Question

Find the total mass and the center of mass of the lamina. The region is\(y = 1 - {x^2},y = 0\). The density is\(\rho (x,y) = ky\).

Short answer

The total mass of lamina is \(\frac{{8k}}{{15}}\,{\rm{units}}\) and the center of mass of lamina is \(\left( {0,\frac{4}{7}} \right).\)

Step-by-step solution

The concept of the total mass and center of the mass

The total mass of the lamina is

The center of mass of the lamina that occupies the given region\(D\)is\((\bar x,\bar y){\rm{. }}\)

Here,

If\(g(x)\)is the function of\(x\)and\(h(y)\)is the function of\(y\)then,

\(\int_a^b {\int_c^d g } (x)h(y)dydx = \int_a^b g (x)dx\int_c^d h (y)dy\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,........\left( 1 \right)\)

Use the above concept to find the mass and center of mass of the lamina

The region\({\rm{D}}\)is bounded by \(y = 1 - {x^2},y = 0\)

The density function is \(\rho (x,y) = ky\)

The total mass of lamina is,

Integrate with respect to\(y\)and apply the corresponding limit,

\(\begin{aligned}{c}m = k\int_{ - 1}^1 {\left( {\frac{{{y^2}}}{2}} \right)_0^{1 - {x^2}}} dx\\ = \int_{ - 1}^1 {\left( {\frac{{{{\left( {1 - {x^2}} \right)}^2}}}{2} - \frac{{{{(0)}^2}}}{2}} \right)} dx\\ = \int_{ - 1}^1 {\left( {\frac{1}{2} - \frac{{2{x^2}}}{2} + \frac{{{x^4}}}{2} - 0} \right)} dx\\ = k\int_{ - 1}^1 {\left( {\frac{1}{2} - {x^2} + \frac{{{x^4}}}{2}} \right)} dx\end{aligned}\)

Integrate with respect to \(x\)and apply the corresponding limit,

\(\begin{aligned}{c}m = k\left( {\frac{x}{2} - \frac{{{x^3}}}{3} + \frac{{{x^5}}}{{2(5)}}} \right)_{ - 1}^1\\ = k\left( {\left( {\frac{{(1)}}{2} - \frac{{{{(1)}^3}}}{3} + \frac{{{{(1)}^5}}}{{10}}} \right) - \left( {\frac{{( - 1)}}{2} - \frac{{{{( - 1)}^3}}}{3} + \frac{{{{( - 1)}^5}}}{{10}}} \right)} \right)\\ = k\left( {\left( {\frac{1}{2} - \frac{1}{3} + \frac{1}{{10}}} \right) - \left( { - \frac{1}{2} + \frac{1}{3} - \frac{1}{{10}}} \right)} \right)\\ = k\left( {1 - \frac{2}{3} + \frac{1}{5}} \right)\\ = \frac{{8k}}{{15}}\end{aligned}\)

In order to get the coordinates of the center of the mass, find\(\bar x{\rm{ and }}\bar y{\rm{. }}\)

Integrate with respect to \(y\)and apply the corresponding limit,

\(\begin{aligned}{c}\bar x = \frac{{15}}{8}\int_{ - 1}^1 {\left( {\frac{{x{y^2}}}{2}} \right)_0^{1 - {x^2}}} dx\\ = \frac{{15}}{8}\int_{ - 1}^1 {\left( {\frac{{x{{\left( {1 - {x^2}} \right)}^2}}}{2} - \frac{{x{{(0)}^2}}}{2}} \right)} dx\\ = \frac{{15}}{8}\int_{ - 1}^1 {\left( {\frac{{x\left( {1 - 2{x^2} + {x^4}} \right)}}{2} - 0} \right)} dx\\ = \frac{{15}}{8}\int_{ - 1}^1 {\left( {\frac{x}{2} - \frac{{2{x^3}}}{2} + \frac{{{x^5}}}{2}} \right)} dx\\ = \frac{{15}}{8}\int_{ - 1}^1 {\left( {\frac{x}{2} - {x^3} + \frac{{{x^5}}}{2}} \right)} dx\end{aligned}\)

