Question 1208389
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The water in a hemi-spherical bowl is 42 cm across the top {{{highlight(cross(is))}}} <U>and</U> 9 cm deep. 
How much more water is needed to fill the bowl to the brim?
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;In his post, &nbsp;Edwin instructs to use the radius of the sphere &nbsp;R = 21 cm.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;It is a strategic error. &nbsp;It shows that &nbsp;Edwin misread the problem.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The radius of the sphere is not given in this problem.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;In this problem, "42 cm across the top" means "42 cm across the water surface".

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Finding the hemi-sphere radius is the first step of the solution.



<pre>
If you make a sketch, you will see a right angled triangle.

Its hypotenuse is the radius of the sphere R.

Its legs are  (R-9) cm and 42/2 = 21 cm.

So, we write the Pythagorean equation

    (R-9)^2 + 21^2 = R^2.


From it, we find

    R^2 - 18R + 81 + 441 = R^2,

    81 + 441 = 18R

    18R = 522

      R = 522/18 = 29.


Thus the sphere radius is 29 cm.


Then we find the volume of the hemi-sphere

    {{{V[hemi-sphere]}}} = {{{(1/2)*(4/3)^pi*R^3}}} = {{{(2/3)*3.14159265*29^3}}} = 51080.202 cm^3.


Next you find the volume of the spherical segment for R = 29 cm and h = 9 cm
using the formula

    {{{V[segment]}}} = {{{(1/3)*pi*h^2*(3R-h)}}} = {{{(1/3)*3.14159265*9^2*(3*29-9)}}} = 6616.194 cm^3.


Now your answer is the difference of the two volumes

    {{{V[hemi-sphere]}}} - {{{V[segment]}}} = 51080.202 - 6616.194 = 44464.008 cm^3,  or 44.464 liters  (rounded).
</pre>

Solved.