# Planar lamina

In mathematics, a planar lamina is a closed set in a plane of mass $m$ and surface density $\rho\ (x,y)$ such that:

$m = \int\int_{}{}\rho\ (x,y)\,dx\,dy$, over the closed set.

The center of mass of the lamina is at the point

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

where $M_y$ moment of the entire lamin about the x-axis and $M_x$ moment of the entire lamin about the y-axis.

$M_y = \lim_{m,n \to \infty}\,\sum_{i=1}^{m}\,\sum_{j=1}^{n}\,x{_{ij}}^{*}\,\rho\ (x{_{ij}}^{*},y{_{ij}}^{*})\,\Delta\Alpha = \iint_{}{} x\, \rho\ (x,y)\,dx\,dy$, over the closed surface.
$M_x = \lim_{m,n \to \infty}\,\sum_{i=1}^{m}\,\sum_{j=1}^{n}\,y{_{ij}}^{*}\,\rho\ (x{_{ij}}^{*},y{_{ij}}^{*})\,\Delta\Alpha = \iint_{}{} y\, \rho\ (x,y)\,dx\,dy$, over the closed surface.

Example 1.

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

$m = \int_0^2{\int_x^{4-x}}_{}{}\,2x+3y+2\,dy\,dx$
$m=\int_0^2 (2xy+\frac{3y^2}{2}+2y)|_x^{4-x}\,dx$
$m=\int_0^2 -4x^2-8x+32\,dx$
$m= (\frac{-4x^3}{3}-4x^2+32x)|_0^2$
$m= \frac{112}{3}$
$M_y = \int_0^2{\int_x^{4-x}}{}{}x\,(2x+3y+2)\,dy\,dx$
$M_y=\int_0^2 (2x^2y+\frac{3xy^2}{2}+2xy)|_x^{4-x}\,dx$
$M_y=\int_0^2 -4x^3-8x^2+32x\,dx$
$M_y= (-x^4-\frac{8x^3}{3}+16x^2)|_0^2$
$M_y= \frac{80}{3}$
$M_x = \int_0^2{\int_x^{4-x}}{}{}y\,(2x+3y+2)\,dy\,dx$
$M_x = \int_0^2 (xy^2+y^3+y^2)|_x^{4-x}\,dx$
$M_x = \int_0^2 (-2x^3+4x^2-40x+80\,dx$
$M_x= \left.\left(\frac{-x^4}{2}+\frac{4x^3}{3}-20x^2+80x\right)\right|_0^2$
$M_x= \frac{248}{3}$

center of mass is at the point

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

Planar laminas can be used to determine moments of inertia, or center of mass.