# Find Center of Mass for Double Integral in Region R

• CaptainEvil
In summary, the problem involves finding the center of mass of a lamina with a density that is inversely proportional to its distance from the origin. The lamina is located inside a circle centered at (0,1) and outside a circle centered at (0,0), with both circles having a radius of 1. The double integral to solve for the center of mass can be simplified by converting to polar coordinates.
CaptainEvil

## Homework Statement

A lamina occupies the region inside the circle x^2 + y^2 = 2y but outside the circle x^2 + y^2 = 1. Find the center of mass if the density at any point is inversely proportional to its distance from the origin.

## Homework Equations

Xcm = double integral of y*f(x,y)
Ycm = double integral of x*f(x,y)

## The Attempt at a Solution

I know my integrand will be k(constant)/r
and I know how to solve for centre of mass, my only trouble is setting up the double integral. I think these are circles centred at 2 and 0, and when I draw them in cartesian coords, my region R is the space in between them that looks like a crescent moon. I don't know how to set up my double integral, because I can't express the first circle in terms of y, since there are two terms of them. Please help?

The first circle is centered at (0, 1), which you can see by completing the square. The second circle is centered at (0, 0). The radius of both circles is 1.

Because of the symmetry of the lamina and of the density function, the center of mass will be on the y-axis, so you only need to find the mass of the region and the moment about the x-axis to find Ycm. The double integral looks like this:
$$Mx =\int \int_A x \frac{k}{\sqrt{x^2 + y^2}} dA$$

Now you need to describe the region A. Whether you have vertical strips that sweep left to right, or horizontal strips that sweep from bottom to top, things are complicated by the fact that the left and right boundaries change with y and the lower and upper boundaries change with x.

This problem is really ripe for changing to polar coordinates, it seems to me, since the region A has a somewhat simpler description in polar coordinates, and your density function is just k/r. The lower circle is just r = 1. The upper circle is a bit more complicated, but you can convert with the formulas
x = r cos(theta)
y = r sin(theta)
r^2 = x^2 + y^2

The only remaining matter is that dA = dx dy = dy dx in Cartesian coordinates, but dA = r dr dtheta in polar coordinates.

Hope that helps.

## 1. How is the center of mass defined?

The center of mass is defined as the point at which the mass of an object or system is evenly distributed in all directions. It is also known as the balance point or centroid.

## 2. What is a double integral?

A double integral is a mathematical concept used to find the volume under a surface in three-dimensional space. It involves integrating a function over a specific region in the x-y plane.

## 3. Why is finding the center of mass important?

Finding the center of mass is important in many scientific fields, including physics, engineering, and astronomy. It helps in understanding the balance and stability of objects, as well as predicting their motion and behavior.

## 4. How is the center of mass calculated using a double integral?

The center of mass for a two-dimensional region R can be calculated by first finding the mass of the region and then finding the weighted average of the x and y coordinates using the double integral formula: x̄ = (1/M)∬xρ(x,y)dA and ȳ = (1/M)∬yρ(x,y)dA, where ρ(x,y) is the mass density function.

## 5. What are some practical applications of finding the center of mass?

Finding the center of mass is used in various real-world applications, such as designing stable structures, calculating the distribution of weight in aircrafts and vehicles, determining the center of gravity of objects, and predicting the motion of celestial bodies.

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