Suppose that we have a convex polygon with $n$ edges. In this case I have drawn a convex polygon with $n=5$.
We know that for a homogeneous (or non homogeneous) solid, there are formulas to calculate the coordinates of the centre of gravity using triple integrals (or double integrals) for my image.
Since I have decomposed my polygon into three triangles each of these triangles will have its centre of gravity $G_i, \text { for } i=1,2,3$.
My question is the following:
Is it possible to create a series or a sequence such that the centre of gravity $G_i$ of each triangle that make up the polygon, converges to the centre of gravity $G$ (where it is possible to find exactly its coordinates) when we use the formulas with the integrals?
Yes.
Label your triangles $T_1, T_2, ...$
Let the centres of masses of the triangles be $(x_1,y_1), (x_2,y_2), ...$
Let the masses of the triangles be $m_1, m_2, ...$
Let the centre of mass of the whole shape after $n$ triangles have been added be $(\bar x_n, \bar y_n)$
Let the total mass of the whole shape after $n$ triangles have been added be $M_n=\Sigma_1^n m_i$
Adding another triangle with mass $m_{n+1}$ and centre of mass $(x_{n+1},y_{n+1})$ means
$M_{n+1}=M_n+m_{n+1}$
$\bar x_{n+1} M_{n+1} = \bar x_{n} M_{n} + x_{n+1} m_{n+1}$
So $\bar x_{n+1} = \frac{\bar x_{n} M_{n} + x_{n+1} m_{n+1}}{M_n+m_{n+1}}$
Similarly $\bar y_{n+1} = \frac{\bar y_{n} M_{n} + y_{n+1} m_{n+1}}{M_n+m_{n+1}}$
Example
For the shape shown above, we have:
$M_1=m_1=126$
$\bar x_1=x_1=126$
$\bar y_1=y_1=28$
$M_2=M_1+m_2=126+225=351$
$\bar x_2=\frac{\bar x_1\times M_1+x_2\times m_2}{M_1+ m_2}=15.41$
$\bar y_2=\frac{\bar y_1\times M_1+y_2\times m_2}{M_1+ m_2}=25.44$
$M_3=M_2+m_3=351+108=459$
$\bar x_3=\frac{\bar x_2\times M_2+x_3\times m_3}{M_2+ m_3}=17.67$
$\bar y_3=\frac{\bar y_2\times M_2+y_3\times m_3}{M_2+ m_3}=23.45$
$M_4=M_3+m_4=459+144=603$
$\bar x_4=\frac{\bar x_3\times M_3+x_4\times m_4}{M_3+ m_4}=18.46$
$\bar y_4=\frac{\bar y_3\times M_3+y_4\times m_4}{M_3+ m_4}=20.24$