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Homework Help: Evaluate the Following Line Integral Part 1

  1. Nov 1, 2011 #1
    1. The problem statement, all variables and given/known data

    [itex]\int_{c}cos (x)dx+sin(y)dy[/itex] where c consist of the top half of the circle x^2+y^2=1 from (1,0) to (-1,0)

    3. The attempt at a solution

    Do I parameterise x=t and then y becomes [itex]y= (1-t^2)^{1/2}[/itex]....? Replace the corresponding dx and dy and then integrate between the limits?
  2. jcsd
  3. Nov 1, 2011 #2

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    Hi bugatti79! :smile:

    Well, yes, you have to parametrize to calculate the result.
    But you can integrate before you parametrize.
  4. Nov 1, 2011 #3


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    That's one way to do it. Another is to think of this in polar coordinates with r= 1. In polar coordinates, [itex]x= r cos(\theta)[/itex] and [itex]y= r sin(\theta)[/itex] so on the unit circle, [itex]x= cos(\theta)[/itex] and [itex]y= sin(\theta)[/itex]. These are the "standard" parameterizations of the unit circle.

    Of course, going form (1, 0) to (-1, 0), [itex]\theta[/itex] goes from 0 to [itex]\pi[/itex]. You will need to determine dx and dy in terms of [itex]d\theta[/itex].
  5. Nov 1, 2011 #4
    Thanks guys,
    I think the notes only show parameterisation etc. I will take note of the other option. I suspect integrating first before parameterising is easier. Will try it later...
  6. Nov 1, 2011 #5
    Ok, I arrive at

    [itex]\displaystyle \int_{0}^{\pi }cos(-r sin(\theta) )d\theta+ sin(r cos(\theta))d\theta[/itex]

    ...this doesnt look right...?
  7. Nov 1, 2011 #6

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    I'm afraid you didn't substitute properly.
    It should be:

    [itex]\displaystyle \int_{0}^{\pi }cos(r \cos(\theta) ) \cdot -r \sin(\theta) d\theta+ \sin(r \sin(\theta)) \cdot r \cos(\theta) d\theta[/itex]

    Furthermore, you should substitute r=1.

    But rather than doing this, I recommend integrating first, then substituting the parameterisation (one or the other), and then calculate the result.

    How would you integrate [itex]\int \cos(x)dx[/itex]?
  8. Nov 1, 2011 #7
    That was an error I made but I continue on and use substitution for the integral

    [itex]\displaystyle \int_{0}^{\pi }-cos^2(\theta) sin(\theta) d\theta + cos (\theta) sin^2(\theta) d\theta[/itex]

    I let [itex]u=cos \theta[/itex] for the first integral and [itex]u=sin \theta[/itex] for second integral. I get an answer of -2/3....?

    If I integrate first then I get sin(x) -cos(y) to be evaluated to some limits...not sure how to proceed..

    In addition to this, in my notes it states not to "calculate"

    [itex]\int f_1(x,y,z)dx+f_2(x,y,z)dy+f_3(x,y,z)dz[/itex] etc with respect to its variable while keeping the other 2 constant.....I suspect it is ok to do in this example because each integral is only a function of one variable...?
  9. Nov 1, 2011 #8

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    Uhhm... did you know that cos(cos(θ))≠cos2(θ)?

    With your original parametrization you would have:
    [tex]\left.sin(x(t)) - cos(y(t))\right|_{t=1}^{t=-1}[/tex]
    or with HoI's parametrization
    [tex]\left.sin(x(θ)) - cos(y(θ))\right|_{θ=0}^{θ=\pi}[/tex]

    If you keep one variable constant, say x, then you'd have [itex]y=\sqrt{1-x^2}[/itex].
    So your integral becomes:
    [tex]\int_{x=1}^{x=-1} \cos(x)dx + \sin(\sqrt{1-x^2}) \cdot d(\sqrt{1-x^2})[/tex]
    [tex]\int_{x=1}^{x=-1} \cos(x)dx + \sin(\sqrt{1-x^2}) \cdot {1 \over 2\sqrt{1-x^2}} \cdot -2x \cdot dx[/tex]

    I recommend solving the second part of the integral with the substitution [itex]y=\sqrt{1-x^2}[/itex]. :wink:
  10. Nov 1, 2011 #9
    Oh, I over looked that substitution mistake!! Thanks!

