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Fundamental Theorem of Line Integrals

  1. Jul 24, 2005 #1
    Quick question, what is the approach to this problem?

    Keep in mind I am supposed to use the Fundamental Theorem of Line Integrals.

    [tex]\int_{C} 2ydx + 2xdy [/tex]

    Where C is the line segment from (0,0) to (4,4).

    Unless I am missing something I need to make that into the form of [tex]\vec{F} \cdot \vec_d{R}[/tex] to use the F.T. of L.I., but I have no clue where to start there. Thanks.
  2. jcsd
  3. Jul 24, 2005 #2
    the fund thm of line integrals states that for a continuous smooth curve C, the following integral depends only on the value of f(x,y) at the END POINTS r1 and r2 of C:

    [tex] \int_{C} (\nabla f) \bullet d \vec{r} \ = \ f(\vec{r_{2}}) - f(\vec{r_{1}})[/tex]

    to apply this thm to your problem, note that:

    [tex] (\nabla f) \bullet d \vec{r} \ = \ \frac{\partial f}{\partial x} dx \ + \ \frac{\partial f}{\partial y} dy [/tex]

    and thus for your problem integral, we must determine f(x,y) such that the following is true and then evaluate {f(4,4) - f(0,0)}:

    [tex] \frac{\partial f}{\partial x} \ = \ 2y [/tex]

    [tex] \frac{\partial f}{\partial y} \ = \ 2x [/tex]

    it follows that:

    [tex] f(x,y) \ = \ 2xy [/tex]

    and hence:

    [tex]\int_{C:(0,0)to(4,4)} (2y)\,dx \ + \ (2x)\,dy \ \, = \, \ \int_{C:(0,0)to(4,4)} \left ( \frac{\partial f}{\partial x} dx \ + \ \frac{\partial f}{\partial y} dy \right ) \qquad \mbox{\{where f(x,y) = 2xy\}} [/tex]

    [tex] \ \, = \, \ \int_{C:(0,0)to(4,4)} (\nabla f) \bullet d\vec{r} \ \, = \, \ f(4,4) \ - \ f(0,0) \ \, = \, \ \color{red} \ ??? \ \ \mbox{(u finish prob solution)} [/tex]
    Last edited: Jul 24, 2005
  4. Jul 24, 2005 #3
    Thanks, I must have missed that in the book!
  5. Jul 24, 2005 #4
    just a minor point:

    when solving exact differential equations like this, the solution should have a " + h(x)" or " + h(y)" term, depending on whether you used separation of variables with df/dx or df/dy. (the function h will depend solely on the OTHER variable, as it is much like a constant of integration--it's the stuff that turns into zero once you differentiate with respect to a specified variable.)

    so let's say we solve df/dx = 2y:

    f(x,y) = 2xy + h(y)

    now to get h(y)...

    df/dy = 2x + h '(y) = 2x

    (the far right hand side was given to us from that first line integral, supposing that Q(x,y) = df/dy)

    so we have that h '(y) = 0. this implies that h(y) = constant.

    as far as this problem is concerned, we can pick any value for that constant, since it will be subtracted out at the end anyway.

    you'll come across this method again in diff eq, so you might as well see it now, i think.
  6. Jul 24, 2005 #5
    Yeah that h(x) or h(y) was in the book. My teacher, however, just said to say the first part is u(x,y) and the second part v(x,y) then take the antiderivative of u(x,y) with respect to x and the antiderivative of v(x,y) with respect to y. Then add the two together, but do not add the same things together. Meaning for this problem, I got the antiderivative of both as 2xy, but it would be just 1 times 2xy not 2 times 2xy.
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