Question 1189060
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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(hallow)}}} <U>HOLLOW</U> sphere of external radius 15cm.
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The solution by &nbsp;@boreal is incorrect: &nbsp;his calculations are &nbsp;WRONG.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;I came to bring a correct solution/answer.



<pre>
The volume of the first sphere is {{{(4/3)*pi*8^3}}} = {{{(4/3)*512*pi}}} cm^3.

the volume of the second sphere is {{{(4/3)*pi*13^3}}} = {{{(4/3)*2197*pi}}} cm^3.


The total metal volume is their sum or {{{(4/3)*2709*pi}}} cm^3.


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

    {{{(4/3)*pi*15^3}}} - {{{(4/3)*pi*r^3}}} = {{{(4/3)*2709*pi}}}.


Cancel the common factors {{{(4/3)*pi}}}  in all the terms of the equation.  You will get then this equation

    {{{15^3}}} - {{{r^3}}} = 2709

    {{{r^3}}} = 3375 - 2709

    {{{r^3}}} = 666

    r = {{{root(3,666)}}} = 8.733 cm  (rounded).


<U>ANSWER</U>.  The thickness of the spherical shell is this difference  15 cm - 8.733 cm = 6.267 cm.
</pre>

Solved (correctly).