Answer to Question #222530 in Calculus for Rohan

Question

Find mass and centre of mass of triangle lamina with vertices(0,0),(2,0)and(0,3).if density is 1+X+5y

Answer 1

Equation of the hypotenuse of a right triangle have form

$$y(x)=3-\dfrac{3}{2}x$$

So mass of the triangular lamina

$$M=\int\int\limits_{D}d(x,y)dxdy=\int\limits_{0}^{2}dx\int\limits_{0}^{3-\dfrac{3}{2}x}d(x,y)dy=$$ $$=\int\limits_{0}^{2}dx\int\limits_{0}^{3-3/2x}(1+ x + 5y)dy=\int\limits_{0}^{2}dx\dfrac{1}{10} (1+ x + 5y)^2\big|_0^{3-\dfrac{3}{2}x}=\int\limits_{0}^{2}dx\dfrac{1}{10} \big[(16 - \dfrac{13}{2}x)^2-(1+x)^2\big]=$$ $$E=mc^2$$ $$=\dfrac{71}{5}$$

centre of gravity

$$M_y=\frac1M\int\int\limits_{D}yd(x,y)dxdy=\frac1M\int\limits_0^{2}\frac16 y^2 (3 + 9 x + 2 y)\big|_{0}^{3-\dfrac{3}{2}x}=$$ $$=\dfrac{5}{71}\int\limits_0^{2}\frac23 (-1 + x)^2 (7 + 5 x)dx=1$$ $$M_x=\frac1M\int\int\limits_{D}xd(x,y)dxdy=\dfrac{5}{71}\int\limits_{0}^{2}dx(255 - 209x + \dfrac{165}{4})x=8.28$$

Answer: centre of gravity $(M_x,M_y)=(8.28, \ 1)$

Mass: $M= \dfrac{71}{5}$