Planar lamina

Properties

The center of mass of the lamina is at the point

${\displaystyle \left(\frac{M_y}{m},\frac{M_x}{m}\right)}$

where ${\displaystyle M_y}$ is the moment of the entire lamina about the y-axis and ${\displaystyle M_x}$ is the moment of the entire lamina about the x-axis:

${\displaystyle M_y=\underset{m,n\to \mathrm{\infty }}{lim}\sum _{i=1}^m\sum _{j=1}^{n}x{{}_{ij}}^{\ast }\rho (x{{}_{ij}}^{\ast },y{{}_{ij}}^{\ast })\mathrm{\Delta }D={\iint }_{D}x\rho (x,y)dxdy}$

${\displaystyle M_x=\underset{m,n\to \mathrm{\infty }}{lim}\sum _{i=1}^m\sum _{j=1}^{n}y{{}_{ij}}^{\ast }\rho (x{{}_{ij}}^{\ast },y{{}_{ij}}^{\ast })\mathrm{\Delta }D={\iint }_{D}y\rho (x,y)dxdy}$

with summation and integration taken over a planar domain ${\displaystyle D}$.

Example

Find the center of mass of a lamina with edges given by the lines ${\displaystyle x=0,}$ ${\displaystyle y=x}$ and ${\displaystyle y=4-x}$ where the density is given as ${\displaystyle \rho (x,y)=2x+3y+2}$.

For this the mass ${\displaystyle m}$ must be found as well as the moments ${\displaystyle M_y}$ and ${\displaystyle M_x}$.

Mass is ${\displaystyle m={\iint }_D\rho (x,y)dxdy}$ which can be equivalently expressed as an iterated integral:

${\displaystyle m={\int }_{x=0}^2{\int }_{y=x}^{4-x}(2x+3y+2)dydx}$

The inner integral is:

${\displaystyle {\int }_{y=x}^{4-x}(2x+3y+2)dy}$

${\displaystyle ={\left(2xy+\frac{3y^2}{2}+2y\right)|}_{y=x}^{4-x}}$

${\displaystyle =\left[2x(4-x)+\frac{3(4-x{)}^2}{2}+2(4-x)\right]-\left[2x(x)+\frac{3(x{)}^2}{2}+2(x)\right]}$

${\displaystyle =-4x^2-8x+32}$

Plugging this into the outer integral results in:

${\displaystyle \begin{array}{cc}m& ={\int }_{x=0}^2\left(-4x^2-8x+32\right)dx\\ & ={\left(-\frac{4x^3}{3}-4x^2+32x\right)|}_{x=0}^2\\ & =\frac{112}{3}\end{array}}$

Similarly are calculated both moments:

${\displaystyle M_y={\iint }_{D}x\rho (x,y)dxdy={\int }_{x=0}^2{\int }_{y=x}^{4-x}x(2x+3y+2)dydx}$

with the inner integral:

${\displaystyle {\int }_{y=x}^{4-x}x(2x+3y+2)dy}$

${\displaystyle ={\left(2x^{2}y+\frac{3x{y}^2}{2}+2xy\right)|}_{y=x}^{4-x}}$

${\displaystyle =-4x^3-8x^2+32x}$

which makes:

${\displaystyle \begin{array}{cc}{M}_y& ={\int }_{x=0}^2(-4x^3-8x^2+32x)dx\\ & ={\left(-x^4-\frac{8x^3}{3}+16x^2\right)|}_{x=0}^2\\ & =\frac{80}{3}\end{array}}$

and

${\displaystyle \begin{array}{cc}{M}_x& ={\iint }_{D}y\rho (x,y)dxdy={\int }_{x=0}^2{\int }_{y=x}^{4-x}y(2x+3y+2)dydx\\ & ={\int }_0^2(x{y}^2+y^3+y^2){|}_{y=x}^{4-x}dx\\ & ={\int }_0^2(-2x^3+4x^2-40x+80)dx\\ & ={\left(-\frac{x^4}{2}+\frac{4x^3}{3}-20x^2+80x\right)|}_{x=0}^2\\ & =\frac{248}{3}\end{array}}$

Finally, the center of mass is

${\displaystyle \left(\frac{M_y}{m},\frac{M_x}{m}\right)=\left(\frac{\frac{80}{3}}{\frac{112}{3}},\frac{\frac{248}{3}}{\frac{112}{3}}\right)=\left(\frac{5}{7},\frac{31}{14}\right)}$