Question

Calcule o centro de massa de uma lâmina triangular de vértices (0,0), (0,1) e (1,0), cuja função densidade $\rho$ é: $\rho(x,y)=xy$

Answer 1

Primeiro calcular a massa total M:
$\displaystyle M=\iint\limits_C\rho dA$
lembrando que:
$A=x y\implies dA=dx dy$
e
$\rho(x,y)=xy$
Como temos os pontos (0,1) e (1,0) pela equação da reta encontramos:
$y=1-x$
a massa é dada por:
$\displaystyle M=\int_{0}^{1}\limits\int\limits^{1-x}_{0}xy\,dA=\int_{0}^{1}\limits\int\limits^{1-x}_{0}xy\,dydx\\\\i)~~~\int_{0}^{1}\limits\int\limits^{1-x}_{0}xy\,dydx=\int\limits_{0}^{1}\left.\frac{1}{2}y^2\right]\limits_{0}^{1-x}dx\\\\ii)~\int\limits_{0}^{1}\left.\frac{1}{2}xy^2\right]\limits_{0}^{1-x}dx=\int\limits_{0}^{1}\frac{1}{2}x(1-x)^2-\frac{1}{2}\cdot0\,dx\\\\iii)\int\limits_{0}^{1}\frac{1}{2}(x-2x^2+x^3)\,dx=\left[\frac{1}{4}x^2-\frac{1}{3}x^3+\frac{1}{4}x^4\right]\limits_{0}^{1}=\right[\frac{3x^4+3x^2-4x^3}{48}\left]\limits_{0}^{1}$
$\displaystyle \\\\\ iv)~\int\limits_{0}^{1}\int\limits_{0}^{1-x}xy\,dydx =\left[\frac{3x^4+3x^2-4x^3}{48}\right]\limits_{0}^{1}=\frac{3+3-4}{48}-\frac{0}{48}\\\\v)~~\int\limits_{0}^{1}\int\limits_{0}^{1-x}xy\,dydx =\frac{2}{48}=\boxed{\frac{1}{24}~~Unidades~de~medida}$

agora para encontrar o centro de massa $(\bar{x},\bar{y})$
usamos a relação:
$\displaystyle \bar{x}=\frac{M_y}{M}\\\\\bar{y}=\frac{M_x}{M}$
onde $M_y,~M_x$ é o primeiro momento do objeto em torno do eixo y e x, respectivamente, e são dados por:
$\displaystyle M_y=\iint\limits_Cx\rho(x,y)\,dA\\\\M_x=\iint\limits_Cy\rho(x,y)\,dA$

então:
$\displaystyle \bar{x}=\frac{\iint\limits_Cx\rho(x,y)\,dA}{\iint\limits_C\rho(x,y) \dA}\\\\\bar{y}=\frac{\iint\limits_Cy\rho(x,y)\,dA}{\iint\limits_C\rho(x,y)\ dA}$

calculando primeiro momento da lâmina:

$\displaystyle \int\limits_{0}^{1}\int\limits_{0}^{1-x}x(xy)dydx=\int\limits_{0}^{1}\int\limits_{0}^{1-x}x^2y\,dydx\\\\i)~~\int\limits_{0}^{1}\int\limits_{0}^{1-x}x^2y\,dydx=\int\limits_{0}^{1}\left.\frac{1}{2}x^2y^2\right]\limits_{0}^{1-x}dx\\\\ii)~~\int\limits_{0}^{1}\frac{1}{2}x^2\left(1-2x+x^2\right)-0\,dx=\int\limits_{0}^{1}\frac{1}{2}x^2-x^3+\frac{1}{2}x^4\,dx\\\\iii)~~\frac{1}{2}\int\limits_{0}^{1}x^2-2x^3+x^4\,dx=\frac{1}{2}\left[\frac{1}{3}x^3-\frac{2}{4}x^4+\frac{1}{5}x^5\right]\limits_{0}^{1}$
$\displaystyle \\\\iv)\int\limits_{0}^{1}\int\limits_{0}^{1-x}x^2y\,dydx=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2}+\frac{1}{5} \right)-0=\frac{1}{2}\cdot\frac{10-15+6}{3\cdot2\cdot5}=\boxed{\frac{1}{60}}$

e

$\displaystyle M_y=\int\limits_{0}^{1}\int\limits_{0}^{1-x}y(xy)dydx=\int\limits_{0}^{1}\int\limits_{0}^{1-x}xy^2\,dydx\\\\i)~~~\int\limits_{0}^{1}\int\limits_{0}^{1-x}xy^2\,dydx=\int_{0}^{1}\left[\frac{1}{3}xy^3\right]\limits_{0}^{1-x}dx\\\\ii)~~\int_{0}^{1}\left(\frac{1}{3}x(x-1)^3-0\right)dx\\\\iii)~~u=x-1\implies du=dx$
$\displaystyle\\\\iv)~~\int_{0}^{1}\left(\frac{1}{3}x(x-1)^3-0\right)dx=\frac{1}{3}\int\limits_{u(0)}^{u(1)}(u+1)u^3\,du\\\\v)~~\frac{1}{3}\int\limits_{-1}^{0}u^4+u^3\,du=\frac{1}{3}\left( \frac{1}{5}u^5+\frac{1}{4}u^4 \right)\limits_{-1}^{0}\\\\vii)\int\limits_{0}^{1}\int\limits_{0}^{1-x}xy^2\,dydx =0-\frac{1}{3}\cdot\left(-\frac{1}{5}-\frac{1}{4}\right)=\boxed{\frac{1}{60}}$

obtemos:
$\displaystyle M=\frac{1}{24}\\\\ M_x=M_y=\frac{1}{60}$
então:
$\displaystyle \bar{x}=\frac{\iint\limits_Cx\rho(x,y)\,dA}{\iint\limits_C\rho(x,y) \dA}=\frac{\frac{1}{60}}{\frac{1}{24}}=\frac{1}{60}\cdot\frac{24}{1}=\frac{24}{60}=\frac{2}{5}\\\\ \bar{y}=\frac{\iint\limits_Cy\rho(x,y)\,dA}{\iint\limits_C\rho(x,y) \dA}=\frac{2}{5}$
O centro de massa dessa lâmina é:
$\boxed{CM=\left(\frac{2}{5},\frac{2}{5}\right)}$
Imagem abaixo: