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Parametric Intergration question

  1. Jun 11, 2008 #1
    1. The problem statement, all variables and given/known data

    a curve has parametric equations:
    x = t - 2sin t
    y = 1 - 2cos t


    R is enclosed by the curve and the x axis. Show that the area of R is given by the integral:

    [tex]\int^{\frac{5\pi}{3}}_{\frac{\pi}{3}} (1-2\cos t)^{2}[/tex]
    2. Relevant equations



    3. The attempt at a solution

    Not sure what to do :\
     
  2. jcsd
  3. Jun 11, 2008 #2

    CompuChip

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    Suppose that the curve was given in the form y = f(x), then how would you calculate the area? Now do the variable substitution from x to t which is given.
     
  4. Jun 11, 2008 #3

    Defennder

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    You might want to make use of this:
    [tex]\int f(x) dx = \int f(t) \frac{dx}{dt} \ dt[/tex]
    By the way what's this mathematical property called?
     
  5. Jun 11, 2008 #4
    ermm mathematical propety???

    dx/dt = 1 - 2cost


    i have to multiply this by f(t) (how do i found that) and then intergrate that with respect to t?

    Thanks
     
  6. Jun 11, 2008 #5

    Defennder

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    Yes, the integration is done wrt t. Isn't that given in the answer? And as for f(t), you should know that f(x) in the original integral f(x)dx represents a particular variable. That variable must now be expressed in terms of t for the integration to work.
     
  7. Jun 11, 2008 #6

    tiny-tim

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    Hi thomas49th! :smile:

    I suspect you're confused by the (x) and the (t).

    Just write it [tex]\int f dx = \int f \frac{dx}{dt} \ dt[/tex]

    or, in this case, [tex]\int y dx = \int y \frac{dx}{dt} \ dt[/tex] :smile:
     
  8. Jun 11, 2008 #7
    ahh yeh it's easy

    got it!

    cheers :)
     
  9. Jun 12, 2008 #8

    CompuChip

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    The property is called substitution of variables. Probably you've seen it in this form:
    Solve [tex]\int \sin(t) \cos(t) \, dt[/tex]; let [itex]u = \sin(t)[/itex], then [itex]du = \cos(t) \, dt[/itex] so [tex]\int \sin(t) \cos(t) \, dt = \int u \, du = \frac12 u^2 + C = \frac12 \sin^2(t) + C[/tex].
    You'll notice that in switching from the variable t to the variable u, I replaced [itex]\frac{du}{dt} dt[/itex] by [itex]du[/itex].

    If you have never seen this, forget it. Otherwise I hope it clarifies what just happened :)
     
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