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

[ZeroSum]

Homework Statement

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

Homework Equations

Arc length formula:
ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2}

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

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:

\frac{dx}{dt} = 1
\frac{dy}{dt} = -\sin{t}
\frac{dz}{dt} = \cos{t}

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

Is this correct?

That would make the remainder of the equations:

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

= \frac{3(2\pi)^4}{32\pi^3} - 0
= \frac{3\pi}{2}

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

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

= \frac{3}{2 \pi^2}

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

Using WA to shorten this...

= -\frac{3}{2 \pi^2}

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.

 

[LCKurtz]

Homework Statement

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

Homework Equations

\int_C \rho(x,y,z)\,ds

Arc length formula:
ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2}

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:

\frac{dx}{dt} = 1
\frac{dy}{dt} = -\sin{t}
\frac{dz}{dt} = \cos{t}

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

Is this correct? Also, how do I go about computing the center of mass over a line integral like this? I can't find any formulas for that in my textbook.

For the mass you want to calculate

m = \int_C \rho(x,y,z)\, ds

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

\overline x = \frac 1 m \int_C x \rho(x,y,z)\, ds

and similarly for the other two coordinates.

 

[ZeroSum]

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 \rho = x^2 + y^2 + z^2 we get \rho = t^2 + \cos^2{t} + \sin^2{t} which is just \rho = t^2 + 1 by the Pyth. ID.

That would yield a mass of:

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

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:

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)

Is this the correct answer for that part as well?