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Line integral

  1. Jul 7, 2016 #1
    1. The problem statement, all variables and given/known data
    i'm not sure what is line integral...


    pcE0VBo.jpg

    2. Relevant equations


    3. The attempt at a solution
    Does it mean total length of line under the curve?
     
  2. jcsd
  3. Jul 7, 2016 #2

    SteamKing

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    If f(x,y,z) = 1, then the line integral of this function over the curve C will give the arc length between a and b. If f(x,y,z) is some other function, this won't be the case.
     
  4. Jul 7, 2016 #3
    If f(x,y,z) is some other function , what would it be?
     
  5. Jul 7, 2016 #4

    SteamKing

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    Who knows?

    Not every line integral is imbued with deep physical meaning.
     
  6. Jul 8, 2016 #5

    LCKurtz

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    If for example ##f(x,y,z)## is the density per unit length of a wire, say in kg/m, then the integral ##\int_C f(x,y,z)~ds## would represent the total kg for the wire.






    c
     
  7. Jul 8, 2016 #6
    Then, how about s( in the first photo in first post) only ? It represents the total length of curve?
     
  8. Jul 8, 2016 #7
    No. s is the cumulative distance along the curve, starting from a specified location.

    What they are doing here is defining the locus of points along a curve in space by specifying each of the three coordinates x, y, and z as a continuous parametric function of a parameter t.
    x = x(t)
    y = y(t)
    z = z (t)
    Specifying a value for t determines the coordinates of any particular point along the curve.

    Consider two closely neighboring points along the curve that are situated at t and at t + dt. The differences in the x, y, and z coordinates of these two points are given by:
    ##dx = x(t+dt)-x(t)=\frac{dx}{dt}dt##
    ##dy = y(t+dt)-y(t)=\frac{dy}{dt}dt##
    ##dz = z(t+dt)-z(t)=\frac{dz}{dt}dt##
    The spatial distance ds between the two neighboring points along the curve is given by the Pythagorean Theorem:
    $$(ds)^2=(dx)^2+(dy)^2+(dz)^2=\left[\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2\right](dt^2)$$
    If you want to find the contour integral of a function f(x,y,z) between two points along the space curve, you want to be integrating f ds. This is the same as$$\int_{t_1}^{t_2}{f\left(x(t),y(t),z(t)\right)\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2}dt}$$
     
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