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Integral over a curve

  1. Oct 5, 2015 #1
    1. The problem statement, all variables and given/known data
    The parametric equations of a circle, center (1,0) and radius 1, can be expressed as x = 2cos^2(theta), y = 2cos(theta)sin(theta).
    Evaluate the integral of {(x+y)dx+x^2dy} along the semicircle for which y >=0 from (0,0) to (2,0).

    2. Relevant equations
    Perhaps Green's Theorem?

    3. The attempt at a solution
    I tried using Green's Theorem, I tried using polar coordinates, I tried using a combination of Green's Theorem and polar coordinates, I tried trig substitution, I tried u substitution, I tried integration by parts, I tried change of variables. I literally tried every technique I know for integration except partial fractions. I am at a complete loss. Any thoughts on the setup?
     
  2. jcsd
  3. Oct 5, 2015 #2

    Simon Bridge

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    Please show working with reasoning for your best attempt.
     
  4. Oct 5, 2015 #3
    @Simon Bridge Here is my cleanest attempt. I got -Pi/2 as my answer, the book says the answer is 2 - Pi/2.
     

    Attached Files:

  5. Oct 5, 2015 #4
    Also, I evaluated it using the vector definition of a line integral using Maple, and got Pi as the answer using the parameters and -Pi/2 as the answer without...
     

    Attached Files:

  6. Oct 6, 2015 #5

    Simon Bridge

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    Green's theorem looks like its complicating things... did you try just going through the usual parameterization stuff first?
    You have an integral of the form ##\int_C Pdx + Qdy : P=x+y,\; Q=x^2##
    The parameterization of the curve is ##x(t)=2\cos^2t,\; y(t)=2\cos t \sin t##

    So the integral becomes:

    Check the limits ... you want x to go from 0 to 2.
    http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsPtII.aspx
     
  7. Oct 6, 2015 #6

    Ray Vickson

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    When I use Maple 11 to evaluate the integral ##J = \int_0^2 (x + y + x^2 y') \, dx## (with ##y = \sqrt{2x - x^2}##) I get ##J = 2 - \pi/2##, which is the book's answer. I did not bother to use the "VectorCalculus" package, because just doing a direct evaluation seemed much faster and easier.

    The images you posted are too small and too blurry to make out, so I am not able to see any of the details of your computations. I would suggest copying and pasting individual lines, rather than attaching an image. For example, here is the indefinite integral (after putting y = sqrt(2*x - x^2) and dy = diff(y,x)):

    > int(x+y+x^2*dy, x); lprint(%);
    (1/2)*x^2-(1/4)*(2-2*x)*(2*x-x^2)^(1/2)-(1/2)*arcsin(-1+x)+(1/3)*x^2*(2*x-x^2)^(1/2)+(1/3)*x*(2*x-x^2)^(1/2)+(2*x-x^2)^(1/2)

    This is copied and pasted directly from the Maple 11 worksheet
     
  8. Oct 6, 2015 #7
    Thank you both for your responses. I see now that I was dropping the x^2 and keeping the y'(X) term. *sigh*
     
  9. Oct 6, 2015 #8

    Simon Bridge

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    Well done - always double check the transcription :)
    You'll find a lot of responders here are impatient with images - best to cut and paste from text editors and terminals, and learn to use LaTeX markup for the handwritten stuff. The approach is good discipline, forces you to pay attention to what you are doing, and is good practise for later.

    Enjoy.
     
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