Question

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

Short answer

The total mass of lamina is \(\frac{3}{{14}}\) and the center of mass of lamina is \(\left( {\frac{{14}}{{27}},\frac{{28}}{{55}}} \right){\rm{. }}\)

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,

\(\int_a^b {\int_c^d g } (x)h(y)dydx = \int_a^b g (x)dx\int_c^d h (y)dy\) …… (1)

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

The regionis bounded by the parabolas \(y = {x^2}{\rm{ and }}x = {y^2}\)

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

The intersection points of the region are found by sketching the parabolas as follows,

Therefore, the points of intersection are \((0,0){\rm{ and }}(1,1).\)

The value of \(x\)and \(y\)ranges between \(0 \le x \le 1\)and \({x^2} \le y \le \sqrt x \)respectively.

The total mass of lamina is,

Integrate with respect to\(y\),

\(\begin{aligned}{l}m = \mathop \smallint \nolimits_0^1 (y\sqrt x )\sqrt x {x^2}dx\\ = \int_0^1 {\left( {\sqrt x - {x^2}} \right)} \sqrt x dx\\ = \int_0^1 {\left( {x - {x^{\frac{5}{2}}}} \right)} dx\\ = \left( {\frac{{{x^2}}}{2} - \frac{{2{x^{\frac{7}{2}}}}}{7}} \right)_0^1\end{aligned}\)

On applying limits, the value of \(m\)is given by,

\(\begin{aligned}{l}m = \left( {\frac{1}{2} - \frac{{2{x^{\frac{7}{2}}}}}{7}} \right)_0^1\\ = \frac{1}{2} - \frac{2}{7}\\ = \frac{3}{{14}}\end{aligned}\)

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,

\(\begin{aligned}{l}\bar x = \frac{{14}}{3}\int_0^1 {\left( {{x^2} - {x^{\frac{7}{2}}}} \right)} dx\\ = \frac{{14}}{3}\left( {\frac{{{x^3}}}{3} - \frac{{2{x^{\frac{9}{2}}}}}{9}} \right)_0^1\\ = \frac{{14}}{3}\left( {\frac{1}{3} - \frac{{2{{(1)}^{\frac{9}{2}}}}}{9}} \right)\\ = \frac{{14}}{3}\left( {\frac{1}{9}} \right)\\ = \frac{{14}}{{27}}\end{aligned}\)

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

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

\(\begin{aligned}{l}\bar y = \frac{7}{3}\left( {\frac{{2{x^3}}}{5} - \frac{{2{x^{\frac{{11}}{2}}}}}{{11}}} \right)_0^1\\ = \frac{7}{3}\left( {\frac{{2{x^{\frac{3}{2}}}}}{5} - \frac{{2{x^{\frac{{11}}{2}}}}}{{11}}} \right)_0^1\\ = \frac{7}{3}\left( {\frac{{12}}{{55}}} \right)\\ = \frac{{28}}{{55}}\end{aligned}\)

The center of mass of lamina is \((\bar x,\bar y) = \left( {\frac{{14}}{{27}},\frac{{28}}{{55}}} \right).\)

Thus, the total mass of lamina is \(\frac{3}{{14}}\) and the center of mass of lamina is \(\left( {\frac{{14}}{{27}},\frac{{28}}{{55}}} \right){\rm{. }}\)