Question

Find the total mass and the center of mass of the lamina. The region is\(x = 0, y = x, 2 x + y = 6\). The density is\(\rho (x,y) = {x^2}\).

Short answer

The total mass of lamina is 4 units and the center of mass of lamina is \(\left( {\frac{6}{5},\frac{{12}}{5}} \right).\)

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 the triangle enclosed by lines \({\rm{x = 0, y = x, 2 x + y = 6}}\)

The density function is \(\rho (x,y) = {x^2}\)

The total mass of lamina is,

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

\(\begin{aligned}{c}m = \int_0^2 {{x^2}} (y)_x^{6 - 2x}dx\\ = \int_0^2 {{x^2}} (6 - 2x - x)dx\\ = \int_0^2 {{x^2}} (6 - 3x)dx\\ = \int_0^2 {\left( {6{x^2} - 3{x^3}} \right)} dx\end{aligned}\)

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

\(\begin{aligned}{c}m = \left( {\frac{{6{x^3}}}{3} - \frac{{3{x^4}}}{4}} \right)_0^2\\ = \left( {2{x^3} - \frac{{3{x^4}}}{4}} \right)_0^2\\ = \left( {2\left( {{2^3}} \right) - \frac{{3{{(2)}^4}}}{4}} \right)\\ = (16 - 12)\\ = 4\end{aligned}\)

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,

\(\begin{aligned}{c}\bar x = \frac{1}{4}\int_0^2 {\left( {6{x^4} - 3{x^4}} \right)} dx\\ = \frac{1}{4}\left( {\frac{{6{x^4}}}{4} - \frac{{3{x^5}}}{5}} \right)_0^2\\ = \frac{1}{4}\left( {\frac{{3{{(2)}^4}}}{2} - \frac{{3{{(2)}^5}}}{5}} \right)\\ = \frac{1}{4}\left( {24 - \frac{{96}}{5}} \right)\end{aligned}\)

On further simplification,

\(\begin{aligned}{c}\bar x = \frac{1}{4}\left( {\frac{{120 - 96}}{5}} \right)\\ = \frac{1}{4}\left( {\frac{{24}}{5}} \right)\\ = \frac{6}{5}\end{aligned}\)

The \(y\)-coordinate of the center of the mass is given by,

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

\(\begin{aligned}{c}\bar y = \frac{1}{8}\int_0^2 {{x^2}} \left( {\left( {36 + 4{x^2} - 24x} \right) - {x^2}} \right)dx\\ = \frac{1}{8}\int_0^2 {\left( {36{x^2} + 3{x^4} - 24{x^3}} \right)} dx\\ = \frac{1}{8}\left( {\frac{{36{x^3}}}{3} + \frac{{3{x^5}}}{5} - \frac{{24{x^4}}}{4}} \right)_0^2\\ = \frac{1}{8}\left( {12{{(2)}^3} + \frac{{3{{(2)}^5}}}{5} - 6{{(2)}^4}} \right)\end{aligned}\)

On further simplification,

\(\begin{aligned}{c}\bar y = \frac{1}{4}\left( {12{{(2)}^2} + \frac{{3{{(2)}^4}}}{5} - 6{{(2)}^3}} \right)\\ = \frac{1}{4}\left( {48 + \frac{{48}}{5} - 48} \right)\\ = \frac{1}{4}\left( {\frac{{48}}{5}} \right)\\ = \frac{{12}}{5}\end{aligned}\)

The center of the mass of lamina is \((\bar x,\bar y) = \left( {\frac{6}{5},\frac{{12}}{5}} \right).\)

Thus, the total mass of lamina is \(4\)units and the center of mass of lamina is \(\left( {\frac{6}{5},\frac{{12}}{5}} \right).\)