Answer with Explanation:
We are given that a lamina occupies the part of the disk
[tex]x^2+y^2\leq 49[/tex] 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
[tex](x-h)^2+(y-k)^2=r^2[/tex]
We have radius =7 and center=(0,0)
[tex]\rho\propto r^2[/tex] Where r= Distance between origin and the point on boundary of circle
Density function [tex]\rho (x,y)=kr^2=K(x^2+y^2)[/tex]
Where K= Proportionality constant
[tex]r=\sqrt{x^2+y^2}[/tex]
[tex]x=rcos\theta, y=rsin\theta[/tex]
Radius varies from 0 to 7 and angle([tex]\theta[/tex]) varies from 0 to [tex]\frac{\pi}{2}[/tex].
Mass of the lamina=m=[tex]\int_{0}^{\frac{\pi}{2}}\int_{0}^{7}kr^2\cdot rdrd\theta[/tex]
[tex]m=k\int_{0}^{\frac{\pi}{2}}[\frac{r^4}{4}]^{7}_{0} d\theta [/tex]
[tex]m=k\int_{0}^{\frac{\pi}{2}}\frac{2401}{4} d\theta[/tex]
[tex]m=\frac{2401k}{4}\times \frac{\pi}{2}[/tex]
[tex]m=\frac{2401\pi}{8}[/tex]
Its first moments is given by
[tex]M_x=\int \int_{region} y\cdot \rho (x,y) dx dy[/tex] ([tex]dxdy=rdr d\theta[/tex])
[tex]M_x=\int_{0}^{\frac{\pi}{2}}\int_{0}^{7}r sin\theta \cdot kr^3 dr d\theta[/tex]
[tex]M_x=\int_{0}^{\frac{\pi}{2}}\int_{0}^{7} kr^4 sin\theta dr d\theta[/tex]
[tex]M_x=K\frac{16807}{5}[-cos\theta]^{\frac{\pi}{2}}_{0}[/tex]
[tex]M_x=\frac{16807k}{5}(0+1)=\frac{16807k}{5}[/tex] ([tex]cos\frac{\pi}{2}=0, cos 0=1[/tex])
[tex]M_y=\int\int_{region}x \cdot \rho(x, y) dx dy [/tex]
[tex]M_y=\int_{0}^{\frac{\pi}{2}}\int_{0}^{7} kr^4 cos\theta drd\theta[/tex]
[tex]M_y=\frac{16807}{5}[sin\theta]^{\frac{\pi}{2}}_{0}[/tex]
[tex]M_y=\frac{16807k}{5}(sin\frac{\pi}{2}-sin0)[/tex]
[tex]M_y=\frac{16807k}{5}(1-0)=\frac{16807k}{5}[/tex] ([tex] sin\frac{\pi}{2}=1, sin 0=0[/tex])
Center of mass is given by
[tex]m_x=\frac{M_x}{m}=\frac{\frac{16807}{5}}{\frac{2401k\pi}{8}}=\frac{56}{5\pi}[/tex]
[tex]m_y=\frac{M_y}{m}=\frac{\frac{16807k}{5}}{\frac{2401k\pi}{8}}=\frac{56}{5\pi}[/tex]
Hence, the center of mass of the lamina=([tex]\frac{56}{5\pi}, \frac{56}{5\pi}[/tex])
