Centre of mass of a wire with line integrals

This solution uses line integrals.

Firstly, we use the parametrization $x=\cos(t), y=\sin(t)$ , $0\leq t \leq \pi$ .

The linear density is $\rho(x,y)=k(1-y)$ , where k is a constant.

The mass of the wire is given by the line integral;

$\displaystyle \int_C k(1-y) ds$ = $\displaystyle \int_{0}^{\pi} k(1-\sin(t)) dt$ = $\displaystyle k(\pi-2)$

We then have the centre mass along $y$ ;

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

$\displaystyle \bar{y} = \frac{1}{k(\pi-2)} \int_C yk(1-y) ds = \frac{1}{\pi -2} \int_{0}^{\pi} y (\sin(t)-\sin^2(t)) dt$

= $\displaystyle \frac{4-\pi}{2(\pi-2)}$

By symmetry we see that $\bar{x}=0$ , the sum of $\bar{x}$ and $\bar{y}$ is 0.38.