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Area of Astroid? Or something like it. Is this right?

  1. Mar 2, 2010 #1
    Find the area of x2/3/9 +y2/3/4 =1
    The parametrizations I used are x=27cos3t y=8sin3t

    Area = [tex]\int xdy = \int_{0}^{2\pi} 27cos^3(t)24sin^2(t)cos(t)= \int_{0}^{2\pi} 27cos^4(t)24sin^2(t)=648((2\pi)/16+sin(4\pi)/64-sin(8\pi)/64-sin(12\pi)/192) = 81\pi[/tex]



    Is this right?
     
    Last edited: Mar 2, 2010
  2. jcsd
  3. Mar 2, 2010 #2

    HallsofIvy

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    Your formula is correct but the result is not at all what I get. Please show how you did the integral.
     
  4. Mar 2, 2010 #3
    [tex]\int xdy = \int_{0}^{2\pi} 27cos^3(t)24sin^2(t)cos(t)= \int_{0}^{2\pi} 27cos^4(t)24sin^2(t)=648((2\pi)/16+sin(4\pi)/64-sin(8\pi)/64-sin(12\pi)/192) = 81\pi[/tex]
     
    Last edited: Mar 2, 2010
  5. Mar 3, 2010 #4
    Can someone help please?
     
  6. Mar 5, 2010 #5
    I get 81pi too. Can anyone confirm this?
     
    Last edited: Mar 6, 2010
  7. Mar 6, 2010 #6
    Anyone, please?
     
  8. Mar 6, 2010 #7

    Dick

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    Yes. 81*pi.
     
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