# How do you find the center of mass if the density at any point is inversely proportional to its distance from the origin of a lamina that occupies the region inside the circle x^2 + y^2 = 10y but outside the circle x^2+y^2=25?

Jan 10, 2017

The enter of mass is $\left(0 , 2.5\right)$

#### Explanation:

Consider the first equation:

${x}^{2} + {y}^{2} = 10 y$

We can put this into standard from by completing the square:

${x}^{2} + {y}^{2} - 10 y = 0$
$\therefore {x}^{2} + {\left(y - 5\right)}^{2} - {5}^{2} = 0$
$\therefore {x}^{2} + {\left(y - 5\right)}^{2} = {5}^{2}$

Which is a circle of radius $5$ and centre $\left(0 , 5\right)$, And now the second equation:

${x}^{2} + {y}^{2} = 25$

Which is a circle of radius $5$ and centre $\left(0 , 0\right)$

We can plot these curves;

For a more complex problem we would need to use integration to find the Centre of Mass, but because of symmetry the density will be evenly distributed about the lines $x = 0$ and $y = 2.5$, these being the lines of symmetry.

Hence the enter of mass is $\left(0 , 2.5\right)$