Calculus, center of mass

[retroglam]

Homework Statement

Find the center of mass of the 2-dimensional plate which occupies the region inside the circle x^2 + y^2 = 2y, but outside the circle x^2 + y^2 = 1, and for which the density is inversely proportional to its distance from the origin.

Homework Equations

m = integral (p(x,y)) dA

The Attempt at a Solution

-> m = integral k/r
x^2 + (y-1)^2 = 1

It's not the integration itself I'm having trouble with just finding the boundaries of integration as they are between the two circles, any help/hints would be appreciated.
Thanks!

 

[Tedjn]

From looking at a graph of the two circles, it appears that the two intersection points occur between the upper half of the circle x² + y² = 1 and the lower half of the circle x² + (y-1)² = 1. That is, between the functions

$$y = \sqrt{1-x^2} \qquad \text{ and } \qquad y = 1-\sqrt{1-x^2}.$$​

From here you can find the intersection points.

 

[HallsofIvy]

First, do you see, from symmetry, that $\overline{x}= 0$?

The region you seek lies outside the unit circle but inside the circle with center at (0, 1) and radius 1. Those intersect where $x^2+ (y-1)^2= 1= x^2+ y^2$ or $y^2- 2y+ 1= y^2$ so $y= 1/2$. Then $x= \pm\sqrt{3}/2$.

There is a slight complication in that the circle actually extends to x=-1 and x= 1 so it is probably best to do this in three separate integrals.

1) x from -1 to $-\sqrt{3}/2$, y from $1- \sqrt{1- x^2}$ to $1+ \sqrt{1- x^2}$ (from solving $x^2+ (y- 1)^2= 1$ for y).

2) x from $-\sqrt{3}/2$, y from $\sqrt{1- x^2}$ (the lower circle) to $1+ \sqrt{1- x^2}$

3) x from $\sqrt{3}/2$ to 1, y from $1- \sqrt{1- x^2}$ to $1+ \sqrt{1- x^2}$.