Baricenter of a region bounded by a closed parametric curve 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] $
 A: 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. 
A: 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

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

