I've always known how to get the center of mass of any region, but now i met a new question with a region bounded by a parametric curve and the question is to get its baricenter! My question is what changes in this case than the normal way of getting the baricenter? Because i dont have any idea now how to solve this question any help please?
The region is bounded by the parametric curve :
$ x(t)=cos(2\pi t)$
$ y(t)=t-t^3 $
$t\in [0,1] $
For any $u\in\left[0,\frac{1}{2}\right]$, the points corresponding to $t=\frac{1}{2}+u$ and $t=\frac{1}{2}-u$ have the same $x$-coordinate. It follows that the section given by $x=x_0\in[-1,1]$ is a segment between the points given by $t=\frac{\text{arccos}(x_0)}{2\pi}$ and $t=1-\frac{\text{arccos}(x_0)}{2\pi}$, having length given by $$ L(x_0)=\frac{\text{arccos}(x_0) \left(2 \pi ^2-3 \pi\,\text{arccos}(x_0)+\text{arccos}(x_0)^2\right)}{4 \pi ^3} $$ and centre located at $$ C(x_0) = \left(x_0, \frac{3 \text{arccos}(x_0) \left(\frac{3 \pi }{2}+\text{arcsin}(x_0)\right)}{8 \pi ^2}\right).$$
It follows that the barycenter $G(x_G,y_G)$ of the region enclosed by your parametric curve is given by:
$$ x_G = \frac{\int_{-1}^{1}x L(x)\,dx}{\int_{-1}^{1}L(x)\,dx},\qquad y_G = \frac{\int_{-1}^{1}C(x) L(x)\,dx}{\int_{-1}^{1}L(x)\,dx}$$ where: $$ \int_{-1}^{1}L(x)\,dx = \frac{3}{2\pi^2},\qquad \int_{-1}^{1}x\,L(x)\,dx = \frac{3}{32\pi^2},\qquad \int_{-1}^{1}C(x)L(x)\,dx = \frac{3(15-\pi^2)}{8\pi^2} $$ give that the centroid lies at