Question

A lamina occupies the part of the disk x2 + y2 ≤ 49 in the first quadrant. Find the center of mass of the lamina if the density at any point is proportional to the square of its distance from the origin.

Answer 1

Answer with Explanation:

We are given that a lamina occupies the part of the disk

$x^2+y^2\leq 49$ in the first quadrant.

Radius is a distance between the center of circle and the point on boundary of circle.

By comparing the equation of circle

$(x-h)^2+(y-k)^2=r^2$

We have radius =7 and center=(0,0)

$\rho\propto r^2$ Where r= Distance between origin and the point on boundary of circle

Density function $\rho (x,y)=kr^2=K(x^2+y^2)$

Where K= Proportionality constant

$r=\sqrt{x^2+y^2}$

$x=rcos\theta, y=rsin\theta$

Radius varies from 0 to 7 and  angle($\theta$) varies from 0 to $\frac{\pi}{2}$.

Mass of the lamina=m=$\int_{0}^{\frac{\pi}{2}}\int_{0}^{7}kr^2\cdot rdrd\theta$

$m=k\int_{0}^{\frac{\pi}{2}}[\frac{r^4}{4}]^{7}_{0} d\theta$

$m=k\int_{0}^{\frac{\pi}{2}}\frac{2401}{4} d\theta$

$m=\frac{2401k}{4}\times \frac{\pi}{2}$

$m=\frac{2401\pi}{8}$

Its first moments is given by

$M_x=\int \int_{region} y\cdot \rho (x,y) dx dy$   ($dxdy=rdr d\theta$)

$M_x=\int_{0}^{\frac{\pi}{2}}\int_{0}^{7}r sin\theta \cdot kr^3 dr d\theta$

$M_x=\int_{0}^{\frac{\pi}{2}}\int_{0}^{7} kr^4 sin\theta dr d\theta$

$M_x=K\frac{16807}{5}[-cos\theta]^{\frac{\pi}{2}}_{0}$

$M_x=\frac{16807k}{5}(0+1)=\frac{16807k}{5}$   ($cos\frac{\pi}{2}=0, cos 0=1$)

$M_y=\int\int_{region}x \cdot \rho(x, y) dx dy$

$M_y=\int_{0}^{\frac{\pi}{2}}\int_{0}^{7} kr^4 cos\theta drd\theta$

$M_y=\frac{16807}{5}[sin\theta]^{\frac{\pi}{2}}_{0}$

$M_y=\frac{16807k}{5}(sin\frac{\pi}{2}-sin0)$

$M_y=\frac{16807k}{5}(1-0)=\frac{16807k}{5}$  ($sin\frac{\pi}{2}=1, sin 0=0$)

Center of mass is given by

$m_x=\frac{M_x}{m}=\frac{\frac{16807}{5}}{\frac{2401k\pi}{8}}=\frac{56}{5\pi}$

$m_y=\frac{M_y}{m}=\frac{\frac{16807k}{5}}{\frac{2401k\pi}{8}}=\frac{56}{5\pi}$

Hence, the center of mass of the lamina=($\frac{56}{5\pi}, \frac{56}{5\pi}$)