Question

Find the center of mass of the lamina. The region is\({x^2} + {y^2} \le 1\). The density is\(\rho (x,y) = k\left( {{x^2} + {y^2}} \right)\).

Short answer

The center of mass of lamina is \(\left( {\frac{8}{{5\pi }},\frac{8}{{5\pi }}} \right)\)

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 mass and center of mass of the lamina

The region\(D\) is the disk \({x^2} + {y^2} \le 1\) in the first quadrant.

The density function is proportional to the square of its distance from the origin, that is \(\rho (x,y) = k\left( {{x^2} + {y^2}} \right)\)

Convert into polar coordinates to make the problem easier. So, from the given conditions it is observed that \(x\) varies from \(0\)to \(1\) and \(\theta \) varies from\(0\)to\(\frac{\pi }{2}\). Then, by (1) the total mass of the lamina is,

Integrate with respect to \(r\) and \(\theta \)by using the property (2),

\(\begin{aligned}{c}m = k\int_0^{\frac{\pi }{2}} d \theta \int_0^1 {{r^3}} dr\\ = k(\theta )_0^{\frac{\pi }{2}}\left( {\frac{{{\mu ^4}}}{4}} \right)_0^1\\ = k\left( {\frac{\pi }{2} - 0} \right)\left( {\frac{{{1^4}}}{4} - \frac{{{0^4}}}{4}} \right)\\ = k\left( {\frac{\pi }{2}} \right)\left( {\frac{1}{4} - 0} \right)\\ = \frac{{k\pi }}{8}\end{aligned}\)

In order to get the coordinates of the center of the mass, find\(\bar x{\rm{ and }}\bar y{\rm{. }}\)Therefore, by equation (1).

Integrate with respect to \(r\) and \(\theta \)by using the property (2),

\(\begin{aligned}{c}\bar x = \frac{8}{\pi }\int_0^{\frac{\pi }{2}} {\cos } \theta d\theta \int_0^1 {{r^4}} dr\\ = \frac{8}{\pi }(\sin \theta )_0^{\frac{\pi }{2}}\left( {\frac{{{r^5}}}{5}} \right)_0^1\\ = \frac{8}{\pi }\left( {\sin \left( {\frac{\pi }{2}} \right) - \sin (0)} \right)_0^{\frac{\pi }{2}}\left( {\frac{{{1^5}}}{5} - \frac{{{0^5}}}{5}} \right)\\ = \frac{8}{\pi }(1 - 0)\left( {\frac{1}{5} - \frac{0}{5}} \right)\end{aligned}\)

Further simplify the terms as shown below,

\(\begin{aligned}{c}\bar x = \frac{8}{\pi }(1)\left( {\frac{1}{5}} \right)\\ = \frac{8}{{5\pi }}\end{aligned}\)

Compute the value of \(\bar y{\rm{. }}\)

Integrate with respect to \(r\) and \(\theta \)by using the property (2),

\(\begin{aligned}{c}\bar y = \frac{8}{\pi }\int_0^{\frac{\pi }{2}} {\sin } \theta d\theta \int_0^1 {{r^4}} dr\\ = \frac{8}{\pi }( - \cos \theta )_0^{\frac{\pi }{2}}\left( {\frac{{{r^5}}}{5}} \right)_0^1\\ = \frac{8}{\pi }\left( { - \cos \left( {\frac{\pi }{2}} \right) + \cos (0)} \right)_0^{\frac{\pi }{2}}\left( {\frac{{{1^5}}}{5} - \frac{{{0^5}}}{5}} \right)\\ = \frac{8}{\pi }( - 0 + 1)\left( {\frac{1}{5} - \frac{0}{5}} \right)\end{aligned}\)

Further, simplify the terms as shown below,

\(\begin{aligned}{c}\bar y = \frac{8}{\pi }(1)\left( {\frac{1}{5}} \right)\\ = \frac{8}{{5\pi }}\end{aligned}\)

The center of mass of lamina is \((\bar x,\bar y) = \left( {\frac{8}{{5\pi }},\frac{8}{{5\pi }}} \right).\)

Thus, the center of mass of lamina is \(\left( {\frac{8}{{5\pi }},\frac{8}{{5\pi }}} \right)\)