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Homework Help: Caclulus with parametric curves

  1. May 24, 2010 #1
    Elipse [0,2pi]
    x=acosT
    y=bsinT\
    by symmetry
    4I(y,x)
    chossing the limits of integration my text is showing
    4I(y,x,0,a)
    T=0=>x=a
    T=pi/2=>x=0
    so wouldnt this be 4I(y,x,a,0)
    proceeding inthe books case gives pi(ab) while in the later case gives -pi(ab)

    Also
    x= (t-1)/t y=(t+1)/t bounded by y=5/2
    5/2=(t+1)/t => t=2/3
    my text is showing t=1/2 or 2
     
    Last edited: May 24, 2010
  2. jcsd
  3. May 24, 2010 #2

    HallsofIvy

    User Avatar
    Science Advisor

    It would have helped if had explained what you are talking about! I think you are trying to find the area of an ellipse and are saying that "by symmetry, the area of the ellipse is 4 times the area the part of the ellipse in the first quadrant".

    Limits of integration for what integral? Since you are using parametic equations with a single parameter, you appear to be integrating around the circumference of the ellipse. But that won't give "area".

    I guessed, above, that "I(y,x)" was the area of the part of the ellipse in the first quadrant but now I have no idea what "I(y,x,a,0)" could mean.

    The are of an ellipse with semi-axes a and b is [math]\pi ab[/math]. Obviously, an area cannot be negative.

    Please tell us what the problem really is!

    I have no idea what you mean by "showing t= 1/2 or 2". Nor do I understand what is meant by "bounded by y= 5/2". Those parametric equations describe a curve. A curve has to be bounded at both ends.
     
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