Answer to Question #176688 in Calculus for Joshua

Question

Find the center of mass of the region bounded by y=x^2 and y=6-x.

Answer 1

the region bounded by the curves $y=x^2$ and $y=6-x$ has the points of intervals where;

$x^2 =6-x$

or

$x^2 + x-6 =0 ...(i)$

the quadratic equation (i) solves to give;

x=-3 and x=2 ...(ii)

for a point $x\in(-3,2)$ say x=0

$y=x^2 \implies y=0 for x=0$

and

$y=6-x \implies y=6 for x=0$

thus the line y=6-x is above the curve $y=x^2 \forall x\in(-3,2)$

we now assume uniform density $\rho,$ of the region.

equation (ii) serves as the limits of our finite integrals for area and moment about the axis

we have

$m=\rho A$

$=\rho \intop ^2 _{-3}[6-x-x^2 ]dx$

$= \rho [6x- \frac {x^2}2 - \frac {x^3}3]^2 _{-3}$

$=\rho [\frac {22}3 + \frac {27}2]= \frac {125}6 \rho$ units of mass.

moment of the region abotu y=axis is

$m_y=\rho \int ^2 _{-3} (x) [ 6-x-x^2 ]dx$

$=\rho \int ^2 _{-3} [6x-x^2 - x^3]dx$

$=\rho [3x^2 -\frac {x^3} 3 - \frac {x^4}4]^2 _{-3}$

$=\rho [\frac {16}3 - \frac {63}4]= (\frac{ -125}{12} )\rho$ units of the length times units of the mass

and the moment abotu the x-axis is

$m_x = \frac12 \rho \int ^2 _{-3} [(6-x)^2 - (x^2)^2]dx$

$= \frac12 \rho \int ^2 _{-3} [36-12x+x^2 - x^4 ]dx$

$=\frac12 \rho [36x-6x^2 + \frac {x^3}3 -\frac {x^5}5]^2_{-3}$

$=\frac12 \rho [\frac {664}{15} -\frac{-612}{5}]= \frac{250}3 \rho$ units of length times units of mass

the center of mass ($\bar x$ ,$\bar y$ ) have

$\bar x= m_y / m =\frac {\frac{-125 \rho}{12} } { \frac{125 \rho} 6 } = -1/2$

$\bar y= m_x /m =\frac { \frac {250\rho}3 } {\frac {125 \rho}6} =4$

$\implies (\frac{-1}2, 4)$ is the center of mass of the region bounded by the y=x^2 and y=6-x under the assumption that mass density is uniform over the region.