Integrate with respect to \(x\)and apply the corresponding limit,

\(\begin{aligned}{c}\bar x = \frac{{15}}{8}\left( {\frac{{{x^2}}}{{2(2)}} - \frac{{{x^4}}}{3} + \frac{{{x^6}}}{{2(6)}}} \right)_{ - 1}^1\\ = \frac{{15}}{8}\left( {\left( {\frac{{{{(1)}^2}}}{4} - \frac{{{{(1)}^4}}}{3} + \frac{{{{(1)}^6}}}{{12}}} \right) - \left( {\frac{{{{( - 1)}^2}}}{4} - \frac{{{{( - 1)}^4}}}{3} + \frac{{{{( - 1)}^6}}}{{12}}} \right)} \right)\\ = \frac{{15}}{8}\left( {\left( {\frac{1}{4} - \frac{1}{3} + \frac{1}{{12}}} \right) - \left( {\frac{1}{4} - \frac{1}{3} + \frac{1}{{12}}} \right)} \right)\\ = \frac{{15}}{8}(0 - 0 + 0)\\ = 0\end{aligned}\)

\(\begin{aligned}{c}\bar y = \frac{1}{m}\int_D^1 y \rho (x,y)dA\\ = \frac{1}{{\left( {\frac{{8k}}{{15}}} \right)}}\int_{ - 1}^{1 - {x^2}} {\int_0^1 k } y(y)dydx\\ = \frac{{15k}}{{8k}}\int_{ - 1}^{1 - {x^2}} {\int_0^1 {{y^2}} } dydx\end{aligned}\)

Integrate with respect to \(y\)and apply the corresponding limit,

\(\begin{aligned}{c}\bar y = \frac{{15}}{8}\int_{ - 1}^1 {\left( {\frac{{{y^3}}}{3}} \right)_0^{1 - {x^2}}} dx\\ = \frac{{15}}{8}\int_{ - 1}^1 {\left( {\left( {\frac{{{{\left( {1 - {x^2}} \right)}^3}}}{3}} \right) - \left( {\frac{{{{(0)}^3}}}{3}} \right)} \right)} dx\\ = \frac{{15}}{8}\int_{ - 1}^1 {\left( {\frac{1}{3} - \frac{{3{x^2}}}{3} + \frac{{3{x^4}}}{3} - \frac{{{x^6}}}{3} - 0} \right)} dx\\ = \frac{{15}}{8}\int_{ - 1}^1 {\left( {\frac{1}{3} - {x^2} + {x^4} - \frac{{{x^6}}}{3}} \right)} dx\end{aligned}\)

Integrate with respect to \(x\)and apply the corresponding limit,

\(\begin{aligned}{c}\bar y = \frac{{15}}{8}\left( {\frac{x}{3} - \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} - \frac{{{x^7}}}{{3(7)}}} \right)_{ - 1}^1\\ = \frac{{15}}{8}\left( {\left( {\frac{{(1)}}{3} - \frac{{{{(1)}^3}}}{3} + \frac{{{{(1)}^5}}}{5} - \frac{{{{(1)}^7}}}{{21}}} \right) - \left( {\frac{{( - 1)}}{3} - \frac{{{{( - 1)}^3}}}{3} + \frac{{{{( - 1)}^5}}}{5} - \frac{{{{( - 1)}^7}}}{{21}}} \right)} \right)\\ = \frac{{15}}{8}\left( {\left( {\frac{1}{3} - \frac{1}{3} + \frac{1}{5} - \frac{1}{{21}}} \right) - \left( { - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{{21}}} \right)} \right)\\ = \frac{{15}}{8}\left( {\frac{2}{5} - \frac{2}{{21}}} \right)\end{aligned}\)

Further, simplify the terms as shown below,

\(\begin{aligned}{c}\bar y = \frac{{15}}{8}\left( {\frac{{32}}{{5(21)}}} \right)\\ = \frac{4}{7}\end{aligned}\)

The center of mass of lamina is \((\bar x,\bar y) = \left( {0,\frac{4}{7}} \right)\)

Thus, the total mass of lamina is \(\frac{{8k}}{{15}}\,{\rm{units}}\) and the center of mass of lamina is \(\left( {0,\frac{4}{7}} \right).\)