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Compute the area under the curve

  1. Dec 26, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi everyone,
    can you give me some hint to solve this? please

    Compute the area under the curve:


    2. Relevant equations
    [tex](x^3+y^3)^2=x^2+y^2[/tex]


    3. The attempt at a solution
    I've been trying to transform it into polar coordinates, but I've finished when I get the equation for the radius:
    [tex]r^2=1/((\cos(\theta))^3+(\sin(\theta))^3)[/tex],
    because then I get the integral which should not be easily integrated (of course I've compute with Jacobian)

    thanks for yout attention
     
  2. jcsd
  3. Dec 27, 2009 #2

    LCKurtz

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    Re: Area

    Have you looked at the graph? It isn't even clear what region the "area under the curve" describes.
     
  4. Dec 28, 2009 #3
    Re: Area

    I am sorry, I forgot add this condition: [tex]x>0, y>0[/tex], so then I will get the integral in this case:
    [tex]
    \int\limits_{0}^{\pi/2}(\int\limits_{0}^{\frac{1}{\sqrt{\cos^3\alpha+\sin^3\alpha}}}r\,dr)\,d\alpha
    [/tex],
    which results to [tex]1/3 (\pi+2 \sqrt{2} \tanh^{-1}(1/\sqrt{2}))[/tex], but it's little bit different form the correct result which is [tex] \frac{\pi}{6}+\frac{\sqrt{2}}{3}\ln(1+\sqrt{2})[/tex]
     
  5. Dec 28, 2009 #4

    LCKurtz

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    Re: Area

    Your integral looks OK but your answer is off by a factor of 1/2. You probably dropped it somewhere in the calculation.
     
  6. Dec 28, 2009 #5
    Re: Area

    Yes, you're right. l forget 1/2 in front of integral, then the results are same:
    1.8781645380502384306795745680939249649333168829204582775232678988974988081580885734995703727615411934.
    Thanks for your time.
     
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