Answer to Question #129355 in Calculus for didi

Question

Find the mass and center of mass of a triangular lamina with vertices
(0, 0), (1, 0)
and
(0, 2)
if the density function is p(x,y)=1+x+y

Answer 1

Equation of the hypotenuse of the right angle triangle is $y = -2x + 2$

Mass of the plate is given by

$M = \int_0^1 \int_0^{(2-2x)} (1+x+y)dydx$

$\int_0^1 (y+xy+\frac{y^2}{2})|_0^{2-2x} dx = \int_0^1 4-4x dx = 2$

Center of mass,

X-coordinate,

$M_x = \frac{1}{M} \int_0^1 \int_0^{2-2x} x(1+y+x)dydx = \frac{1}{M} \int_0^1 [xy+x^2y+\frac{xy^2}{2} ]_0^{2-2x} dx$

$= \frac{1}{M} \int_0^1 (4x-4x^2) dx = \frac{1}{3}$

Y-coordinate,

$M_y=\frac{1}{M} \int_0^1 \int_0^{2-2x} y(1+y+x)dydx$

$= \frac{1}{M} \int_0^1[ \frac{(2-2x)^2}{2}+\frac{x(2-2x)^2}{2} + \frac{(2-2x)^3}{3} ]dx$

$= \frac{3}{4}$