A thin wire of constant linear mass density $k$ takes the shape of an arch of the cycloid $$x = a(t − \sin t),\quad y = a(1 − \cos t), \quad 0 ≤ t ≤ 2π.$$
Determine the mass $m$ of the wire, and find the location of its center of mass.
I am assuming I am supposed to use line integrals and vector fields to solve this. Any help would be appreciated. :)
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The object is to find the arc length $s$ of the curve
$$x = a(t − \sin t),\quad y = a(1 − \cos t), \quad 0 ≤ t ≤ 2π$$
We can set $a=1$ since it's just a scaling factor.
The arc length is given by
$$s=\int_0^{2\pi}\sqrt{1+\left(\frac{dy}{dx} \right)^2 }\ dx$$
Now, we have
$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{\sin t}{1-\cos t}$$
and
$$s=\int_0^{2\pi}\sqrt{1+\left(\frac{dy}{dx} \right)^2 }\ \frac{dx}{dt}dt$$
After some straightforward manipulation we obtain
$$s=\int_0^{2\pi}\sqrt{(1-\cos t)^2+\sin^2 t}\ dt=\int_0^{2\pi}\sqrt{2-2\cos t}\ dt$$
With the help of WolframAlpha we get
$$s=2\sqrt{2-2\cos t}\cdot\cot(t/2)\big|_0^{2\pi}=8$$
Hints:
To find the mass, find the arc-length and multiply with the density.
To find the center of mass, you know that, due to symmetry, it must lie halfway through the arc in the $x$-direction. Now you only need to find the mean value of the $y$-values present in the arc.