Question #303837

Find the center of mass of a triangular lamina with vertices (0,0),(1,0) and (0,2)

if the density function is ρ(x,y)=1+3x+3y



1
Expert's answer
2022-03-01T04:12:27-0500

Let A=(0,2),B=(1,0),O=(0,0).A=(0,2), B=(1, 0), O=(0,0).

Line OA: x=0,0y2x=0, 0\leq y\leq2

LineAB: y=2x+2,0x1y=-2x+2, 0\leq x\leq1

Line OB: y=0,0x1y=0, 0\leq x\leq 1


Given ρ(x,y)=1+3x+3yρ(x,y)=1+3x+3y

Find the mass of the lamina


m=Dρ(x,y)dA=01022x(1+3x+3y)dydxm=\iint_D\rho(x,y)dA=\displaystyle\int_{0}^1\displaystyle\int_{0}^{2-2x}(1+3x+3y)dydx




=01[y+3xy+3y22]22x0dx=\displaystyle\int_{0}^1\big[y+3xy+{3y^2\over 2}\big]\begin{matrix} 2-2x \\ 0 \end{matrix}dx




=01(22x+6x6x2+612x+6x2)dx=\displaystyle\int_{0}^1\big(2-2x+6x-6x^2+6-12x+6x^2\big)dx


=[4x2+8x]10=4(units of mass)=\big[-4x^2+8x\big]\begin{matrix} 1 \\ 0 \end{matrix}=4(units\ of\ mass)

Mass of the lamina is 44 units of mas.


Find the coordinates of the center of mass


xˉ=1mDxδ(x,y)dA\bar{x}={1\over m}\iint_Dx\delta(x,y)dA




=140101xx(1+3x+3y)dydx={1\over 4}\displaystyle\int_{0}^1\displaystyle\int_{0}^{1-x}x(1+3x+3y)dydx

=1401x[y+3xy+3y22]22x0dx={1\over 4}\displaystyle\int_{0}^1x\big[y+3xy+{3y^2\over 2}\big]\begin{matrix} 2-2x \\ 0 \end{matrix}dx




=1401(2x2x2+6x26x3)dx={1\over 4}\displaystyle\int_{0}^1\big(2x-2x^2+6x^2-6x^3\big)dx

+1401(6x12x2+6x3)dx+{1\over 4}\displaystyle\int_{0}^1\big(6x-12x^2+6x^3\big)dx


=14[83x3+4x2]10=13={1\over 4}\big[-{8\over 3}x^3+4x^2\big]\begin{matrix} 1 \\ 0 \end{matrix}={1\over 3}

yˉ=1mDyρ(x,y)dA\bar{y}={1\over m}\iint_Dy\rho(x,y)dA




=140101xy(1+3x+3y)dydx={1\over 4}\displaystyle\int_{0}^1\displaystyle\int_{0}^{1-x}y(1+3x+3y)dydx




=1401[y22+3xy22+y3]22x0dx={1\over 4}\displaystyle\int_{0}^1\big[{y^2\over 2}+{3xy^2\over 2}+y^3\big]\begin{matrix} 2-2x \\ 0 \end{matrix}dx




=1401(24x+2x2+6x12x2+6x3)dx={1\over 4}\displaystyle\int_{0}^1\big(2-4x+2x^2+6x-12x^2+6x^3\big)dx

+1401(824x+24x28x3)dx+{1\over 4}\displaystyle\int_{0}^1\big(8-24x+24x^2-8x^3\big)dx

=14[12x4+143x311x2+10x]10=1924={1\over 4}\big[-{1\over 2}x^4+{14\over 3}x^3-11x^2+10x\big]\begin{matrix} 1 \\ 0 \end{matrix}={19\over 24}

Center of mass (13,1924)\big(\dfrac{1}{3},\dfrac{19}{24}\big)

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