Baricenter of a region bounded by a parametric curve I just want to ask if there exists a general rule to get the baricenter of a region bounded by a parametric curve? 

To my knowledge a "wide" approach (don't dare to call it "general") would be as follows:  


*

*Determine if the curve does not intersect itself and does not retrace; in case, separate/limit the parameter excursion appropriately.  

*Call  $
\mathbf{r} = \left( \begin{gathered}
  x \hfill \\
  y \hfill \\ 
\end{gathered}  \right)$
the position vector, $C$ the curve, $G$ the region enclosed by it, $A$ its area
$A = \iint\limits_{\left( {x,y} \right)\, \in \;G} {dx\,dy}$
then the baricenter (centroid) will be given by
$\mathbf{b} = \frac{1}
{A}\iint\limits_{\left( {x,y} \right)\, \in \;G} {\mathbf{r}\,dx\,dy}$

*The baricentre of a composite figure is the "area-weighted" average of the baricenters of each portion.
The same holds true in the limit so
$$
\mathbf{b} = \frac{1}
{A}\int\limits_G {\mathbf{\beta }\,dA} 
$$
where $\mathbf{\beta }(x)$ is the baricenter of each infinitesimal portion of area $dA$, into which $G$ has been exhaustively decomposed.

*Now, if the expressions for $x(t)$ and $y(t)$ allow to be converted into easily-integrable $y(x)$ or $x(y)$, then use the double integral,.
That is equivalent  to convert into stripes parallel to the $x$ (or $y$) axis, whose $\mathbf{\beta }(x)$ and $dA(x)$
(or expressed in terms of $y$ or $t$) are easily determined, i.e. the integral in $dA$.

*Instead, if the expressions for $x(t)$ and $y(t)$ allow to be converted into easily-integrable polar coordinates
$\rho$ and $\alpha$, then splitting the area into circular sectors, we can write
$$
\mathbf{b} = \frac{1}
{A}\int\limits_{\left( {\rho ,\alpha } \right)\, \in \;C} {\left( \begin{gathered}
  2/3\,\rho \cos \alpha  \hfill \\
  2/3\,\rho \sin \alpha  \hfill \\ 
\end{gathered}  \right)\frac{1}
{2}\rho ^{\,2} \,d\alpha }  = \frac{1}
{{3A}}\int\limits_{\left( {\rho ,\alpha } \right)\, \in \;C} {\left( \begin{gathered}
  \cos \alpha  \hfill \\
  \sin \alpha  \hfill \\ 
\end{gathered}  \right)\rho ^{\,3} \,d\alpha } 
$$
6.The integral in $dA$ can also be put in an alternative way, considering that the area spanned by $ \mathbf{r} $
can be written as
$$
dA = \frac{1}
{2}\left| {\,\mathbf{r} \times d\mathbf{r}\,} \right|\quad \text{(with}\;\text{sign)} = \frac{1}
{2}\left| {\,\begin{array}{*{20}c}
   x & y  \\
   {\frac{{dx}}
{{dt}}} & {\frac{{dy}}
{{dt}}}  \\
 \end{array} \,} \right|dt
$$
with baricenter $\mathbf{\beta } = \frac{2}{3}\mathbf{r}$
Thus
$$
\mathbf{b} = \frac{1}
{{3A}}\oint\limits_C {\mathbf{r}\left| {\,\begin{array}{*{20}c}
   x & y  \\
   {dx/dt} & {dy/dt}  \\
 \end{array} \,} \right|dt} 
$$
$ -  -  -  -  -  -  -  -  -  - $


