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] $

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. 
