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Homework Help: Line Integral and swept area

  1. Jan 10, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the area swpet out by the line from the origin to the ellipse x=cos t and y=sin t as t varies from 0 to t_0 where t_0 is constant between 0 and 2 pi


    2. Relevant equations

    Use [itex]A= \frac{1}{2} \oint_{C} y dx -xdy[/itex]


    3. The attempt at a solution

    How does one determine how to graph this in order to aid calculation?

    Thanks
     
  2. jcsd
  3. Jan 10, 2012 #2

    LCKurtz

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    You could eliminate the parameter t if you want the xy equation, but you don't really need to for this problem. You have x and y given in terms of t; just plug x, y, dx, and dy in the integral and integrate with respect to t.
     
  4. Jan 11, 2012 #3
    Ok,

    1) I calculate

    [itex] \displaystyle A= \frac{1}{2} \int_{0}^{2 \pi} a sin t (-a sin t) dt - a cos t (a cos t)dt = \frac{-a^2}{2} \int_{0}^{ 2 \pi} (sin^2 t +cos^2 t) dt = -a^2 \pi [/itex]..? A negative area is not right..

    2)I am interested in how you would eliminate the parameter t as another method...can you advise?

    Thanks
     
  5. Jan 11, 2012 #4

    LCKurtz

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    That is because your formula I have highlighted in red is wrong. It should have a minus sign in front of it. And where did the ##a## come from in your answer. There is no ##a## in your problem.
    What do you get if you calculate ##x^2+y^2## in terms of your parameter ##t##?
     
  6. Jan 11, 2012 #5
    Well spotted, thanks. I left that out by mistake.The question has x= a cos t and y = a sin t
    I guess there isn't enough info to calculate the area numerically, just in terms of 'a' like I have seen in similar example elsewhere.

    Well, equation of ellipse is x^2/a^2 +y^2/b^2 = 1. But based on this, shouldn't the above y=a sin t be y = b sin t?

    I think we get a more difficult integral using the 'eliminate t' method........?
     
  7. Jan 11, 2012 #6

    LCKurtz

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    There isn't any ##b## in your problem so why are you bringing that up? What do you get if you calculate ##x^2+y^2## in your example?
     
  8. Jan 11, 2012 #7
    I calculate x^2 +y^2 =a^2. So I am guessing we could rearrange for x and y and put back into original 'corrected' integral in post 1...?
     
  9. Jan 11, 2012 #8

    LCKurtz

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    We are now just going around in circles. You asked in your original post how you could graph the curve. I said one way was to eliminate the parameter but I told you that was unnecessary to work the problem. But if you want to see the curve, graph ##x^2+y^2=a^2## and look at it. I presume you do know what that graph is and how to graph it, right?
     
  10. Jan 11, 2012 #9



    Sorry, I mis-interpreted your first quote above. I thought you meant 'evaluating the integral' by eliminating the variable t instead of 'plotting it'.

    Yes I do know how to plot it. In the original question, I was thrown off by the word 'line' I thought it was more than just an ellipse,ie some combination of a revolved line with the ellipse...?

    Other than that, I'm pretty happy :-)
     
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