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!

Line integrals

  1. May 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Let F=(3x+2y)i+(2x-y)j Evaluate [itex] \int_{C}F . dr [/itex] where C is the line segment from (0,0) to (1,1)

    2. Relevant equations

    [itex]\int_{a}^{b}[f(x(t),y(t))x'(t) + g(x(t),y(t))y'(t)]dt [/itex]

    3. The attempt at a solution

    How do I choose the correct values of x and y as a function of t?
    And do I just integrate over a = 0 to b = 1 ?

    regards
     
  2. jcsd
  3. May 19, 2009 #2

    Cyosis

    User Avatar
    Homework Helper

    What parametrization have you found for x(t) and y(t)?
     
  4. May 19, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    i.e. what is a parameterization for the line from (0,0) to (1,1)?
     
  5. May 19, 2009 #4
    I'm using the formula to find
    [itex]
    r(t) = (1-t)(x_{0},y_{0})+t(x_{1},y_{1})
    =(1-t)(0,0)+t(1,1)\\

    \text{ so } x = t y = t [/itex] for 0<t<1
     
  6. May 20, 2009 #5
    So is the integral to evaluate

    [itex]

    \int_{0}^{1} [(3t+2t)3+(2t-t)]dt

    [/itex]

    regards
     
  7. May 20, 2009 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    So x= t, y= t. Yes, that goes from (0,0) to (1,1)

    where did you get the "3" multiplying (3t+ 2t)?
     
  8. May 21, 2009 #7
    Let F=(3x+2y)i+(2x-y)j

    [itex]\int_{a}^{b}[f(x(t),y(t))x'(t) + g(x(t),y(t))y'(t)]dt [/itex]


    I got the three from the partial derivative dx of 3x+2y = 3

    But I think I may have stuffed up because the formula
    says x'(t) which would be 1.

    So the integral would be

    [itex]\int_{0}^{1} [(3t+2t)+(2t-t)]dt[/itex]


    Is that better?
     
  9. May 21, 2009 #8

    yes your limits would be from 0 to 1
     
  10. May 22, 2009 #9
    Hi Guys

    If C instead of a line segment is y= x^2 can I substitute x = t so that C is now just
    [latex}
    fy(t)= t^2
    [/itex]

    regards
     
  11. May 22, 2009 #10
    Sorry,

    [itex]
    fy(t)= t^2
    [/itex]
     
  12. May 22, 2009 #11
    I've used the new parameterization of x = t^2 y = t^2 ( for C y=x^2)

    On the original function

    So is the integral to evaluate now

    [itex]\int_{0}^{1} [(3t+2t)2t+(2t-t)2t]dt[/itex]

    Where x'(t) = x'(t^2) = 2t

    and

    Where y'(t) = y'(t^2) = 2t

    Is this right so far?

    regards
    Brendan
     
  13. May 22, 2009 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You need to put x=t^2 and y=t^2 into the definition of the vector F as well, right? Or if you are doing y=x^2, x=t, y=t^2, dx=dt, dy=2tdt. You seem to be mixing up your parametrizations.
     
  14. May 22, 2009 #13
    So I need to put it as
    [itex]\int_{0}^{1} [(3t^2+2t^2)2t+(2t^2-t^2)2t]dt[/itex]

    regards
     
  15. May 22, 2009 #14

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, that's right for parametrizing along the line between (0,0) and (1,1) using x=t^2 and y=t^2. And you should get the same answer as using x=t and y=t.
     
  16. May 23, 2009 #15
    Thanks for all your help guys
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Line integrals
  1. Line Integral (Replies: 1)

  2. Line integral (Replies: 2)

  3. Line Integral (Replies: 4)

  4. Line Integrals (Replies: 4)

  5. Line Integrals (Replies: 4)

Loading...