Expert: Paul Klarreich - 9/12/2007

Question
Hello.
Can you help me please to find the integral of the center of mass of the area between this functions, but just in the first cuadrant

the functions are:

1=x^2 + y^2

4=x^2 + y^2

The problem i have is that i do not know how to evaluate the integral, because it is supouse to be evaluated between 0 and 2m but the first function i gave you is not defined from 1 to infinity, or am i wrong?

Thanks.

mp

Answer
Questioner:   Muricio
Category:  Advanced Math
Private:  No
 
Subject:  Center of mass
Question:  Hello.
Can you help me please to find the integral of the center of mass of the area between these functions, but just in the first quadrant

the functions are:

1 = x^2 + y^2

4 = x^2 + y^2

The problem I have is that I do not know how to evaluate the integral, because it is supposed to be evaluated between 0 and 2m but the first function i gave you is not defined from 1 to infinity, or am i wrong?

Thanks.

mp
............................
Hi, Muricio,

Your two functions are simply circles with radius 1 and 2:

1 = x^2 + y^2  >> circle with radius 1, center at origin.
4 = x^2 + y^2  >> circle with radius 2, center at origin.

I think this is best done using polar coordinates.  In that case, the equations are simply  r = 1 and r = 2.

In the first quadrant, the only issue is: how far from the origin is the center of mass?  It is certainly along the line  theta = pi/2.

So we take as our standard 'piece' a 'quarter-circle' that looks like this:

1. Take the circles with radius  r and  r + dr.  They define an 'annulus' or 'ring' whose area is:

dA = pi((r + dr)^2 - r^2)

dA = pi(r^2 + 2 r dr + dr^2 - r^2)

dA = pi(2 r dr + dr^2)

In the limit, as  dr -> 0, we can ignore the  dr^2 term, and have:

dA = 2 pi r dr

2. Chop out only the quadrant of that circle in the first quadrant.

dA = 2 pi r dr  / 4 = pi r dr / 2

Now integrate that from  r = 1 to r = 2:

{2
|   pi/2 r dr =
}1

pi r^2/4   from 1 to 2  =

(pi/4)( 2^2 - 1^2) = 3pi/4

OKAY, NOW.  That is the area of the figure.  To find the center of mass, we find the moment, which involves an extra factor of r:

{2
|   pi/2 r^2 dr =
}1


pi r^3/6   from 1 to 2  =

(pi/6)( 2^3 - 1^3) = 7pi/6

Finally, divide the moment by the area:

7pi/6    7*4    28     14
----- =  --- = ---- = ----
3pi/4    3*6    18      9

That's it, basically.  Your c.of.m is at  r = 14/9,  theta = pi/4.  In rectangular coordinates, that would be at:

x = y = 7 sqrt(2)/9