The region\({\rm{D}}\)is enclosed by the lines \(y = {x^2},y = x + 2\)
The density function is \(\rho (x,y) = kx{\rm{. }}\)
The intersection points of the region are found as follows.
The equations are set equal to solve for \(x.\)
\(\begin{aligned}{c}x + 2 = {x^2}\\{x^2} - x - 2 = 0\\(x - 2)(x + 1) = 0\\x = - 1,\,2\end{aligned}\)
The points of intersection are \(\left( { - 1,{{( - 1)}^2}} \right) = ( - 1,1)\) and \(\left( {2,{2^2}} \right) = (2,4).\)
The value of \(x\)and \(y\)ranges between \( - 1 \le x \le 2\) and \({x^2} \le y \le x + 2\) respectively.
The total mass of lamina is,
Integrate with respect to \(y,\)
\(\begin{aligned}{c}m = \int_{ - 1}^2 k x(y)_{{x^2}}^{x + 2}dx\\ = \int_{ - 1}^2 k x\left( {x + 2 - {x^2}} \right)dx\\ = k\int_{ - 1}^2 {\left( {{x^2} + 2x - {x^3}} \right)} dx\\ = k\left( {\frac{{{x^3}}}{3} + {x^2} - \frac{{{x^4}}}{4}} \right)_{ - 1}^2\end{aligned}\)
On applying limits, the value of\(m\)is given by,
\(\begin{aligned}{c}m = k\left( {\left( {\frac{{{2^3}}}{3} + {2^2} - \frac{{{2^4}}}{4}} \right) - \left( {\frac{{{{( - 1)}^3}}}{3} + {{( - 1)}^2} - \frac{{{{( - 1)}^4}}}{4}} \right)} \right)\\ = k\left( {\frac{8}{3} + 4 - \frac{{16}}{4}} \right) - k\left( { - \frac{1}{3} + 1 - \frac{1}{4}} \right)\\ = k\left( {\frac{{8 + 1}}{3} + 4 - 1 - \frac{{16 - 1}}{4}} \right)\\ = k\left( {6 - \frac{{15}}{4}} \right)\end{aligned}\)
On further simplification, the total mass becomes,
\(\begin{aligned}{c}m = \frac{{k(24 - 15)}}{4}\\ = \frac{{9k}}{4}\,{\rm{units}}\end{aligned}\)
In order to get the coordinates of the center of the mass, find\(\bar x{\rm{ and }}\bar y{\rm{. }}\)
Integrate with respect to \(x\)and apply the corresponding limit,
\(\begin{aligned}{c}\bar x = \frac{k}{{\frac{{9k}}{4}}}\int_{ - 1}^2 {\left( {{x^3} + 2{x^2} - {x^4}} \right)} dx\\ = \frac{{4k}}{{9k}}\left( {\frac{{{x^4}}}{4} + \frac{{2{x^3}}}{3} - \frac{{{x^5}}}{5}} \right)_{ - 1}^2\\ = \frac{4}{9}\left( {\left( {\frac{{{{(2)}^4}}}{4} + \frac{{2{{(2)}^3}}}{3} - \frac{{{2^5}}}{5}} \right) - \left( {\frac{{{{( - 1)}^4}}}{4} + \frac{{2{{( - 1)}^3}}}{3} - \frac{{{{( - 1)}^5}}}{5}} \right)} \right)\\ = \frac{4}{9}\left( {\frac{{16}}{4} + \frac{{16}}{3} - \frac{{32}}{5}} \right) - k\left( {\frac{1}{4} - \frac{2}{3} + \frac{1}{5}} \right)\end{aligned}\)
On further simplification,
\(\begin{aligned}{c}\bar x = \frac{4}{9}\left( {\left( {\frac{{16 - 1}}{4} + \frac{{16 + 2}}{3} - \frac{{32 + 1}}{5}} \right)} \right)\\ = \frac{4}{9}\left( {\frac{{15}}{4} + \frac{{18}}{3} - \frac{{33}}{5}} \right)\\ = \frac{4}{9}\left( {\frac{{189}}{{60}}} \right)\\ = \frac{{21}}{{15}}\\ = \frac{7}{5}\end{aligned}\)
The \(y\)-coordinate of the center of the mass is given by,
Integrate with respect to \(x\)and apply the corresponding limit,
\(\begin{aligned}{c}\bar y = \frac{4}{{18}}\int_{ - 1}^2 {\left( {x\left( {{x^2} + 4x + 4} \right) - {x^5}} \right)} dx\\ = \frac{4}{{18}}\int_{ - 1}^2 {\left( {{x^3} + 4{x^2} + 4x - {x^5}} \right)} dx\\ = \frac{4}{{18}}\left( {\frac{{{x^4}}}{4} + \frac{{4{x^3}}}{3} + 2{x^2} - \frac{{{x^6}}}{6}} \right)_{ - 1}^2\\ = \frac{4}{{18}}\left( {\left( {\frac{{{2^4}}}{4} + \frac{{4{{(2)}^3}}}{3} + 2{{(2)}^2} - \frac{{{{(2)}^6}}}{6}} \right) - \left( {\frac{{{{( - 1)}^4}}}{4} + \frac{{4{{( - 1)}^3}}}{3} + 2{{( - 1)}^2} - \frac{{{{( - 1)}^6}}}{6}} \right)} \right)\end{aligned}\)
On further simplification,
\(\begin{aligned}{c}\bar y = \frac{4}{{18}}\left( {\frac{{16}}{4} + \frac{{32}}{3} + 8} \right.\left. { - \frac{{64}}{6} - \frac{1}{4} + \frac{4}{3} - 2 + \frac{1}{6}} \right)\\ = \frac{4}{{18}}\left( {\frac{{45}}{4}} \right)\\ = \frac{{45}}{{18}}\\ = \frac{5}{2}\end{aligned}\)
The center of mass of lamina is \((\bar x,\bar y) = \left( {\frac{7}{5},\frac{5}{2}} \right).\)
Thus, the total mass of lamina is \(\frac{{9k}}{4}\,{\rm{units}}\) and the center of mass of lamina is \(\left( {\frac{7}{5},\frac{5}{2}} \right){\rm{. }}\)
