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Finding the mass and center of mass of a wire using a line integral.

  1. Apr 12, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the mass and center of mass of a wire in the shape of the helix [tex]x=t[/tex], [tex]y=\cos{t}[/tex], [tex]z = \sin{t}[/tex], [tex]0 \le t \le 2 \pi[/tex], if the density at any point is equal to the square of the distance from the origin.

    2. Relevant equations
    Arc length formula:
    [tex]ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2}[/tex]

    [tex]m = \int_C \rho(x,y,z)\,ds[/tex]
    [tex]\bar{x} = \frac{1}{m} \int_C x \rho(x,y,z)\,ds[/tex]
    [tex]\bar{y} = \frac{1}{m} \int_C y \rho(x,y,z)\,ds[/tex]
    [tex]\bar{z} = \frac{1}{m} \int_C z \rho(x,y,z)\,ds[/tex]

    3. The attempt at a solution

    I've tried using the parametric equations given to find the value for ds using the arc length formula. I cannot be sure if what I've done yields the proper answer, though. Here is my attempt at an answer:

    [tex]\frac{dx}{dt} = 1[/tex]
    [tex]\frac{dy}{dt} = -\sin{t}[/tex]
    [tex]\frac{dz}{dt} = \cos{t}[/tex]

    [tex]m = \int_{0}^{2\pi} t^2 \sqrt{\left(1\right)^2 + \left(-sin{t}\right)^2 + \left(cos{t}\right)^2}\, dt[/tex]
    [tex]=\int_{0}^{2\pi} t^2 \sqrt{2} \, dt[/tex]
    [tex]=\frac{\sqrt{2} \, t^3}{3}\right \bigg{|}^{2\pi}_{0}[/tex]
    [tex]=\frac{8 \sqrt{2}\, \pi^3}{3}[/tex]

    Is this correct?

    That would make the remainder of the equations:

    [tex]\bar{x} = \frac{3}{8 \sqrt{2}\, \pi^3} \int_C t(t^2) \sqrt{2}\,ds[/tex]
    [tex] = \frac{3}{8 \, \pi^3} \int_C t^3\,ds[/tex]
    [tex] = \frac{3}{8 \, \pi^3} \frac{t^4}{4} \bigg{|}^{2\pi}_{0}[/tex]

    [tex] = \frac{3(2\pi)^4}{32\pi^3} - 0[/tex]
    [tex] = \frac{3\pi}{2} [/tex]

    [tex]\bar{y} = \frac{3}{8 \sqrt{2}\, \pi^3} \int_C t^2 \cos{t} \sqrt{2}\,ds[/tex]
    [tex]\bar{y} = \frac{3}{8 \, \pi^3} \int_C t^2 \cos{t}\,ds[/tex]

    Bunch of ugly integration by parts here, which I'll have Wolfram Alpha do...

    [tex] = \frac{3}{2 \pi^2}[/tex]

    [tex]\bar{z} = \frac{3}{8 \sqrt{2}\, \pi^3} \int_C z \sqrt{2}\,ds[/tex]

    Using WA to shorten this...

    [tex] = -\frac{3}{2 \pi^2}[/tex]

    Since my book does not have answers for even problems, I can't tell if these are correct or not. Could anyone check them to see if I know what I'm doing with these?

    Edit: I found the equation for the center of mass and updated everything.
    Last edited: Apr 12, 2011
  2. jcsd
  3. Apr 12, 2011 #2


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    Gold Member

    For the mass you want to calculate

    [tex]m = \int_C \rho(x,y,z)\, ds[/tex]

    which is apparently what you tried, but you didn't label it as m and t2 is not the distance from (x,y,z) to the origin, squared. You need x2+y2+z2 in the integrand.

    Then to get the center of mass you need

    [tex]\overline x = \frac 1 m \int_C x \rho(x,y,z)\, ds[/tex]

    and similarly for the other two coordinates.
  4. Apr 12, 2011 #3
    Thank you, LCKurtz.

    I edited my post right as you were replying, I guess. I found and added the formulas and added the m = to the front of the equation for mass along the line integral.

    So, if we're using [tex]\rho = x^2 + y^2 + z^2 [/tex] we get [tex]\rho = t^2 + \cos^2{t} + \sin^2{t}[/tex] which is just [tex]\rho = t^2 + 1[/tex] by the Pyth. ID.

    That would yield a mass of:

    [tex]m = \frac{\sqrt{2} (8\pi^3 + 6\pi)}{3}[/tex]

    I left out the steps since I expect you can use Wolfram Alpha as well as I. Is this what you get as well?

    Then subbing in 1/m and solving the other equation in the same way I get:

    [tex]cm = \left(\frac{3 (\pi+2 \pi^3)}{3+4 pi^2},\frac{6}{3+4 \pi^2}, -\frac{6 \pi}{3+4 \pi^2}\right)[/tex]

    Is this the correct answer for that part as well?
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