Question

. (15.4) A thin lamina is formed by considering the region inside the circle x 2 + y 2 = 6y and outside the circle x 2+y 2 = 9. Find the center of mass of the lamina if the density at any point (in grams per meter squared) is inversely proportional to its distance from the origin. Follow up question: why do we not care what the constant of proportionality is?

Answer 1

Answer:

Center of mass=$(\bar{x},\bar{y})=(0,\frac{M_x}{m})=(0,\frac{3\sqrt{3}k}{2\sqrt{3}-\pi})$

Step-by-step explanation:

Given equation of circles are,

$x^2+y^2=6y$

$x^2+y^2=9$

To convert into parametric equation let $x=r\cos\theta,y=r\sin\theta$

we get,

$r=6\sin\theta$ and $r=\pm 3$

when,

$r=3, \theta=\frac{\pi}{6}$

$r=-3,\theta=\frac{7\pi}{6}$

Therefore density function is,

$\roh (x,y)=\frac{k}{\sqrt{x^2+y^2}}=\frac{k}{r}$

where k is constant of proportionality.

Then,

mass=m=$\int_{\frac{\pi}{6}}^{\frac{7\pi}{6}}\int_{3}^{6\sin\theta}\frac{k}{r}rdrd\theta$

$=k\int_{\frac{\pi}{6}}^{\frac{7\pi}{6}}(6\sin\theta-3)d\theta$

$=3k(2\sqrt{3}-\pi)$

By the symmetry of the domain and the function f(x)=x, $M_y=0$ and,

$M_x=\int_{\frac{\pi}{6}}^{\frac{7\pi}{6}}\int_{3}^{6\sin\theta}krdrd\theta$

$=\frac{9k}{2}\int_{\frac{\pi}{6}}^{\frac{7\pi}{6}}(4\sin^3\theta-\sin\theta)d\theta$

$=18k\int_{\frac{\pi}{6}}^{\frac{7\pi}{6}}36\sin^2\theta\sin\theta d\theta-\frac{9k}{2}\int_{\frac{\pi}{6}}^{\frac{7\pi}{6}}\sin\theta d\theta$

$=-18k\int_{\frac{\pi}{6}}^{\frac{7\pi}{6}}36(1-\cos^2\theta) d\cos\theta-\frac{9k}{2}\int_{\frac{\pi}{6}}^{\frac{7\pi}{6}}\sin\theta d\theta$

$=\frac{18\sqrt{3}k}{2}$

Hence,

Center of mass=$(\bar{x},\bar{y})=(0,\frac{M_x}{m})=(0,\frac{3\sqrt{3}k}{2\sqrt{3}-\pi})$