Question

Find the mass and center of mass of the thin plate that occupies the triangle with vertices (0,0), (2,1), and (0,3) and has density function p(x,y) = x + y.

Answer 1

$$m=\int\int_Rdm=\int\int_R\rho dA=\displaystyle\int_{0}^{2}\displaystyle\int_{x/2}^{-x+3}(x+y)dydx$$

$$=\displaystyle\int_{0}^{2}\bigg[xy+\dfrac{y^2}{2}\bigg]\begin{matrix} -x+3 \\ x/2 \end{matrix}dx$$

$$=\displaystyle\int_{0}^{2}(-x^2+3x+x^2/2-3x+9/2-x^2/2-x^2/8)dx$$

$$=\displaystyle\int_{0}^{2}(-9x^2/8+9/2)dx=[-3x^3/8+9x/2]\begin{matrix} 2 \\ 0 \end{matrix}$$

$$=-3+9=6$$

$$M_x=\int\int_Rx\rho dA=\displaystyle\int_{0}^{2}\displaystyle\int_{x/2}^{-x+3}x(x+y)dydx$$

$$=\displaystyle\int_{0}^{2}\bigg[x^2y+\dfrac{xy^2}{2}\bigg]\begin{matrix} -x+3 \\ x/2 \end{matrix}dx$$

$$=\displaystyle\int_{0}^{2}(-x^3+3x^2+x^3/2-3x^2+9x/2-x^3/2-x^3/8)dx$$

$$=\displaystyle\int_{0}^{2}(-9x^3/8+9x/2)dx=[-9x^4/32+9x^2/4]\begin{matrix} 2 \\ 0 \end{matrix}$$

$$=-9/2+9=9/2$$

$$M_y=\int\int_Ry\rho dA=\displaystyle\int_{0}^{2}\displaystyle\int_{x/2}^{-x+3}y(x+y)dydx$$

$$=\displaystyle\int_{0}^{2}\bigg[xy^2/2+\dfrac{y^3}{3}\bigg]\begin{matrix} -x+3 \\ x/2 \end{matrix}dx$$

$$=\displaystyle\int_{0}^{2}(x^3/2-3x^2+9x/2-x^3/3+3x^2-9x+9)dx$$

$$-\displaystyle\int_{0}^{2}(x^3/8+x^3/24)dx$$

$$=\displaystyle\int_{0}^{2}(-9x/2+9)dx$$

$$=[-9x^2/4+9x]\begin{matrix} 2 \\ 0 \end{matrix}$$

$$=-9+18=9$$

$$\bar{x}=\dfrac{M_y}{m}=\dfrac{9}{6}=\dfrac{3}{2}$$

$$\bar{y}=\dfrac{M_x}{m}=\dfrac{9/2}{6}=\dfrac{3}{4}$$

$$m=6, C\bigg(\dfrac{3}{2}, \dfrac{3}{4}\bigg)$$