# Centre of mass of region using density (multivariable calculus)

A lamina (two–dimensional plate) occupies the region inside the circle $x^2 + y^2 = 2y$, but outside the circle $x^2 +y^2 = 1$. Find the centre of mass if the density = $\frac{k}{r}$ (inversely proportional to its distance from the origin). The formulae for centre of mass for $x,y$ are below:

the graph of the two circles is below ($x^2 + y^2 = 2y$ is $x^2 + (y - 1)^2 = 1$ ):

I don't understand how to find the boundaries of region D. I know that it has to be in the form of polar co-ords $(r, \theta)$ but I can't figure it out. Plz help!

• By symmetry the $x$-coordinate of the centre of mass is $0$. For the $y$-coordinate, we will need to integrate. Polar coordinates sound good. The point of intersection in the first quadrant is $(\sqrt{3}/2,1/2)$ so for the integral we will go $\theta=\pi/6$ to $\theta=5\pi/6$, or better $\pi/6$ to $\pi/2$ then double. – André Nicolas May 2 '15 at 3:28

For $m$, we have that

\begin{align} m&=\int_{\pi/6}^{5\pi/6} \int_1^{2\sin \phi} \frac{k}{\rho}\rho d\rho d\phi\\\\ &=k\int_{\pi/6}^{5\pi/6} (2\sin \phi -1)d\phi\\\\ & =(2\sqrt{3}-2\pi/3)k \end{align}

For the term $M_x$, we have that

\begin{align} M_x&=\int_{\pi/6}^{5\pi/6} \int_1^{2\sin \phi} \frac{k\rho \cos \phi}{\rho}\rho d\rho d\phi\\\\ &=k\int_{\pi/6}^{5\pi/6} \cos \phi \int_1^{2\sin \phi} \rho d\rho \\\\ &=k\int_{\pi/6}^{5\pi/6}\cos \phi \left(\frac12 (4\sin^2\phi -1)\right)d\phi\\\\ &=0 \end{align}

For the term $M_y$, we have that

\begin{align} M_y&=\int_{\pi/6}^{5\pi/6} \int_1^{2\sin \phi} \frac{k\rho \sin \phi}{\rho}\rho d\rho d\phi\\\\ &=k\int_{\pi/6}^{5\pi/6} \sin \phi \int_1^{2\sin \phi} \rho d\rho \\\\ &=k\int_{\pi/6}^{5\pi/6}\sin \phi \left(\frac12 (4\sin^2\phi -1)\right)d\phi\\\\ &=\sqrt{3}k \end{align}

You drew a good picture of the region of integration, though it might be helpful to you if you make your drawing larger. On a larger drawing it's easier to add details like these:

The possible values of $\theta$ for integration will range from the angle of the ray $\overrightarrow{OA}$ to the angle of $\overrightarrow{OB}$. We know that the length $OA = OB = 1$, and it should be clear by symmetry that the $y$-coordinate of $A$ is $\frac12,$ and the same for $B$. The ray $\overrightarrow{OA}$ therefore has $\theta = \arcsin \frac12,$ and you should recognize that $\arcsin \frac12 = \frac\pi6.$ The ray $\overrightarrow{OB}$ is in a symmetric position with $\theta = \pi - \frac\pi6.$

Within that range of $\theta$ values, consider an arbitrary line $\overleftrightarrow{OP}$. The radius of any point on that line within in the region of integration is at least $1$, but no greater than the distance $OP$. That distance is $r$ for a point with polar coordinates $(r,\theta)$ on the circle $x^2 + y^2 = 2y.$

Subsituting $x^2 + y^2 = r^2$ and $y = r\sin\theta,$ the equation of the circle is $r^2 = 2 r\sin\theta,$ or more simply $r = 2 \sin\theta.$ You should now have all you need to set up the boundaries of the double integral.