20, 2010 #5
    Well, I don't know if this helps, but the answer given by the textbook is (1/8)(pi - 2). I'm not so much looking for the answer as I am looking for the method. I'm completely clueless about how to set up this problem.
     
  7. Mar 20, 2010 #6
    I am hesitant to say how I did since I have +2 not minus 2
     
  8. Mar 20, 2010 #7
    I just checked my answer in maple and it is plus 2.
     
  9. Mar 20, 2010 #8
    Why do you say cos goes from pi/4 to pi/2?
    If you graph the function, it goes from 0 to 1 on the r axis and pi/4 to negative pi/4.
     
  10. Mar 20, 2010 #9
    That's weird... I was looking at a website earlier today that did the problem by a different method involving two single integrals, and it also came upon the answer (1/8)(pi - 2).
     
  11. Mar 20, 2010 #10
    Now we can integrate the whole function or use symmetry. How would you prefer to do it?
     
  12. Mar 20, 2010 #11
    Because the region bounded by sin(theta) and cos(theta) is the cats-eye shape shared by the two circles. The lower boundary of this region is drawn by sin(theta) as it goes from theta=0 to theta=pi/4. The upper boundary of this region is drawn by cos(theta) as it goes from theta=pi/4 to theta=pi/2.
     
  13. Mar 20, 2010 #12
    It was unclear to me you wanted to integrate their intersection. I will look it at it again now hold on.
     
  14. Mar 21, 2010 #13
    I have the answer but I don't like how I obtained it.
     
  15. Mar 21, 2010 #14
    Ok ok ok. So when looking at the intersection we notice, by looking at half it, that we can track sin from 0 to pi/4 to arrive at the intersection. Do you agree?
     
  16. Mar 21, 2010 #15
    Now we can start establishing the bounds of integration. We have already arrived at 0 to pi/4 for d theta. What do you think the bounds for r are?
     
  17. Mar 21, 2010 #16
    Let me make sure I'm thinking about this correctly... the integration of sin(theta) from 0 to pi/4 would be equal to exactly half of the area of the region, correct?
     
  18. Mar 21, 2010 #17
    Yes, but sin(theta) 0 to pi/4 is the bounds not the integration. You will be integrating r dr d(theta). What do you think the r bounds are or should be?
     
  19. Mar 21, 2010 #18

    ideasrule

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    It seems that the intersected area is just twice the area occupied by r=sin(theta) from theta=0 to theta=45 degrees. Why not find that integral and multiply it by 2?

    EDIT: oops, somebody beat me to it.
     
  20. Mar 21, 2010 #19
    I recognize that the lower limit of integration with respect to r is 0, out of common sense. I'm tempted to say that the upper limit of integration with respect to r is sin(theta). That way, the integral allows a tracing from 0 to pi/4, and it allows the radius to be as large as sin(pi/4)=sqrt(2)/2, but no larger.

    Am I correct in thinking that the setup would be:

    integral(theta=0, theta=pi/4) integral(r=0,r=sin(theta)) r dr d(theta)?
     
  21. Mar 21, 2010 #20
    Times 2 since that is only half of if.
     
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