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Line integral problem

  1. Dec 11, 2007 #1
    1. The problem statement, all variables and given/known data

    Let F=x[tex]^{2}[/tex]i+2xyj, and let C be the lower half of the unit circle, with perametrization r(t)=<cos(t),sin(t)>,[tex]\pi[/tex][tex]\leq[/tex]t[tex]\leq[/tex][tex]\pi[/tex]. Evaluate [tex]\oint[/tex]F[tex]\cdot[/tex]dr.


    2. Relevant equations



    3. The attempt at a solution

    The first thing I tried to do was to find a function f(x,y) so that F=[tex]\nabla[/tex]f

    In order to do this I integrated the i portion of F with respect to x which gave me f(x,y)=[tex]\frac{1}{3}[/tex]x[tex]^{3}[/tex]+g(y).

    Then to find out what g was I took the derivative with respect to y which just left me with g'(y). I could not think of how to get past this part. I tried doing it with the j first as well in which I first take the integral of the j part of F with respect to y and then the derivative with respect to x. This left me with f[tex]_{x}[/tex](x,y)=y[tex]^{2}[/tex]+f'(x).

    Basically I was trying to use the Fundamental Theorem of Line Integrals but I could not find a function f(x,y) so that F=[tex]\nabla[/tex]f. Help would be appreciated.
     
    Last edited: Dec 11, 2007
  2. jcsd
  3. Dec 12, 2007 #2

    Avodyne

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    Science Advisor

    It's not necessarily the case that F is the gradient of anything. If if happens to be, then the line integral is independent of the path, which makes it particularly easy; but if not, then you just have to do the integral.
     
  4. Dec 12, 2007 #3

    HallsofIvy

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    Do you mean [itex]-\pi\le x\le 0[/itex]? Or [itex]\pi\le x\le 2\pi[/itex] which would work just as well.

    Evaluate [tex]\oint[/tex]F[tex]\cdot[/tex]dr.


    2. Relevant equations



    3. The attempt at a solution

    The first thing I tried to do was to find a function f(x,y) so that F=[tex]\nabla[/tex]f

    In order to do this I integrated the i portion of F with respect to x which gave me f(x,y)=[tex]\frac{1}{3}[/tex]x[tex]^{3}[/tex]+g(y).

    Then to find out what g was I took the derivative with respect to y which just left me with g'(y). I could not think of how to get past this part. I tried doing it with the j first as well in which I first take the integral of the j part of F with respect to y and then the derivative with respect to x. This left me with f[tex]_{x}[/tex](x,y)=y[tex]^{2}[/tex]+f'(x).

    Basically I was trying to use the Fundamental Theorem of Line Integrals but I could not find a function f(x,y) so that F=[tex]\nabla[/tex]f. Help would be appreciated.[/QUOTE]
    You can't find such an f because [itex]x^2 dx+ 2xy dy[/itex] is not an exact differential: [itex](x^2)_y= 0[/itex] while [itex](2xy)_x[/itex]= 2y. Since those are not the same, there is no F such that grad F= [itex]x^2\vec{i}+ 2xy\vec{j}[/itex]

    With x= cos(t) and y= sin(t), dx= -sin(t)dt and dy= cos(t)dt [itex]x^2 dx+ 2xy dy= cos^2(t)(-sint dt)+ 2cos(t)sin(t)(cos(t) dt)= cos^2(t) sin(t) dt. It should be easy to integrate that directly.
     
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