Answer to Question #243135 in Calculus for brittany

Question

Find the center of mass of a triangular sheet whose vertices are (8, 0), (2, 6) and (2, 0), whose density function is

p (x, y) = x + y + 1

Answer 1

Vertices:

$(0,8),(2.6),(2,0)$

Graphs:

$(0,8),(2.6)$ :

$8=b$

$6=2a+b\implies a=-1$

$y=8-x$

$(2.6),(2,0)$

$x=2$

$(0,8),(2,0)$

$b=8$

$0=2a+b\implies a=-4$

$y=8-4x$

mass:

$m=\iint \rho(x,y)dA=\int^{x=2}_{x=0}\int^{y=8-x}_{y=8-4x}(x+y+1)dydx=$

$=\int^{x=2}_{x=0}(xy+y^2/2+y)|^{y=8-x}_{y=8-4x}dx=$

$=\int^{x=2}_{x=0}(x(8-x)+(8-x)^2/2+8-x-x(8-4x)-(8-4x)^2/2-8+4x)dx=$

$=\int^{x=2}_{x=0}(-4.5x^2+27x)dx=(27x^2/2-1.5x^3)|^2_0=42$

$M_x=\iint y\rho(x.y)dA=\int^{x=2}_{x=0}\int^{y=8-x}_{y=8-4x}y(x+y+1)dydx=$

$=\int^{x=2}_{x=0}(y^2x/2+y^3/3+y^2/2)|^{y=8-x}_{y=8-4x}dx=$

$=\int^{x=2}_{x=0}(x(8-x)^2/2+(8-x)^3/3+(8-x)^2/2-x(8-4x)^2/2-(8-4x)^3/3-$

$-(8-4x)^2/2)dx=\int^{x=2}_{x=0}(32x-8x^2+x^3/2+512/3-64x+8x^2-x^3/3+$

$+32-8x+x^2/2-32x+32x^2-8x^3-512/3+256x-128x^2+64x^3/3-$

$-32+32x-8x^2)dx=\int^{x=2}_{x=0}(216x-103.5x^2+13.5x^3)dx=$

$=(108x^2-34.5x^3+3.375x^4)|^2_0=210$

$M_y=\iint x\rho(x.y)dA=\int^{x=2}_{x=0}\int^{y=8-x}_{y=8-4x}x(x+y+1)dydx=$

$=\int^{x=2}_{x=0}(y^2x/2+x^2y+xy)|^{y=8-x}_{y=8-4x}dx=$

$=\int^{x=2}_{x=0}((8-x)^2x/2+x^2(8-x)+x(8-x)-(8-4x)^2x/2-$

$-x^2(8-4x)-x(8-4x))dx=\int^{x=2}_{x=0}(32x-8x^2+x^3/2+8x^2-x^3+$

$+8x-x^2-32x+32x^2-8x^3-8x^2+32x^3-8x+4x^2)dx=$

$=(9x^3+5.75x^4)|^2_0=164$

The center of mass:

$\overline{x}=\frac{M_y}{m}=\frac{164}{42}=\frac{82}{21}=3.90$

$\overline{y}=\frac{M_x}{m}=\frac{210}{42}=5.00$