Find the centroid of a lamina Region is bounded by : $y=x+4$, $y=5$, $y=-x-4$, $x=3$
Graph is here: https://www.desmos.com/calculator/cwmlwknywe
Can I just split the shape at the x-axis, find the individual centroids, and then add them or average them? I can't seem to find an example like this in my book.  
 A: This problem can be simplified by reminding that the centroid of any isosceles triangle divides the height in two parts with ratio $2:1$ (with the longest part towards the vertex). In this problem we have two right isosceles triangles. The height of the large one is included between $(-4,0)$ and $(3,0)$, so the centroid is in $(-\frac{1}{2},0)$. The height of the small one is included between $(3,5)$ and $(2,6)$, so the centroid is in $(\frac{7}{3},\frac{17}{3})$.
Now simply apply the standard method of averaging the two centroid, taking into account that the two areas are $49$ and $2$ and that the area of the small triangle has to be considered negative.
A: If $n$ masses $m_1,\;m_2,\ldots,\,m_n$ are placed in points $(x_1,y_1),\,(x_2,y_2),\ldots,(x_n,y_n)$, respectively, then the centroid of the systems is $(\overline{x},\overline{y})$ where
$$\overline{x}=\dfrac{\sum_{i=1}^{n}{m_ix_i}}{\sum_{i=1}^{n}{m_i}}\qquad\text{and}\qquad \overline{y}=\dfrac{\sum_{i=1}^{n}{m_iy_i}}{\sum_{i=1}^{n}{m_i}}$$
In the present case, if $m_1$ is the area(mass) of the big triangle and $m_2$ is the mass(area) of the little one, we have
$$\overline{x}=\dfrac{m_1x_1-m_2x_2}{m_1-m_2}\qquad\text{and}\qquad \overline{y}=\dfrac{m_1y_1-m_2y_2}{m_1-m_2}$$
where $(x_1,y_1)$ and $(x_2,y_2)$ are the centroid of the triangles big and little one, respectively.
