SOLUTION: Find the thickness of the spherical shell. Two metal spheres of radii 8cm and 13cm , respectively are melted down and a cast into a hallow sphere of external radius 15cm.

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Question 1189060: Find the thickness of the spherical shell. Two metal spheres of radii 8cm and 13cm , respectively are melted down and a cast into a hallow sphere of external radius 15cm.
Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
The volume of the first is pi*8^3=512 π; the second is π*13^3=2197 π units cm^3.
The total volume is their sum or 2709 π cm^3.
The external radius of the resulting sphere has area 4πr^2 or 900 π cm^2, since r^2=225 cm^2
The thickness is the total volume of 2709π cm^3/900π cm^2
or 3.01 cm.

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
Find the thickness of the spherical shell. Two metal spheres of radii 8cm and 13cm ,
respectively are melted down and a cast into a cross%28hallow%29 HOLLOW sphere of external radius 15cm.
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            The solution by  @boreal is incorrect:  his calculations are  WRONG.

            I came to bring a correct solution/answer. 


The volume of the first sphere is %284%2F3%29%2Api%2A8%5E3 = %284%2F3%29%2A512%2Api cm^3.

the volume of the second sphere is %284%2F3%29%2Api%2A13%5E3 = %284%2F3%29%2A2197%2Api cm^3.


The total metal volume is their sum or %284%2F3%29%2A2709%2Api cm^3.


We should find the radius  "r"  of the spherical hollow part from the equation

    %284%2F3%29%2Api%2A15%5E3 - %284%2F3%29%2Api%2Ar%5E3 = %284%2F3%29%2A2709%2Api.


Cancel the common factors %284%2F3%29%2Api  in all the terms of the equation.  You will get then this equation

    15%5E3 - r%5E3 = 2709

    r%5E3 = 3375 - 2709

    r%5E3 = 666

    r = root%283%2C666%29 = 8.733 cm  (rounded).


ANSWER.  The thickness of the spherical shell is this difference  15 cm - 8.733 cm = 6.267 cm.

Solved (correctly).