Answer to Question #126224 in Calculus for nur shafiqqah

Question

find the mass and center of mass of the triangular lamina with vertices (0,0), (0,1) and (1,0) and density function  𝛿(x,y)=xy

Answer 1

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

Line OA: $x=0, 0\leq y\leq1$

LineAB: $y=-x+1, 0\leq x\leq1$

Line OB: $y=0, 0\leq x\leq 1$

Given $\delta(x,y)=xy$

Find the mass of the lamina

$$m=\iint_D\delta(x,y)dA=\displaystyle\int_{0}^1\displaystyle\int_{0}^{1-x}xydydx=$$

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

$$=\big[{x^2 \over 4}-{x^3 \over 3}+{x^4 \over 8}\big]\begin{matrix} 1 \\ 0 \end{matrix}={1 \over 24} (units\ of\ mass)$$

Mass of the lamina is $\dfrac{1}{24}$ units of mas.

Find the coordinates of the center of mass

$$\bar{x}={1\over m}\iint_Dx\delta(x,y)dA=24\displaystyle\int_{0}^1\displaystyle\int_{0}^{1-x}x^2ydydx=$$

$$=24\displaystyle\int_{0}^1x^2\big[{y^2\over 2}\big]\begin{matrix} 1-x \\ 0 \end{matrix}dx=12\displaystyle\int_{0}^1(x^2-2x^3+x^4 )dx=$$

$$=12\big[{x^3 \over 3}-{2x^4 \over 4}+{x^5 \over 5}\big]\begin{matrix} 1 \\ 0 \end{matrix}=4-6+{12 \over5}={2 \over5}$$

$$\bar{y}={1\over m}\iint_Dy\delta(x,y)dA=24\displaystyle\int_{0}^1\displaystyle\int_{0}^{1-x}xy^2dydx=$$

$$=24\displaystyle\int_{0}^1x\big[{y^3\over 3}\big]\begin{matrix} 1-x \\ 0 \end{matrix}dx=8\displaystyle\int_{0}^1(x-3x^2+3x^3-x^4 )dx=$$

$$=8\big[{x^2 \over 2}-{3x^3 \over 3}+{3x^4 \over 4}-{x^5 \over 5}\big]\begin{matrix} 1 \\ 0 \end{matrix}=4-8+6-{8 \over5}={2 \over5}$$

Center of mass $\big(\dfrac{2}{5},\dfrac{2}{5}\big)$