1. Not finding help here? Sign up for a free 30min 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 integral

  1. Jun 13, 2010 #1
    1. The problem statement, all variables and given/known data
    hello again, sorry for asking so many questions, i just want to make sure if im correct or not

    calculate the line integral y^2dx+x^2dy where line C is the triangle with sides x=1, y=0 and y=x



    3. The attempt at a solution

    first of all i tried to find a customization of the line

    we know that x = 1 hence it will be like this

    r(t) = (1,t) but im not sure if it's correct

    then i said that the integral would be this

    [URL]http://www2.wolframalpha.com/Calculate/MSP/MSP470019b34815c34hai2900000d7egii1beg78b2i?MSPStoreType=image/gif&s=9&w=88&h=37[/URL]

    could I just use Green's theorem?

    I mean using Greens theorem I get the same result

    [URL]http://www2.wolframalpha.com/Calculate/MSP/MSP167119b34a9g72bg50ia000021ai53gc37dac65a?MSPStoreType=image/gif&s=3&w=186&h=37[/URL]

    im 99% sure that greens theorem is correct, i mean the way i implemented it, but is the first way i showed also correct?

    thanks in advance
     
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. Jun 13, 2010 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    [/URL]
    "we know that x=1"? Did you leave out much of the problem? We know that x= 1, y= any number from 0 to 1 is one side of triangle and so one part of the path over which we want to integrate. Taking x= 1, y= t as parametric equations, dx= 0, dy= dt so the integral becomes
    [tex]\int_0^1 1 dx= 1[/itex]

    But you still have to do the other two sides of the triangle.

    On the line y= 0, we can use parametric equations x= t, y= 0 with t from 0 to 1. Then dx= dt, dy= 0 but [itex]y^2 dx= 0dt[/itex] so the integral is
    [tex]\int_0^1 0dt= 0[/itex].

    On the line y= x, where we are integrating from (1, 1) to (0, 0) (we got counterclockwise around the closed path), we can take x= t, y= t so that dx= dt, dy= dt and the integral is
    [tex]\int_1^0 2t^2 dt= -\int_0^1 2t^2 dt= -2/3[/itex] and the entire integral is 1- 2/3= 1/3.

     
    Last edited by a moderator: Apr 25, 2017
  4. Jun 13, 2010 #3
    thanks a lot for your help

    that cleared up everything in my mind
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Line integral
  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...