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Integrating the generator lines of an elliptical orbit

  1. May 1, 2012 #1
    Hi,
    I am having difficulty understanding the following:

    [itex]\int^{2π}_{0}(x+y)\,dθ[/itex] = [itex]\int^{2π}_{0} 2a\,dθ[/itex] = [itex]\textbf{4}πa[/itex]

    where x and y are the generator lines of an elipse, a is the semimajor axis and θ is the angle formed by x and the major axis.

    I understand that x+y = 2a. However I don't understand how to integrate with respect to θ without expressing (x+y) or 2a in terms of θ.

    Any help would be much appreciated.
     
  2. jcsd
  3. May 1, 2012 #2

    HallsofIvy

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    The integral of a constant, such as 2a, with respect to any differential, dx, is that constant times the variable x. Normally, [itex]\int C dx= Cx[/itex] (plus a constant of integration) is one of the first integral facts one learns in Calculus.
     
  4. May 1, 2012 #3
    Because 2a and (x + y) are not the variables of integration, they are treated as constants and can thus be pulled out of the integral.

    ##\displaystyle \int_{0}^{2 \pi} 2a \ d\theta = 2a \int_{0}^{2 \pi} \ d\theta = \Bigl.2a \theta \ \Bigr|_{\theta = 0}^{\theta = 2 \pi} = 4 \pi a##
     
  5. May 1, 2012 #4
    Thanks for the help. I stumbled into this whilst reading "Gamma, exploring Euler's constant". I was inspired to get this after reading John Derbyshire's "Unknown Quantity, a real and imagined history of Algebra" which I found hugely illuminating. However as someone who only studied maths until 16 I feel I may have bitten off more than I can chew with this new book.
     
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