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Math Help - Integration of parametric equations.

  1. #1
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    Integration of parametric equations.

    Hi, I'm having trouble understanding how it is possible that one can integrate a parametric equation y=g(t); x=h(t) when y is not a function of x.


    for instance x=2\cost-\cos2t, y=2sint-sin2t


    As I understand \int_a^bf(x)dx=\int_{\alpha}^{\beta}g(t)h'(t)dt where both integrals measure the area under the curve.

    But somehow, even if no f(x) exists, one can still use the parametric integral to measure the area within the shape it describes.
    Can somebody explain to me why that is?
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  2. #2
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    Re: Integration of parametric equations.

    Hey HK47.

    Have you done line-integrals and multi-variable calculus before?
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    Re: Integration of parametric equations.

    No, I havent, I had this exercise in my calculus 1 class. The other parametric integrations were area below the line, that one was different, however using the same principle seemed to give a reasonable answer.
    Last edited by HK47; July 3rd 2013 at 12:09 AM.
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    Re: Integration of parametric equations.

    remember the parametric equations of some known curves are:
    circle- x = a cos t , y = a sin t
    parabola: x = at^2 , y = 2at
    Ellipse : x = acos t , y = bsin t etc.
    The shape of the curve remains as it is but it is the way we write its equation
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    Re: Integration of parametric equations.

    The thing is, that I understand how to integrate all of the above-mentioned shapes.
    I can integrate the circle, although it is not y=f(x), because I notice that it is 2 times the integral from 0 to pi, which can be shown as y=f(x), because only one y point corresponds to each x. Or I see that the integral is the sum of two integrals of y=f(x), one of which corresponds to the area above the curve and the other to the area below it.
    \int_\pi^0f(x)-\int_\pi^{2\pi}f(x)=\int_{2\pi}^0f(x)
    The same with an ellipse.
    And a parabola is y=f(x) whatever the parameter values.
    So that is why I can integrate those shapes over the cycle of the parameters.

    But with the problem i mentioned, that rationalization does not work, because I don't see how the parametric equation can be represented by y=f(x)
    Last edited by HK47; July 3rd 2013 at 03:41 AM.
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  6. #6
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    Re: Integration of parametric equations.

    Ok, I figured it out. The parametric equation is essentially divided into functions, y=f(x) and it is integrated in parts, but the parts add up in a convenient way. The substracted areas will have opposite paths, which means the paths will be reversed due to the minus sign and the integrals can be added. The parts under the x axis will also get their paths reversed, and added, so for any parametric equation the area within the graph can be found integrating through the whole parameter change and plugging the x and y into y=f(x) even though the y=f(x) in that case isnt a function through all of the parameters values, but rather represents several functions, that can be integrated and summed up to that value.
    Last edited by HK47; July 4th 2013 at 09:29 AM.
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  7. #7
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    Re: Integration of parametric equations.

    Hi,
    As you have convinced yourself, the formula for the area of parametric curves is correct. However, you have to be careful to make certain a given curve satisfies the implicit hypotheses of the formula. That is, each area problem needs to be analyzed carefully.

    The attachment shows a curve where blind application of the formula gives an absurd answer, but the formula still works if you're careful.

    Integration of parametric equations.-mhfparametricarea.png
    Thanks from HK47
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