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

## 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.

Last edited:

LCKurtz
Homework Helper
Gold Member

## 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.

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?