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Vector Calc-Line Integral

  1. Nov 7, 2006 #1
    A wire lies along the piecewise linear curve extending from the point (2,4) to the point (8,6) to the point (8,11). If the density of the wire is given by p(x,y)=2xy+6x, use a line integral to find the mass of the wire.

    Here is what I have tried:

    C1: (1-t)<2,4>+t<8,6>
    C1: x=2+6t y=4+2t

    [tex]\int_{0}^{1} 2xy+6y dt [/tex]
    [tex]\int_{0}^{1} 2(2+6t)(4+2t)+6(2+6t) dt = 82[/tex]

    C2: (1-t)<8,6> + t<8,11>
    C2: x=8 y=6+5t

    [tex]\int_{0}^{1} 2xy+6y dt [/tex]
    [tex]\int_{0}^{1} 2(8)(6+5t)+6(8) dt = 184[/tex]

    82+184 = 266

    However this is incorrect :(
  2. jcsd
  3. Nov 7, 2006 #2


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    I think the integrand should be [tex]\rho ds=\rho \frac{ds}{dt}dt[/tex]
    where ds is a piece of length from your curve. So if you parametrized the line by [tex]\vec r(t)[/tex], then [tex]ds=|\frac{d\vec r}{dt}|dt[/tex]
  4. Nov 7, 2006 #3
    Are you saying I need to do

    [tex] \int_{0}^{1} (2xy+6x)* \sqrt{(2y+6)^2+(2x)^2} [/tex]

    C1: (1-t)<2,4>+t<8,6>
    C1: x=(2+6t) y=(4+2t)

    [tex]\int_{0}^{1} (2(2+6t)(4+2t)+6(2+6t)) * \sqrt{(2(4+2t)+6)^2+(2(2+6t))^2} [/tex]

    C2: (1-t)<8,6> + t<8,11>
    C2: x=(8) y=(6+5t)

    [tex]\int_{0}^{1} 2xy+6y dt [/tex]
    [tex]\int_{0}^{1} (2(8)(6+5t)+6(8)) * \sqrt{(2(6+5t)+6)^2+(2(8))^2} = [/tex]

    Then add the results of the two integrals?

    However this is incorrect :(
  5. Nov 7, 2006 #4
    Any ideas... I was not quite sure what you meanat by this post...
  6. Nov 7, 2006 #5
    Ahhh I figured it out... I was doing the wrong dervatives....
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