Answer to Question #176668 in Calculus for Joshua

Question

Find the center of mass for the region bounded by y=4-x^2 that is in the first quadrant.

Answer 1

the region whose center of mass is desired is the region bounded by the curves $y=4-x^2$, the positive x-axis and positive y-axis

since we are not given the mass density relation $\rho(x,y)$ , we may take it uniform over the region.

thus we have

mass, $m= \int ^{2}_{0}\rho[4-x^2]dx= \rho [4x-\frac {x^2}{2}]^{2}_{0}$

$=6\rho$ units of mass

moments about x=axis,$m_x= \frac 12 \rho \int ^{2}_{0}(4-x^2)^2dx=\frac12 \rho\int^{2}_{0}(16-8x^2+x^4)dx$

$=\frac 12\rho[16x-\frac83x^3+\frac 15x^5]^{2}_{0}=\frac12 \rho(\frac{256}{15})$

$=\frac{128}{15}\rho$ units of the mass times units of length

and

moments about y-axis

$m_y= \rho\int^{2}_{0}x(4-x^2)dx= \rho\int^{2}_{0}(4x-x^3)dx$

$=\rho[2x^2-\frac {x^4}{ 4}]^{2}_{0}= 4\rho$ units of mass times units of length

the center of mass $(\bar x, \bar y)$ is such that $\bar x= \frac {m_y}{m} = \frac {4\rho}{6\rho}= \frac 23, \bar y = \frac {m_x}{m}= \frac {\frac {128\rho}{15}}{6\rho}=\frac {64}{45}$

$\implies (\frac 23, \frac {64}{45})$ is the center of mass $y=4-x^2$ in the first quadrant