Question

Find the center of mass of the lamina. The length is \(a.\)the density is \(\rho (x,y) = k\left( {{x^2} + {y^2}} \right).\)

Short answer

The center of mass of lamina is \(\left( {\frac{{2a}}{5},\frac{{2a}}{5}} \right).\)

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\(f\)is a polar rectangle\(R\)given by\(0 \le a \le r \le b,\alpha \le \theta \le \beta {\rm{,}}\)where\(0 \le \beta - \alpha \le 2\pi {\rm{,}}\)then,

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

\(\int_a^b {\int_c^d g } (x)h(y)dydx = \int_a^b g (x)dx\int_c^d h (y)dy\,\,\,\,\,\,\,\,\,\,\,\,..........\left( 2 \right)\)

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

From the given condition it is observed that\(x\)varies from \(0\)to \(a\)and \(y\)varies from \(0\)to \(a - x.\) then, the total mass of the lamina is,

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

\(\begin{aligned}{c}m = k\int_0^a {\left( {{x^2}y + \frac{{{y^3}}}{3}} \right)_0^{a - x}} dx\\ = k\int_0^a {\left( {\left( {{x^2}(a - x) + \frac{{{{(a - x)}^3}}}{3}} \right) - \left( {{x^2}(0) + \frac{{{{(0)}^3}}}{3}} \right)} \right)} dx\\ = k\int_0^a {\left( {a{x^2} - {x^3} + \frac{{{a^3}}}{3} - \frac{{3{a^2}x}}{3} + \frac{{3a{x^2}}}{3} - \frac{{{x^3}}}{3} - 0 - 0} \right)} dx\\ = k\int_0^a {\left( {a{x^2} - {x^3} + \frac{{{a^3}}}{3} - {a^2}x + a{x^2} - \frac{{{x^3}}}{3}} \right)} dx\end{aligned}\)

Therefore,\(m = k\int_0^a {\left( {2a{x^2} - {a^2}x + \frac{{{a^3}}}{3} - \frac{{4{x^3}}}{3}} \right)} dx\)

Integrate with respect to \(x\)using the technique of integration by parts,

Let \(u = x.\)

Then \(dv = {e^{ - 2x}}dx\)

Thus, the total mass of lamina is obtained as follows,

\(\begin{aligned}{c}m = k\left( {\frac{{2a{x^3}}}{3} - \frac{{{a^2}{x^2}}}{2} + \frac{{{a^3}x}}{3} - \frac{{4{x^4}}}{{3(4)}}} \right)_0^a\\ = k\left( {\left( {\frac{{2a{{(a)}^3}}}{3} - \frac{{{a^2}{{(a)}^2}}}{2} + \frac{{{a^3}(a)}}{3} - \frac{{{{(a)}^4}}}{3}} \right) - \left( {\frac{{2a{{(0)}^3}}}{3} - \frac{{{a^2}{{(0)}^2}}}{2} + \frac{{{a^3}(0)}}{3} - \frac{{{{(0)}^4}}}{3}} \right)} \right)\\ = k\left( {\left( {\frac{{2{a^4}}}{3} - \frac{{{a^4}}}{2} + \frac{{{a^4}}}{3} - \frac{{{a^4}}}{3}} \right) - (0 - 0 + 0 - 0)} \right)\\ = k\left( {\frac{{2{a^4}}}{3} - \frac{{{a^4}}}{2}} \right)\end{aligned}\)

Apply the limit of \(x,\)

\(\begin{aligned}{c}m = k\left( {\frac{{4{a^4} - 3{a^4}}}{6}} \right)\\ = k\frac{{{a^4}}}{6}\\ = \frac{{k{a^4}}}{6}\end{aligned}\)

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,

\(\begin{aligned}{c}\bar x = \frac{6}{{{a^4}}}\int_0^a {\left( {{x^3}y + \frac{{x{y^3}}}{3}} \right)_0^{a - x}} dx\\ = \frac{6}{{{a^4}}}\int_0^a {\left( {\left( {{x^3}(a - x) + \frac{{x{{(a - x)}^3}}}{3}} \right) - \left( {{x^3}(0) + \frac{{x{{(0)}^3}}}{3}} \right)} \right)} dx\\ = \frac{6}{{{a^4}}}\int_0^a {\left( {\left( {a{x^3} - {x^4} + \frac{{x\left( {{a^3} - 3{a^2}x + 3a{x^2} - {x^3}} \right)}}{3}} \right) - (0 + 0)} \right)} dx\\ = \frac{6}{{{a^4}}}\int_0^a {\left( {a{x^3} - {x^4} + \frac{{{a^3}x}}{3} - \frac{{3{a^2}{x^2}}}{3} + \frac{{3a{x^3}}}{3} - \frac{{{x^4}}}{3}} \right)} dx\end{aligned}\)

Further simplify the terms as shown below,

\(\begin{aligned}{c}\bar x = \frac{6}{{{a^4}}}\int_0^a {\left( {a{x^3} - {x^4} + \frac{{{a^3}x}}{3} - {a^2}{x^2} + a{x^3} - \frac{{{x^4}}}{3}} \right)} dx\\ = \frac{6}{{{a^4}}}\int_0^a {\left( {2a{x^3} - \frac{{4{x^4}}}{3} - {a^2}{x^2} + \frac{{{a^3}x}}{3}} \right)} dx\end{aligned}\)

