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Area of parametric equation

  1. Apr 14, 2005 #1
    Hi,

    I've been trying to do this one question: Let R be the region enclosed by the graph x=t^2-2 y=t^3-2t. Set up the integral for the area of R.

    I know that if y is continous function of x on an interval a ≤ x ≤ b where x=f(t) and y=g(t) then [tex]\int_{a}^{b}[/tex] y dx =[tex]\int_{t1}^{t2}[/tex] g(t)f'(t)dt provided that f(t1)=a and f(t2)=b and both g and f' are continous on [t1,t2].

    But everytime i equate and solve for t i get sqrt(2) and 1. This is where i get lost I cant seem to find the interval for parametric equations (any of them :mad: ). Other then the interval I do belive i have the rest of the integral:

    [tex]\int[/tex] (t^3-2t)(2t)dt

    Thanks in advance (sorry for long post as well as horrid use of that latex typing first time)
     
  2. jcsd
  3. Apr 14, 2005 #2

    learningphysics

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    Is there any more information given?

    The graph translated to [tex]y=x\sqrt{x+2}[/tex]. There doesn't seem to be any enclosed region.
     
  4. Apr 14, 2005 #3
    the question copied and pasted from the pdf file is as follows:

    4. Let R be the region enclosed by the graph of x = t^2 − 2, y = t^3 − 2t . (The graph was sketch for the final in 2004)

    a. Set up, BUT DO NOT EVALUATE an integral for the area of R .

    Thats all the info that was given :/
     
    Last edited: Apr 14, 2005
  5. Apr 14, 2005 #4

    HallsofIvy

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    Sorry, there is NO area "enclosed" by that graph.
     
  6. Apr 14, 2005 #5
    Actually, the parametrization can be broken into the 3 implicit functions y = [itex]x\sqrt{x+2}[/itex] for [itex]t\in(0, \infty)[/itex], y = [itex]-x\sqrt{x+2}[/itex] for [itex]t\in(-\infty, 0)[/itex], and y = 0 for t = 0. The graphs of these functions enclose a region between x=0 and x=-2. Given the symmetry of the graphs of the functions (one is just a reflection in the x-axis of the other), all one needs to do is integrate the positive function on this interval and multiply by 2. :smile:
     
  7. Apr 14, 2005 #6
    Thanks for the help got was able to get the answer now.
     
  8. Apr 14, 2005 #7

    learningphysics

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    Oops. Yes... the minus sign. Sorry about that.
     
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