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Arc Length and Line Integrals

  1. Apr 1, 2009 #1
    When I take the line integral around a square shape path "C" as follows:

    From A to B to C to D to A

    C1 = A(0, 0) to B (4, 0)

    t i
    0 <= t <= 4

    C2 = B (4, 0) to C (4, 7)

    4 i + (t - 4) j
    4 <= t <= 11

    C3 = C (4, 7) to D (0, 7)

    (15 - t) i + 7 j

    11 <= t <= 15

    C4 = D (0, 7) to A (0, 0)

    (22 - t) j

    15 <= t <= 22

    Why is that when I take the line integral around this path using [tex]\int_{C }||r'(t)|| dt[/tex] and the above parameterization, I end up with 0, when I should be getting the arc length of the path, which is the perimeter of the square?

    Thanks!
     
  2. jcsd
  3. Apr 2, 2009 #2

    HallsofIvy

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    When you take the integral of what around that path?

    If f(x,y)dx+ g(x,y)dy is an exact differential, that is if there exist F(x,y) such that dF= f(x,y)dx+ g(x,y)dy, then the integral from any point [itex](x_0,y_0)[/itex] to [itex](x_1,y_1)[/itex],
    [tex]\int_{(x_0,y_0)}^{(x_1,y_1)} f(x,y)dx+ g(x,y)dy= F(x_1,y_1)- F(x_0,y_0)[/tex]
    and, in particular, if [itex](x_1,y_1)= (x_0,y_0)[/itex] then [itex]F(x_1,y_1)= F(x_0,y_0)[/itex] and that difference is 0.

    Now, what is ||r'(t)||?
     
  4. Apr 2, 2009 #3
    I was taking the arc length of f(x, y) = 1 along ds.

    Since "x" is a potential function of 1, it seems I should then be getting 0.

    Yet I thought taking a line integral where f(x, y) = 1 should be giving me the arc length of the path, yet this turns out to be 0.

    Where am I mistaken?
     
  5. Apr 3, 2009 #4
    Also [tex]||r'(t)||= \frac{ds}{dt}[/tex]
     
  6. Apr 3, 2009 #5

    HallsofIvy

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    This makes no sense. You don't find the "arc length" of a function, you find the length of an arc. In particular, integrating the dervative of any differentiable function around a closed path will give 0.

    You are not taking direction into account. On the line (0,0) to (4,0), ds= dx and you have
    [tex]\int_0^4 dx= 4[/itex]

    On the line segment (4, 0) to (4, 7), ds= dy and you have
    [tex]\int_0^7 dy= 7[/itex]

    But on the line segment (4, 7) to (0, 7), x is decreasing from 4 to 0 so ds= -dx and you integrate
    [tex]\int_4^0 (-dx)= -(-4)= 4[/tex]

    On the line integral (0,7) to (0,0), y is decreasing from 7 to 0 so ds= - dy and you integrate
    [tex]\int_7^0 (-dy)= -(-7)= 7[/tex]

    The integral around the entire rectangle is 4+ 7+ 4+ 7= 22.
     
    Last edited: Apr 3, 2009
  7. Apr 3, 2009 #6
    Hmm, I see your point. I meant we are taking the line integral along [tex]ds[/tex] with [tex]f(x,y) =1[/tex]

    I was computing it wrong, as the perimeter can be calculated with such a line integral.

    [tex]\int_C f(x, y)||r'(t)|| dt =[/tex]

    [tex]C_{1}=\int_0^4 dt = 4[/tex]

    [tex]C_{2}=\int_4^{11} dt = 7[/tex]

    [tex]C_{3}=\int_{11}^{15} ||-1|| dt = 4[/tex]

    [tex]C_{4}=\int_{15}^{22} ||-1|| dt = 7[/tex]

    [tex]C_{1} + C_{2} + C_{3} + C_{4} = 22[/tex]
     
  8. Apr 3, 2009 #7

    HallsofIvy

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    How did you get the values t= 11, t= 15, t= 22?:wink:
     
  9. Apr 3, 2009 #8
    By constructing a piece-wise vector valued function around the rectangle.
     
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