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Find Mass and C.O.M using a line integral

  1. May 14, 2013 #1
    1. The problem statement, all variables and given/known data

    Using line integrals, find the mass and the position of the center of mass of a thin wire in the shape of a half-circle [itex]x^{2} + y^{2} = r^{2}, x ≥ 0 [/itex] and [itex]-r ≤ y≤ r[/itex] if the linear density is [itex]ρ(x,y) = x^{2} + y^{2}[/itex]

    The mass is given by the integral of the density along the curve while the center of mass is defined as

    [itex]\frac{∫_{C} x ρ(x,y)ds}{∫_{C}ρ(x,y)ds}[/itex]

    for the x co-ordinate and similarly for the y co-ordinate.

    3. The attempt at a solution

    My attempt was to substitute [itex]x^{2} = y^{2} - r^{2}[/itex] and [itex]y^{2} = x^{2} - r^{2}[/itex] into the density function to get [itex]ρ(x,y) = x^{2} + y^{2} - 2r^{2}[/itex] then do the double integral

    [itex]∫^{x}_{0}∫^{r}_{-r} x^{2} + y^{2} - 2r^{2} dy dx[/itex]

    from that I got an answer of [itex]\frac{2rx^{2}}{3}[/itex]

    EDIT: I get [itex]\frac{2rx^{2}}{3} - \frac{10r^3x}{3}[/itex] rather then [itex]\frac{2rx^{2}}{3}[/itex]

    Is that correct? Or am I supposed to follow Example 1 on this page, http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsPtI.aspx ?
     
    Last edited: May 14, 2013
  2. jcsd
  3. May 14, 2013 #2

    LCKurtz

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    Stop right there! Those are line integrals and you don't evaluate them with double integrals. Do the line integrals. Polar coordinates might be a good idea.
     
  4. May 14, 2013 #3
    Hah, physics 211 I assume?
     
  5. May 15, 2013 #4
    Physics 211 indeed, ParoxysmX =]. LCKrutz, ok so,

    [itex]x = r cos(t), y = r sin(t)[/itex]

    [itex]ds = \sqrt{((dx/dt)^{2} + (dy/dt)^{2})}
    = \sqrt{(r^{2}sin^{2}(t) +r^{2}cos^{2}(t))}dt
    = \sqrt{r^2}dt = r dt[/itex]

    Then substitute x and y values into the density function to get

    [itex]Mass = ∫_{C}(r^{2}cos^{2}(t) + r^{2}sin^{2}(t))(r)dt[/itex]
    [itex]Mass =∫_{C}r^{3}(cos^{2}(t)+sin^{2}(t))dt = ∫_{C}r^{3}dt [/itex]
    [itex]Mass =r^{3}t |^{\pi}_{0} = r^{3}\pi[/itex]

    Correct?

    Then for the x and y, do the above process but multiply by [itex]r cost[/itex] and [itex]r sint[/itex] and divide by [itex]r^{3}\pi[/itex] to get x and y which are [itex]x = 0[/itex] and [itex]y = 2r^{4}/r^{3}\pi[/itex].
     
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