Planar lamina

In mathematics, a planar lamina (or plane lamina^[1] ) is a figure representing a thin, usually uniform, flat layer of the solid. It serves also as an idealized model of a planar cross section of a solid body in integration.

Planar laminas can be used to determine moments of inertia, or center of mass of flat figures, as well as an aid in corresponding calculations for 3D bodies.

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 ${\displaystyle \rho (x,y),}${\displaystyle \rho (x,y),} the mass ${\displaystyle m}${\displaystyle m} of the planar lamina D is a planar integral of ρ over the figure:^[3]

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

The center of mass of the lamina is at the point

${\displaystyle \left({\frac {M_{y}}{m}},{\frac {M_{x}}{m}}\right)}${\displaystyle \left({\frac {M_{y}}{m}},{\frac {M_{x}}{m}}\right)}

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

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

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

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

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

For this the mass ${\displaystyle m}${\displaystyle m} must be found as well as the moments ${\displaystyle M_{y}}${\displaystyle M_{y}} and ${\displaystyle M_{x}}${\displaystyle M_{x}}.

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

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

The inner integral is:

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

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

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

Plugging this into the outer integral results in:

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

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

with the inner integral:

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

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

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

which makes:

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

and

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

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

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