Answer to Question #254223 in Calculus for Susan

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

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

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

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

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

$=-3+9=6$

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

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

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

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

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

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

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

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

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

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

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

$=-9+18=9$

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

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

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