Integrate with respect to \(x,\)

\(\begin{aligned}{c}\bar x = \frac{6}{{{a^4}}}\left( {\frac{{2a{x^4}}}{4} - \frac{{4{x^5}}}{{3(5)}} - \frac{{{a^2}{x^3}}}{3} + \frac{{{a^3}{x^2}}}{{3(2)}}} \right)_0^a\\ = \frac{6}{{{a^4}}}\left( {\left( {\frac{{a{{(a)}^4}}}{2} - \frac{{4{{(a)}^5}}}{{15}} - \frac{{{a^2}{{(a)}^3}}}{3} + \frac{{{a^3}{{(a)}^2}}}{6}} \right) - \left( {\frac{{a{{(0)}^4}}}{2} - \frac{{4{{(0)}^5}}}{{15}} - \frac{{{a^2}{{(0)}^3}}}{3} + \frac{{{a^3}{{(0)}^2}}}{6}} \right)} \right)\\ = \frac{6}{{{a^4}}}\left( {\left( {\frac{{{a^3}}}{2} - \frac{{4{a^5}}}{{15}} - \frac{{{a^5}}}{3} + \frac{{{a^3}}}{6}} \right) - (0 - 0 + 0 + 0)} \right)\\ = \frac{6}{{{a^4}}}\left( {\frac{{45{a^3} - 24{a^5} - 30{a^5} + 15{a^3}}}{{90}}} \right)\end{aligned}\)

Further simplify the terms as shown below,

\(\begin{aligned}{c}\vec x = \frac{6}{{{a^4}}}\left( {\frac{{6{a^5}}}{{90}}} \right)\\ = \frac{{2a}}{5}\end{aligned}\)

Compute the value of \(\bar y.\)

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

\(\begin{aligned}{c}\bar y = \frac{6}{{{a^4}}}\int_0^a {\left( {\frac{{{y^4}}}{4} + \frac{{{y^2}{x^2}}}{2}} \right)_0^{a - x}} dx\\ = \frac{6}{{{a^4}}}\int_0^a {\left( {\left( {\frac{{{{(a - x)}^4}}}{4} + \frac{{{x^2}{{(a - x)}^2}}}{2}} \right) - \left( {\frac{{{{(0)}^4}}}{4} + \frac{{{x^2}{{(0)}^2}}}{2}} \right)} \right)} dx\\ = \frac{6}{{{a^4}}}\int_0^a {\left( {\left( {\frac{{{{(a - x)}^4}}}{4} + \frac{{{x^2}\left( {{a^2} - 2ax + {x^2}} \right)}}{2}} \right) - (0 + 0)} \right)} dx\\ = \frac{6}{{{a^4}}}\int_0^a {\left( {\frac{{{{(a - x)}^4}}}{4} + \frac{{{a^2}{x^2}}}{2} - \frac{{2a{x^3}}}{2} + \frac{{{x^4}}}{2}} \right)} dx\end{aligned}\)

Integrate with respect to\(x,\)

\(\begin{aligned}{c}\bar y = \frac{6}{{{a^4}}}\left( { - \frac{{{{(a - x)}^5}}}{{4(5)}} + \frac{{{a^2}{x^3}}}{{2(3)}} - \frac{{2a{x^4}}}{{2(4)}} + \frac{{{x^5}}}{{2(5)}}} \right)_0^a\\ = \frac{6}{{{a^4}}}\left( {\left( { - \frac{{{{(a - a)}^5}}}{{20}} + \frac{{{a^2}{a^3}}}{6} - \frac{{2a{a^4}}}{8} + \frac{{{a^5}}}{{10}}} \right) - \left( { - \frac{{{{(a - 0)}^5}}}{{20}} + \frac{{{a^2}{{(0)}^3}}}{6} - \frac{{2a{{(0)}^4}}}{8} + \frac{{{{(0)}^5}}}{{10}}} \right)} \right)\\ = \frac{6}{{{a^4}}}\left( {\left( { - 0 + \frac{{{a^5}}}{6} - \frac{{{a^5}}}{4} + \frac{{{a^5}}}{{10}}} \right) - \left( { - \frac{{{a^5}}}{{20}} + 0 - 0 + 0} \right)} \right)\\ = \frac{6}{{{a^4}}}\left( {\frac{{10{a^5} - 15{a^5} + 6\left( {{a^3}} \right.}}{{60}} + \frac{{{a^5}}}{{20}}} \right)\end{aligned}\)’

Further simplify the terms as shown below,

\(\begin{aligned}{c}\bar y = \frac{6}{{{a^4}}}\left( {\frac{{10{a^3} - 15{a^3} + 6{a^5} + 3{a^3}}}{{60}}} \right)\\ = \frac{6}{{{a^4}}}\left( {\frac{{4{a^5}}}{{60}}} \right)\\ = \frac{{2a}}{5}\end{aligned}\)

The center of mass of the lamina is\((\bar x,\bar y) = \left( {\frac{{2a}}{5},\frac{{2a}}{5}} \right).\)

Thus, the center of mass of lamina is \(\left( {\frac{{2a}}{5},\frac{{2a}}{5}} \right).\)