Calculus 3?

Question

O1: z = (x^2+y^2)^.5
O2: z = a, a constant and a>0
Define W to be the solid below O2 and above O1, and define the mass-density of W to be O: W-->R
o(x,y,z) = k(x^2+y^2+z^2)^.5, k a positive constant.
Find the mass of W, mass(W).

Answer 1

Note that
r^2 = x^2 + y^2
O1: z = r
O2: z = a, a constant and a>0
o(x,y,z) = k(r^2+z^2)^0.5, k a positive constant
Then
mass(W) = int[o(x,y,z) dV] over volume of the solid
Integrate over r and z with dV = 2pi r dr dz (a double integral)
mass(W) = int[J(z) dz)] from 0 to a
where
J(z) = int[k(r^2+z^2)^0.5 (2pi r dr)] from 0 to z (limit is from O1)
= 2pi*k*{[(r^2+z^2)^1.5/3 ]at r=z - [r^2+z^2)^1.5/3]at r=0}
= (2/3)pi*k*(2^1.5-1)*z^3
mass(W) = (2/3)pi*k*(2^1.5 -1)*int[(z^3 dz)] from 0 to a
= (2/3)pi*k*(2^1.5 -1)*a^4/4

mass(W) = (pi/6)*(2*sqrt(2) -1)*k*a^4 = 0.9574*k*a^4

I hope you can follow this. I have checked the integrals with Maple. It is awkward to do much math in Answers.

Answer 2

The volume is a conical section (if I interpret your problem properly). At any value of z, the equation of O1 is z^2 = x^2 + y^2, which is the equation of a circle of radius z. The density function is a sphere. It seems it would be easier to solve this in polar coordinates. Check this out

http://img102.imageshack.us/img102/7637/conemassve6.png

and see if that works for you.

EDIT: I have changed the file a bit for clarity.