Answer to Question #85732 in Calculus for o

Question

Find the mass and centre of gravity of a triangular lamina with vertices (0,0), (1,0)and (0,2) if the density function is given by d(x, y) =1+ 3x + y.

Answer 1

Solution

Equation of the hypotenuse of a right triangle have form

$$y(x)=2-2x$$

So mass of the triangular lamina

$$M=\int\int\limits_{D}d(x,y)dxdy=\int\limits_{0}^{1}dx\int\limits_{0}^{2-2x}d(x,y)dy=$$ $$=\int\limits_{0}^{1}dx\int\limits_{0}^{2-2x}(1+ 3x + y)dy=\int\limits_{0}^{1}dx\frac12 (1+ 3x + y)^2\big|_0^{2-2x}=\int\limits_{0}^{1}dx\frac12 \big[(3-x)^2-(1+3x)^2\big]=$$ $$E=mc^2$$ $$=\int\limits_{0}^{1}dx(4-4x^2)=\frac83$$

centre of gravity

$$M_y=\frac1M\int\int\limits_{D}yd(x,y)dxdy=\frac1M\int\limits_0^{1}\frac16 y^2 (3 + 9 x + 2 y)\big|_{0}^{2-2x}=$$

$$=\frac38\int\limits_0^{1}\frac23 (-1 + x)^2 (7 + 5 x)dx=\frac38\frac{11}{6}=\frac{11}{16}$$

$$M_x=\frac1M\int\int\limits_{D}xd(x,y)dxdy=\frac38\int\limits_{0}^{1}dx(4-4x^2)x=\frac38$$

Answer: centre of gravity $(M_x,M_y)=(\frac38,\frac{11}{16})$

Mass: $M=\frac83$