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Line integral examples in book

  1. Nov 22, 2005 #1
    Hi, I've just started working on line integrals and I don't understand one of the examples in my book.

    \int\limits_C {y^2 dx + xdy}

    Where C is the arc of the parabola x = 4 - y^2 from (-5,-3) to (0,2). The book proceeds by suggesting that y is taken as the parameter so that the arc C is represented by x = 4 - y^2, y = y and -3 <= y <= 2.

    Then dx = -2ydy so that [tex]\int\limits_C {y^2 dx + xdy} = \int\limits_{ - 3}^2 {y^2 \left( { - 2y} \right)dy} + \left( {4 - y^2 } \right)dy = ... = 40\frac{5}{6}[/tex].

    So one of the variables (in this case y) has been taken as the parameter. I am wondering why that decision was made. Why note take x as the parameter? Is it because taking y as the parameter leads to a simpler calculation.

    I am also having having trouble understanding the difference between a line integral wrt to an arc length and a line integral wrt x or y. I have the following formula for the line integral with respect to arc length.

    \int\limits_C {f\left( {x,y} \right)} ds = \int\limits_a^b {f\left( {x\left( t \right),y\left( t \right)} \right)} \sqrt {\left( {\frac{{dx}}{{dt}}} \right)^2 + \left( {\frac{{dy}}{{dt}}} \right)^2 } dt
    [/tex]...equation (1)

    The formula for the integral with respect to the y is:

    \int\limits_C {f\left( {x,y} \right)} dy = \int\limits_a^b f \left( {x\left( t \right),y\left( t \right)} \right)y'\left( t \right)dt
    [/tex]...equation (2)

    My confusion basically comes down to not understanding the difference between equations (1) and (2). In equation (1) is the integration along some curve? In equation (2) is the integration along the y-axis? If so then how come to example above, y is chosen as the parameter rather than using t as the parameter as equation (2) suggests it should be done.

    I know that in this case using y, or t or pretty much any other variable as the parameter makes no difference to the value of the parameter. It just seems strange that the example is not consistent with equation (2).

    The above is probably quite confusing by any help would be great thanks.
  2. jcsd
  3. Nov 22, 2005 #2
    If you start with a curve C of arc lenght s, given in parametric form

    t is the parameter. If you don't start with the curve s given in parametric form, but rather you have y=y(x) ( or x=x(y) in the case of the parabola in your proble x=y^2-4) is up to you to parametrize the equation in the way you find more convenient.
    In your example y is used as the parameter. They could have used x as the parameter, or introduce another parametrization for the parabola in terms of some t,w,u,v or whatever. It is just simpler to use y as the parameter, but not obligatory.
    Last edited: Nov 22, 2005
  4. Nov 22, 2005 #3
    Although you may write the line integrals in terms of x or interms of y, doing so in terms of y drastically simplifies the calculation. If you were to write it in terms of x, you would have to break up your integral into 2, since solving for y yields 2 solutions. y=+-sqrt(4-x). You must deal with each case separately. Chosing the negative value and integrating from -5 to 4, and then chosing the positive value and integrating from 0 to 4. For any contour/path that is not one to one, it would be better to write things in terms of y.
  5. Nov 22, 2005 #4
    equation 1 as you call it above, is just the equivalence of a line integral of a function f(x,y) over a contour C, given by the parameterization C:=(x(t), y(t)). equation 2 says something different, and I think you meant dr not dy. Eq 2 is the line integral of a vector field, defined on a contour C, given by the parameterization r(t)=(x(t),y(t)). So in summary, one is the line integral of a function, the other of a vector field.
  6. Nov 22, 2005 #5
    I don't know if I was clear, eq 2 should read: int(f(x,y)dr,C)=int(f(x(t),y(t))*r'(t)dt)
  7. Nov 22, 2005 #6
    About your other question, i think that you are confusing the line integral of a scalar field f(x,y) over the curve C


    with the line integral of a vector field lets say F
    int(Fods )
    where o denotes the dot product and ds here if the differential VECTOR lenght over the curve. In 2D, F=F(x,y)=F1(x,y) i + F2(x,y) j , i and j are unit vector in the x and y direction. Hence

    int_C(Fods)=int( F1 dx + F2 dy)

    In this case if the curve can be parametrized by and equation r=r(t) then

    int_C(Fods)=int_{a}^{b} ( F(r(t))o(dr/dt) dt)
  8. Nov 22, 2005 #7
    Hmm...thanks for the input mathphys and BerkMath. I'll go over the theory section and see if I can spot any misconceptions I might have.
  9. Nov 22, 2005 #8
    If you look carefully line integral w.r.t. dx,dy and dz is a result of dotting the line integral with a vector field. Thats why some problems are in terms of dx,dy and dz. I thought the same thing when I first came across it.
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