Answer to Question #236118 in Calculus for naji

Question

A triangular lamina in the xy -plane such that its vertices are (0,0), (0,1) and (1,0). Suppose that the density function of the lamina is defined by p(x,y)=30xy gram per cubic centimetre. What is the total mass of the lamina and the center of gravity. The moment about the y-axis of the lamina.the moment about the x axis of the lamina

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 $\rho(x,y)=30xy$

Find the mass of the lamina

$$m=\iint_D\rho(x,y)dA=\displaystyle\int_{0}^1\displaystyle\int_{0}^{1-x}30xydydx=$$ $$=30\displaystyle\int_{0}^1x\big[{y^2\over 2}\big]\begin{matrix} 1-x \\ 0 \end{matrix}dx=15\displaystyle\int_{0}^1(x-2x^2+x^3 )dx=$$ $$=15\big[{x^2 \over 2}-{2x^3 \over 3}+{x^4 \over 4}\big]\begin{matrix} 1 \\ 0 \end{matrix}={5 \over 4} (units\ of\ mass)$$

Mass of the lamina is $\dfrac{5}{4}$ units of mass.

The moment of the lamina about the x-axis

$$M_x=\iint_Dy\rho(x,y)dA=30\displaystyle\int_{0}^1\displaystyle\int_{0}^{1-x}xy^2dydx=$$

$$=30\displaystyle\int_{0}^1x\big[{y^3\over 3}\big]\begin{matrix} 1-x \\ 0 \end{matrix}dx$$

$$=10\displaystyle\int_{0}^1(x-3x^2+3x^3-x^4 )dx$$

$$=10\big[{x^2 \over 2}-{3x^3 \over 3}+{3x^4 \over 4}-{x^5 \over 5}\big]\begin{matrix} 1 \\ 0 \end{matrix}$$

$$=5-10+\dfrac{15}{2}-2={1 \over2}$$

The moment of the lamina about the y-axis

$$M_y=\iint_Dx\rho(x,y)dA=\displaystyle\int_{0}^1\displaystyle\int_{0}^{1-x}30x^2ydydx$$

$$=30\displaystyle\int_{0}^1x^2\big[{y^2\over 2}\big]\begin{matrix} 1-x \\ 0 \end{matrix}dx$$

$$=15\displaystyle\int_{0}^1(x^2-2x^3+x^4 )dx$$

$$=15\big[{x^3 \over 3}-{2x^4 \over 4}+{x^5 \over 5}\big]\begin{matrix} 1 \\ 0 \end{matrix}$$

$$=5-\dfrac{15}{2}+3=\dfrac{1}{2}$$

$$M_x=\dfrac{1}{2}, M_y=\dfrac{1}{2}$$

Find the coordinates of the center of mass

$$\bar{x}={M_y\over m}=\dfrac{\dfrac{1}{2}}{\dfrac{5}{4}}=\dfrac{2}{5}$$

$$\bar{y}={M_x\over m}=\dfrac{\dfrac{1}{2}}{\dfrac{5}{4}}=\dfrac{2}{5}$$

Center of gravity is$\bigg(\dfrac{2}{5},\dfrac{2}{5}\bigg).$