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Integration of polar curves

  1. Jan 31, 2008 #1
    1. The problem statement, all variables and given/known data

    find the area bounded by one of the four loops of: [itex] (x^2 + y^2)^3 = 4x^2y^2 [/itex]

    2. Relevant equations

    3. The attempt at a solution

    I converted to polar coordinates and got [itex] r^{3/2} = sin^2(2\theta) [/itex]
    The typical formula for polar integration for area would imply that I integrate sin to the power of some fraction which I don't know how to do. It would be great if there was a relationship between the area for r^1.5 and r though. Also this question might be related to line integrals (not sure how) which we have been doing. thnx for any help.
  2. jcsd
  3. Jan 31, 2008 #2
    Are you sure you did this conversion correct? Or are the loops described by
    [tex] \left(\sqrt{x^{2}+y^{2}}\right)^{3} = 4x^{2}y^{2}.[/tex]?

    What's the integral for determining area in cartesian coordinates? What's the effect on an integral by substituting to polar coordinates? (What's the Jacobian Determeninant of the transformation to polar coordinates?)
  4. Jan 31, 2008 #3
    I strongly believe the conversion is correct. I don't even know how to set up the integral in cartesian coordinates! What's the integrand?
  5. Feb 1, 2008 #4
    Sorry, I didn't mean to give you the impression that you should try this in cartesian coordinates. But can you show me how you did the conversion? Because I get:
    [itex](x^2 + y^2)^3 = 4x^2y^2[/itex]
    Substituting [itex]x=r\cos \theta, y=r\sin \theta[/itex]
    [itex]r^2 = \sin^{2}2\theta.[/itex]
  6. Feb 1, 2008 #5
    ah? you have r^6 on the left and r^4 on the right...
  7. Feb 1, 2008 #6
    Such equations are extremly hard, interesting do realy everybody who finishing universities realy good understand it?..
  8. Feb 1, 2008 #7
    Yes, because
    [itex]\left(x^2+y^2\right)^3 = \left(r^2\cos^2\theta+r^2\sin^2\theta\right)^3 = r^2\left(\cos^2\theta+\sin^2\theta\right)^3 = \left(r^2\right)^3,[/itex]
    [itex]4x^2y^2 = 4\left(r^2\cos^2\theta\right) \left(r^2\sin^2\theta\right) = r^4\left(4\cos^2\theta\sin^2\theta\right) = r^4\sin^2 2 \theta.[/itex]
    Do you see how you can proceed further now?
  9. Feb 1, 2008 #8


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    Certainly anyone who is majoring or minoring in mathematics should be able to do it! Believe me, upper level math courses get a lot harder, and more abstract, than this!

    Teleport just made a slight numerical error in his calculations. Looks like penguino has it cleared up now.
    Last edited: Feb 1, 2008
  10. Feb 1, 2008 #9
    penguino: no as we are where we were last time. not to mention there's a mistake in ur derivation.
  11. Feb 1, 2008 #10


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    Science Advisor

    No, penguino is exactly right.
    [tex](x^2+ y^2)^3= 4x^2 y^2[/tex]
    [tex](r^2)^3= [(2 r sin(\theta)cos(\theta))(2 r sin(\theta) cos(\theta))]= r^4 sin^2(2\theta)[/tex]
    [tex]r^2= sin^2(2\theta)[/tex]

    Put that into your integral.
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