Answer to Question #303837 in Calculus for MEIMEI

Question

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

Answer 1

Let $A=(0,2), B=(1, 0), O=(0,0).$

Line OA: $x=0, 0\leq y\leq2$

LineAB: $y=-2x+2, 0\leq x\leq1$

Line OB: $y=0, 0\leq x\leq 1$

Given $ρ(x,y)=1+3x+3y$

Find the mass of the lamina

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

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

$$=\displaystyle\int_{0}^1\big(2-2x+6x-6x^2+6-12x+6x^2\big)dx$$

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

Mass of the lamina is $4$ units of mas.

Find the coordinates of the center of mass

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

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

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

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

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

$$={1\over 4}\big[-{8\over 3}x^3+4x^2\big]\begin{matrix} 1 \\ 0 \end{matrix}={1\over 3}$$

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

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

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

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

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

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

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