Baricenter of a region bounded by a closed parametric curve

Question

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] $

Answer 1

So we are looking for the baricenter (centroid) of the region enclosed by the curve $$ C:\left\{ \begin{gathered} 0 \leqslant t \leqslant 1 \hfill \\ x(t) = \cos \left( {2\pi t} \right) \hfill \\ y(t) = t - t^{\,3} \hfill \\ \end{gathered} \right. $$

Among the methods exposed in answering to Baricenter of a region bounded by a parametric curve we can in this case profitably choose the last.

Thus the area of the region enclosed by $C$ will be: $$ \begin{gathered} A = \frac{1} {2}\oint\limits_C {\left| {\,\begin{array}{*{20}c} x & y \\ {dx/dt} & {dy/dt} \\ \end{array} \,} \right|dt} = \frac{1} {2}\int_{t\, = \,0}^{\;1} {\left| {\,\begin{array}{*{20}c} {\cos \left( {2\pi t} \right)} & {t - t^{\,3} } \\ { - 2\pi \sin \left( {2\pi t} \right)} & {1 - 3t^{\,2} } \\ \end{array} \,} \right|dt} = \hfill \\ = \frac{1} {2}\int_{t\, = \,0}^{\;1} {\left( {\left( {1 - 3t^{\,2} } \right)\cos \left( {2\pi t} \right) + 2\pi \left( {t - t^{\,3} } \right)\sin \left( {2\pi t} \right)} \right)dt} = - \frac{3} {{2\,\pi ^{\,2} }} \hfill \\ \end{gathered} $$

which tells us that, for $t$ going from $0$ to $1$, we are tracing the curve clock-wise.
Then the baricenter $\mathbf{b}$ will be: $$ \begin{gathered} \mathbf{b} = \frac{1} {{3A}}\oint\limits_C {\mathbf{r}\left| {\,\begin{array}{*{20}c} x & y \\ {dx/dt} & {dy/dt} \\ \end{array} \,} \right|dt} = \hfill \\ = - 2\left( {\frac{\pi } {3}} \right)^{\,2} \int_{t\, = \,0}^{\;1} {\left( \begin{gathered} \cos \left( {2\pi t} \right) \\ t - t^{\,3} \\ \end{gathered} \right)\left( {\left( {1 - 3t^{\,2} } \right)\cos \left( {2\pi t} \right) + 2\pi \left( {t - t^{\,3} } \right)\sin \left( {2\pi t} \right)} \right)dt} = \hfill \\ = - 2\left( {\frac{\pi } {3}} \right)^{\,2} \left( \begin{gathered} - \frac{9} {{32\,\pi ^{\,2} }} \\ \frac{{9\,\pi ^{\,2} - 135}} {{4\,\pi ^{\,4} }} \\ \end{gathered} \right) = \left( \begin{gathered} \frac{1} {{16}} \\ \frac{{15}} {{2\,\pi ^{\,2} }} - \frac{1} {2} \\ \end{gathered} \right) \hfill \\ \end{gathered} $$

which matches with the answer above.

Answer 2

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

$$ \color{red}{G\left(\frac{1}{16},\frac{15}{2\pi^2}-\frac{1}{2}\right)}\approx(0.0625,0.26). $$