Question helpa. a hemispherical bowl of radius a contains water to a depth h. find the volume of the water in the bowl.

The volume of the hemisphere is:
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" \frac{2}{3} [/tex] [tex] \pi [/tex] * h³ " ; 
 
or; write as:  " [tex] \frac{2 \pi h^3}{3} [/tex] " .
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Explanation:
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The formula/equation for the volume, "V", of a sphere:

V = ([tex] \frac{4}{3}[/tex]) * [tex] \pi [/tex] * r³ ;

The depth; or, height, "h"; would represent the "radius", "r" .

So, the volume would be:

V = [tex] \frac{1}{2} [/tex] * [tex] \frac{4}{3} [/tex] * [tex] \pi [/tex] * h³ ;
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Note:  " [tex] \frac{1}{2} [/tex] * [tex] \frac{4}{3} [/tex] " ; 

                 = [tex] \frac{(1*4)}{(2*3)} = \frac{4}{6} = \frac{(4/2)}{(6/2)} = \frac{2}{3} [/tex] ;
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The volume of the hemisphere is:

" \frac{2}{3} [/tex] [tex] \pi [/tex] * h³ . "

or; write as:  " [tex] \frac{2 \pi h^3}{3} [/tex] " .
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