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Area Inside Lemniscate Vector Calculus

  1. Sep 17, 2007 #1
    Find the area inside the lemniscate ,which is described by the equation (x^2 + y^2) = 2a^2 (x^2 - y^2)

    I have no idea where to start so I found out what a lemniscate is.. but this didn't help me much. I have rearranged the equation for a:

    a = sqrt ( (x^2 + y^2)^2 / 2(x^2 - y^2) )
    a = x^2 + y^2 / sqrt (2(x^2 - y^2))

    How do I proceed from here?
     
  2. jcsd
  3. Sep 17, 2007 #2

    EnumaElish

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    First, the equation should have been (x^2 + y^2)^2 = 2a^2 (x^2 - y^2).
     
  4. Sep 17, 2007 #3
    oh yes, when i isolated for a i did remember the ^2
    but i have no idea how to proceed from there
     
  5. Sep 18, 2007 #4
    Hint: convert to Lemniscate equation into polar form and calculate area from there
     
  6. Sep 19, 2007 #5
    what does the a even mean..
    my prof did not say anything about a lemniscate, I am confused out of my mind
     
  7. Sep 19, 2007 #6
    okay so i can only put it in polar coordinates, but where are the limits of integration in which i am suppose to integrate??

    (r^2)^2 = 2a^2 (rcos^2theta - rsin^2theta) r
    r^4 = 2a^2 (rcos^2theta - rsin^2theta) r
    r^3 = 2a^2 (r^2cos^2theta - r^2sin^2theta)
    r = 2a^2 (cos^2theta - sin^2theta)
    r = 2a^2 (1-sin^2theta - sin^2theta)
    r = 2a^2 (1-2sin^2theta)
    Using half angle rule:
    r = 2a^2 (1-2(1-cos^2theta / 2))
    r = 2a^2 (1-(1-cos^2theta))
    r = 2a^2 (cos^2theta)

    i still do not know the meaning of a.. and the boundaries
     
  8. Sep 19, 2007 #7
    should be r^2 = 2a^2 (cos^2theta);

    sketch the curve to find the limits of intergration, I'd say for the right loop you integrate
    from -pi/4 to pi/4.
    Constant a controls the form of the curve.

    Try sketching it here http://graph.seriesmathstudy.com/
     
  9. Sep 19, 2007 #8
    so how do i get rid of the constant?
     
  10. Sep 19, 2007 #9
    Well, you don't.
    What is the integral of y = ax ? ax^2 / 2, right? Can you integrate r^2 = 2a^2 (cos^2theta) then, integrate wrt to theta not a?
     
  11. Sep 19, 2007 #10
    i was doing the same problem and i got 2a^2 as my answer


    r^2 = 2a^2cos(2theta)

    the integral of r^2 should be the area hence integrating the right side gives u an asnwer wrt a
     
  12. Sep 19, 2007 #11
    anyone know how to compute Integral 0 to 1 ; integral y to 1 (1+x^2)^(1/2) dxdy

    its a double integral problem, im stuck....
     
  13. Sep 19, 2007 #12
    wait its a^2 not 2a^2
     
  14. Sep 19, 2007 #13
    i found examples in class where it is integrating the double integral w.r.t dr dtheta
    so do i rearrange for a to get rid of a... or do as u say and integrate w.r.t dtheta
    i still have not found the limits of integration yet, i dont see how i am suppose to figure all this stuff out with no information and no knowledge of a lemniscate
     
  15. Sep 19, 2007 #14
    yeh hahaha ethans class
     
  16. Sep 23, 2007 #15
    i still need help on this,
    anyone have any ideas?
    is hanhan correct that r^2 is the area so u can integrate ??
     
  17. Sep 23, 2007 #16
  18. Sep 23, 2007 #17
    but i am learning about double integrals and it is only a single integration on the website link
     
  19. Sep 24, 2007 #18
    anyone know how i can change it into a double integral to solve for the area inside the lemniscate?
     
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