Basically, a planar lamina is defined as a figure (a closed set) D of a finite area in a plane, with some mass m.^[2]

This is useful in calculating moments of inertia or center of mass for a constant density, because the mass of a lamina is proportional to its area. In a case of a variable density, given by some (non-negative) surface density function ${\displaystyle \rho (x,y),}$  the mass ${\displaystyle m}$  of the planar lamina D is a planar integral of ρ over the figure:^[3]

${\displaystyle m=\iint _{D}\rho (x,y)\,dx\,dy}$ 

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:

${\displaystyle M_{y}=\lim _{m,n\to \infty }\,\sum _{i=1}^{m}\,\sum _{j=1}^{n}\,x{_{ij}}^{*}\,\rho \ (x{_{ij}}^{*},y{_{ij}}^{*})\,\Delta D=\iint _{D}x\,\rho \ (x,y)\,dx\,dy}$ 

${\displaystyle M_{x}=\lim _{m,n\to \infty }\,\sum _{i=1}^{m}\,\sum _{j=1}^{n}\,y{_{ij}}^{*}\,\rho \ (x{_{ij}}^{*},y{_{ij}}^{*})\,\Delta D=\iint _{D}y\,\rho \ (x,y)\,dx\,dy}$ 

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

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 ${\displaystyle m=\iint _{D}\rho (x,y)\,dx\,dy}$  which can be equivalently expressed as an iterated integral:

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

The inner integral is:

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

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

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

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

Plugging this into the outer integral results in:

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

Similarly are calculated both moments:

${\displaystyle M_{y}=\iint _{D}x\,\rho (x,y)\,dx\,dy=\int _{x=0}^{2}\int _{y=x}^{4-x}x\,(2x+3y+2)\,dy\,dx}$ 

with the inner integral:

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

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

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

which makes:

${\displaystyle {\begin{aligned}M_{y}&=\int _{x=0}^{2}(-4x^{3}-8x^{2}+32x)\,dx\\&=\left.\left(-x^{4}-{\frac {8x^{3}}{3}}+16x^{2}\right)\right|_{x=0}^{2}\\&={\frac {80}{3}}\end{aligned}}}$ 

and

${\displaystyle {\begin{aligned}M_{x}&=\iint _{D}y\,\rho (x,y)\,dx\,dy=\int _{x=0}^{2}\int _{y=x}^{4-x}y\,(2x+3y+2)\,dy\,dx\\&=\int _{0}^{2}(xy^{2}+y^{3}+y^{2}){\Big |}_{y=x}^{4-x}\,dx\\&=\int _{0}^{2}(-2x^{3}+4x^{2}-40x+80)\,dx\\&=\left.\left(-{\frac {x^{4}}{2}}+{\frac {4x^{3}}{3}}-20x^{2}+80x\right)\right|_{x=0}^{2}\\&={\frac {248}{3}}\end{aligned}}}$ 

Finally, the center of mass is

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