Answer to Question #186546 in Calculus for Jethro

Question

Consider a lamina that occupies the region D bounded by the parabola x = 1 - y² and the coordinate axes in the first quadrant with density function p (x, y) = y. Find the center of mass.

Answer 1

Given, the region D is bounded by the parabola x = 1 - y² and the coordinate axes in the first quadrant with density function p (x, y) = y.

Moments,

$M_x=\int\int yp(x,y)dA\newline=\int_{x=0}^1 \int_{y=0}^1 yp(x,y)dA\newline=\int_{x=0}^1 \int_{y=0}^1 y^2dydx\newline=\int_{x=0}^1[ \frac{y^3}{3}]_{0}^1dydx\newline=\frac{1}{3}\int_{x=0}^1dx\newline=\frac{1}{3}[ x]_{0}^1\newline=\frac{1}{3}\newline$

$M_y=\int\int xp(x,y)dA\newline=\int_{x=0}^1 \int_{y=0}^1 xydydx\newline=\int_{x=0}^1x[ \frac{y^2}{2}]_{0}^1dydx\newline=\frac{1}{2}\int_{x=0}^1xdx\newline=\frac{1}{2}[ \frac{x^2}{2}]_{0}^1\newline=\frac{1}{4}$

And,

Mass,m is given by,

$m=\int\int p(x,y)dA\newline=\int_{x=0}^1 \int_{y=0}^1 ydydx\newline=\int_{x=0}^1[ \frac{y^2}{2}]_{0}^1dydx\newline=\frac{1}{2}\int_{x=0}^1dx\newline=\frac{1}{2}[ x]_{0}^1\newline=\frac{1}{2}$

Let (x,y) be the center of mass.

$y=\frac{M_y}{m}\newline=\frac{\frac{1}{4}}{\frac{1}{2}}\newline=\frac{1}{2}\newline x=\frac{M_x}{m}\newline=\frac{\frac{1}{3}}{\frac{1}{2}}\newline=\frac{2}{3}$

Thus, center mass is $(\frac{2}{3},\frac{1}{2}).$