Question

A lamina occupies the part of the disk x2 + y2 ≤ 36 in the first quadrant. Find its center of mass if the density at any point is proportional to its distance from the x-axis.

Answer 1

Answer:$\left ( \frac{9}{4},\frac{9\pi }{8} \right )$

Explanation:

The density at a point is proportional to its distance from the x axis

Writing equation in polar form

$x=rcos\theta ,y=rsin\theta x^2+y^2=r^2,dA=rdrd\theta$

and $\theta is (0,\frac{\pi }{2})$

$\rho =ky=k\cdot rsin\theta$

where k is proportionality constant

Total mass ig given by

$m=\iint_{}^{}\rho dA$

$m=\int_{0}^{\frac{\pi }{2}}\int_{0}^{6}kyrdrd\theta$

$m=\int_{0}^{\frac{\pi }{2}}\int_{0}^{6}krsin\theta rdrd\theta$

$m=k\int_{0}^{\frac{\pi }{2}}sin\theta d\theta \int_{0}^{6}r^2dr$

$m=k\left ( -cos\theta \right )_0^{\frac{\pi }{2}}\left ( \frac{r^3}{3}\right )_{0}^{6}$

m=72k

Also

$M_y=\int \int x\rho dA$

$M_y=\int_{0}^{\frac{\pi }{2}}\int_{0}^{6}rcos\theta ky\cdot rdr d\theta$

$M_y=\int_{0}^{\frac{\pi }{2}}\int_{0}^{6}rcos\theta \cdot k\cdot rsin\theta \cdot rdr\cdot d\theta$

$M_y=k\int_{0}^{\frac{\pi }{2}}sin\theta cos\theta d\theta \int_{0}^{6}r^3dr$

$M_y=81k(2)=162k$

Similarly

$M_x=\int \int y\rho dA$

$M_x=\int_{0}^{\frac{\pi }{2}}\int_{0}^{6}rsin\theta krsin\theta \cdot rdr d\theta  $

$M_x=k\int_{0}^{\frac{\pi }{2}}sin^2\theta d\theta \int_{0}^{6}r^3dr$

$M_x=k\int_{0}^{\frac{\pi }{2}}\frac{1-cos(2\theta )}{2}d\theta \left (\frac{r^4}{4} \right )_0^6$

$M_x=\frac{k}{2}\left [ \theta -\frac{sin2\theta }{2} \right ]_0^\frac{\pi }{2}\cdot 324$

$M_x=81k\pi$

Centre of mass co-ordinates is $\bar{x},\bar{y}=\left ( \frac{M_x}{m},\frac{M_y}{m} \right )$

$\left ( \frac{162k}{72k},\frac{81k\pi }{72k} \right )=\left ( \frac{9}{4},\frac{9\pi }{8} \right )$