Expert: Paul Klarreich - 2/21/2007

Question
My teacher and I approached this problem two different ways that we both feel SHOULD work, but which give different answers. A spherical snowball is packed most densely near the center, and its density x cm from the center snowball (in g/cm^3) is given by p(x)=1/(1+x^.5).  The diameter is 12cm.  I suggested using the average value theorem to find the average density throughout the entire snowball, then multiplying that number by volume, 4/3*pi*x^3.  He went about the problem by integrating the function p(x) times volume given by the surface area of each shell times average thickness of each shell, or integrating p(x)*4*pi*x^2*dx.  The two methods give respective answers of 365g and 295g.  Which method (if either) is correct and why isn't the other?  Thank you.

Answer
Questioner:   andy lane
Category:  Calculus
Private:  No
 
Subject:  snowball application
Question:  My teacher and I approached this problem two different ways that we both

feel SHOULD work, but which give different answers. A spherical snowball is packed

most densely near the center, and its density x cm from the center snowball (in g/cm^3) is given by p(x)=1/(1+x^.5).  The diameter is 12cm.  I suggested using the average value theorem

>> Sorry.  Exactly what theorem is that?  How did you compute the average density?
Did you average it over every cubical volume element?
>> ..........
to find the average density throughout the entire snowball, then multiplying that number by volume, 4/3*pi*x^3.  He went about the problem by integrating the function p(x) times volume given by the surface area of each shell times average thickness of each shell, or integrating p(x)*4*pi*x^2*dx.  The two methods give respective answers of 365g and 295g.  Which method (if either) is correct and why isn't the other?  

Thank you.
......................................................
Hi,  Andy,

Sorry, but I think I like HIS solution.  Each spherical shell has volume equal to its surface area (4 pi r^2) times its thickness, which is going to be  dr, times its density, which will be  p(x).  

[I am using r for the distance from the center, rather than x.]

dV = 4 pi x^2 p(x)dx  -- that's his volume element, as you wrote.

Integrating:

{ 4 pi x^2 dx
| -----------
} 1 + sqrt(x)

which does come out around 295.  

If you computed average density just along the line x = 0 to x = 6, then you assumed that values all count the same, but they don't.  Values nearer to the outside count more -- there are more 'little cubes' near the outside, so the values near x = 6 must be 'weighted' more than the values near x = 6.