Question

A triangular lamina in the xy plane such that its vertices are (0 0) (0 1) and (1 0). suppose that the density of the lamina is defined by p(x y)=45xy gram per cubic centimetre. find the total mass of the lamina and the center of the lamina

Answer 1

Mass of lamina:

$m=\iint_Sp(x,y)dxdy=\int^1_0dx\int^{1-x}_045xydy=45\int^1_0xdx\int^{1-x}_0ydy=$

$=\frac {45}{2}\int^1_0x(1-x)^2dx=\frac {45}{2}(\frac {x^2}{2}-\frac {2x^3}{3}+\frac {x^4}{4})|^1_0=$

$=\frac {45}{2}(\frac {1}{2}-\frac {2}{3}+\frac {1}{4})=\frac {45}{2}\cdot\frac{6-4\cdot2+3}{12}=\frac {45}{24}$ gram

The center of lamina:

$x_0=\frac {\iint_Sxp(x,y)dxdy}{m}=24\intop^1_0x^2dx\intop^{1-x}_0ydy=12\intop^1_0x^2(1-x)^2dx=$

$=12(\frac {x^3}{3}-\frac {x^4}{2}+\frac {x^5}{5})|^1_0=12(\frac {1}{3}-\frac {1}{2}+\frac {1}{5})=$

$=12(\frac {10-15+6}{30})=\frac {2}{5}$

$y_0=\frac {\iint_Syp(x,y)dxdy}{m}=24\intop^1_0xdx\intop^{1-x}_0y^2dy=8\intop^1_0x(1-x)^3dx=$

$=8(\frac {x^2}{2}-x^3+\frac {3x^4}{4}-\frac{x^5}{5})|^1_0=8(\frac {1}{2}-1+\frac {3}{4}-\frac {1}{5})=$

$=8(\frac {10-20+15-4}{20})=\frac {2}{5}$