1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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]


    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].
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted