Cfrgtkky

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!

For $m$, we have that

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

For the term $M_x$, we have that

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

For the term $M_y$, we have that

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

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 $arcsin frac12 = fracpi6.$
The ray $overrightarrow{OB}$ is in a symmetric position
with $theta = pi - fracpi6.$

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 $y = rsintheta,$
the equation of the circle is
$r^2 = 2 rsintheta,$
or more simply $r = 2 sintheta.$
You should now have all you need to set up the boundaries of the
double integral.