    How is yours evaluated? The sine or cos of 1 is not correct...?

    So based on my comment on the way NOT to calculate...is it okay to do here because we are only dealing with one variable per integral?

    I get the answer 2 using HoI's method.
    Attempt on difficult integral will follow.
  11. Nov 1, 2011 #10

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    You have:
    [tex]\left.\sin(x(t)) - \cos(y(t))\right|_{t=1}^{t=-1}[/tex]
    with [itex]x(t)=t[/itex] and [itex]y(t)=\sqrt{1-t^2}[/itex]

    [tex]\left.\sin(t) - \cos(\sqrt{1-t^2})\right|_{t=1}^{t=-1}[/tex]

    So yes, you would get sin(1) in there.

    You would get:
    [tex](\sin(-1) - \cos(\sqrt{1-(-1)^2})) - (\sin(1) - \cos(\sqrt{1-1^2})) = -2 \sin(1)[/tex]

    I'm not sure what it means NOT to calculate.
    Does it mean you should not actually integrate or something?

  12. Nov 1, 2011 #11
    How would -2sin(1) evaluate to 2?

    Yes, it says

    "you should never calculate

    [itex]\int f_1(x,y,z)dx+f_2(x,y,z)dy+f_3(x,y,z)dz[/itex] by integrating f_1(x,y,z) wrt x (treating y+z like constants), same for f_2(x,y,z) etc. THis will give wrong answer....."
  13. Nov 1, 2011 #12

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    It doesn't. Why should it? :confused:

    Ah, now I get it.
    Indeed you can not do the integration wrt to 1 variable and expect it to work out.

    And yes, in this case it was possible, because the integral could be separated into separate integrals with only 1 variable.
  14. Nov 1, 2011 #13
    But arent the 3 different methods suppose to yield the saem answer...ie your method as in post #2 should yield the same as HoI..right? :-)
  15. Nov 1, 2011 #14

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    Of course.
    Any reason to think that it doesn't?
  16. Nov 1, 2011 #15


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    There is a much easier way to do this problem. Call C1 the given curve, C2 the straight line segment from [-1,0] to [1,0], and let C be [itex]C_1\cup C_2[/itex]. Since you have the hypotheses for Green's Theorem you have:

    [tex]0 = \int_C \cos x\, dx + \sin y\, dy = \left(\int_{C_1}+\int_{C_2}\right)( \cos x\, dx + \sin x\, dy)[/tex]

    [tex]\int_{C_1} \cos x\, dx + \sin y\, dy = - \int_{C_2} \cos x\, dx + \sin y\, dy
    = -\int_{-1}^1\cos x\, dx[/tex]
  17. Nov 1, 2011 #16

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    Uhh... :uhh: Aren't you glossing over the fact the the curl of the function over the surface enclosed by C is zero in this case? :confused:
  18. Nov 1, 2011 #17
    Im afraid I dont know. In your method, we have integrated wrt x and used the x coordinates...so it seems right is it not?

    For the difficult integral I end up with

    [itex]sin(x)|_1^{-1} + (-cos(1-x^2)^{1/2})|_1^{-1}[/itex]
  19. Nov 1, 2011 #18

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    Yes..... :rolleyes:
  20. Nov 1, 2011 #19
    This doesnt give HoI answer of 2 either. I give up. I least I get nearly full marks :-) Thanks.

    BTW..when I say I nearly get full marks...'if' it was an exam. It is just exercises :-)
  21. Nov 1, 2011 #20

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    Ah well, with HoI's parametrization you get:
    [tex]\left.\sin(x(θ)) - \cos(y(θ))\right|_{θ=0}^{θ=\pi}[/tex]
    [tex]\left.\sin(\cos(θ)) - \cos(\sin(θ))\right|_{θ=0}^{θ=\pi}[/tex]
    [tex](\sin(\cos(\pi)) - \cos(\sin(\pi))) - (\sin(\cos(0)) - \cos(\sin(0))) = -2\sin(1)[/tex]
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