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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.
     
  22. Mar 21, 2010 #21
    Ah. Other than that, though, the integral is correct?
     
  23. Mar 21, 2010 #22
    Yup. You understand why you are multiplying 2 though right. That integral is like the lower half with a 45 degree line cutting the top.
     
  24. Mar 21, 2010 #23
    Yes. It's the same reason that I would multiply by two if I were simply going to integrate sin(theta) from theta=0 to theta=pi/4.

    Now that I think about it, haven't I really just expanded upon the idea of a single integral for r=sin(theta) from 0 to pi/4? Instead of leaving it as a single integral, I expanded it by putting a second integral up and set up the bounds from 0 to sin(theta) to ensure that evaluating that integral would lead to just sin(theta)?

    Does this mean that any single integral can be expanded by setting up limits of integration that include the function of interest?
     
  25. Mar 21, 2010 #24
    Remember when you learned to that if two function intersect you can subtract the top function from the lower and integrate to get the region of intersection? You can prove that a double integral is equal to that process.
     
  26. Mar 21, 2010 #25
    I think one of us must be missing something... the answer given by the partial solutions key in the back of the book says that the answer is (1/8)*(pi - 2). How can we possibly get that answer from this?
     
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