Example:
Consider the semiparabolic sector comprised among the axes and the curve $C$
$$
C:\left\{ \begin{gathered}
  0 \leqslant t \leqslant 1 \hfill \\
  x = t \hfill \\
  y = 1 - t^{\,2}  \hfill \\ 
\end{gathered}  \right.\quad  \to \quad C:\left\{ \begin{gathered}
  0 \leqslant x \leqslant 1 \hfill \\
  y = 1 - x^{\,2}  \hfill \\ 
\end{gathered}  \right.\quad  \to \quad G:\left\{ \begin{gathered}
  0 \leqslant x \leqslant 1 \hfill \\
  0 \leqslant y \leqslant 1 - x^{\,2}  \hfill \\ 
\end{gathered}  \right.
$$
Its area is 
$
A = \iint\limits_G {dx\,dy} = \int_{x = 0}^{\;1} {\left( {\int_{y = 0}^{\;1 - x^{\,2} } {dy} } \right)dx}  = \frac{2}{3}
$  
and the baricenter will be
$$
\mathbf{b} = \frac{1}
{A}\iint\limits_G {\left( \begin{gathered}
  x \\ 
  y \\ 
\end{gathered}  \right)dx\,dy} = \frac{3}
{2}\int_{x = 0}^{\;1} {\left( \begin{gathered}
  x\int_{y = 0}^{\;1 - x^{\,2} } {dy}  \\ 
  \int_{y = 0}^{\;1 - x^{\,2} } {y\,dy}  \\ 
\end{gathered}  \right)dx}  = \frac{3}
{2}\int_{x = 0}^{\;1} {\left( \begin{gathered}
  x - x^{\,3}  \\ 
  \frac{{\left( {1 - x^{\,2} } \right)^{\,2} }}
{2} \\ 
\end{gathered}  \right)dx}  = \left( \begin{gathered}
  \frac{3}
{8} \hfill \\
  \frac{2}
{5} \hfill \\ 
\end{gathered}  \right)
$$
or, otherwise
$$
\mathbf{b} = \frac{1}
{A}\int\limits_G {\mathbf{\beta }\,dA}  = \frac{3}
{2}\int_{x\, = \,0}^{\;1} {\left( \begin{gathered}
  x \\ 
  \left( {1 - x^{\,2} } \right)/2 \\ 
\end{gathered}  \right)\,\left( {1 - x^{\,2} } \right)dx}  = \left( \begin{gathered}
  \frac{3}
{8} \\ 
  \frac{2}
{5} \\ 
\end{gathered}  \right)
$$
and with the curve integral
$$
\begin{gathered}
  \mathbf{b} = \frac{1}
{{3A}}\oint\limits_C {\mathbf{r}\left| {\,\begin{array}{*{20}c}
   x & y  \\
   {dx/dt} & {dy/dt}  \\
 \end{array} \,} \right|dt}  = \frac{1}
{2}\oint\limits_C {\left( \begin{gathered}
  t \\ 
  1 - t^{\,2}  \\ 
\end{gathered}  \right)\left| {\,\begin{array}{*{20}c}
   t & {1 - t^{\,2} }  \\
   1 & { - 2t}  \\
 \end{array} \,} \right|dt}  =  \hfill \\
   = \frac{1}
{2}\oint\limits_C {\left( \begin{gathered}
  t \\ 
  1 - t^{\,2}  \\ 
\end{gathered}  \right)\left( { - t^{\,2}  - 1} \right)dt}  = \frac{1}
{2}\left( {\int\limits_{y\, = \,0} {0\,dt}  + \int_{t\, = \,1}^{0\,} {\left( \begin{gathered}
  t\left( { - t^{\,2}  - 1} \right) \\ 
  \left( {1 - t^{\,2} } \right)\left( { - t^{\,2}  - 1} \right) \\ 
\end{gathered}  \right)dt}  + \int\limits_{x\, = \,0} {0\,dt} } \right) =  \hfill \\
   = \frac{1}
{2}\int_{t\, = \,0}^{\,1\,} {\left( \begin{gathered}
  t\left( {1 + t^{\,2} } \right) \\ 
  \left( {1 - t^{\,4} } \right) \\ 
\end{gathered}  \right)dt}  = \frac{1}
{2}\left( \begin{gathered}
  \frac{1}
{2} + \frac{1}
{4} \\ 
  \left( {1 - \frac{1}
{5}} \right) \\ 
\end{gathered}  \right) = \left( \begin{gathered}
  \frac{3}
{8} \\ 
  \frac{2}
{5} \\ 
\end{gathered}  \right) \hfill \\ 
\end{gathered} 
$$
