1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    I like Serena

    User Avatar
    Homework Helper

    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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    I like Serena

    User Avatar
    Homework Helper

    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

    I like Serena

    User Avatar
    Homework Helper

    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

    I like Serena

    User Avatar
    Homework Helper

    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]

    Substituting:
    [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

    I like Serena

    User Avatar
    Homework Helper

    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

    I like Serena

    User Avatar
    Homework Helper

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

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    I like Serena

    User Avatar
    Homework Helper

    @LCKurtz:

    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

    I like Serena

    User Avatar
    Homework Helper


    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

    I like Serena

    User Avatar
    Homework Helper

    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Evaluate the Following Line Integral Part 1
Loading...