# An ellipsoid has radii with lengths of 5 , 5 , and 8 . A portion the size of a hemisphere with a radius of 3  is removed form the ellipsoid. What is the remaining volume of the ellipsoid?

Feb 22, 2018

see a solution step below;

#### Explanation:

Please note that we are dealing with an Ellipsoid and Hemisphere..

The formula for the volume of an ellipsoid is given below;

$\text{Volume of an ellipsoid} \Rightarrow {V}_{e} = \frac{4}{3} \pi a b c$

The formula for the volume of a hemisphere is given below;

$\text{Volume of a hemisphere} \Rightarrow {V}_{h} = \frac{2}{3} \pi {r}^{3}$

Since we are removing a portion of a Hemisphere from an Ellipsoid, it means that we need to subtract the volume of a hemisphere from the ellipsoid which is given below;

$\text{Remaining volume of an ellipsoid} \textcolor{w h i t e}{x} {R}_{v e} = {V}_{e} - {V}_{h}$

${R}_{v e} = {V}_{e} - {V}_{h}$

Where;

${R}_{v e} = \text{Remaining volume of an ellipsoid}$

${V}_{e} = \text{Volume of an ellipsoid}$

${V}_{h} = \text{Volume of a hemisphere}$

$a , b , c = \text{lengths}$

$r = \text{radius}$

${R}_{v e} = {V}_{e} - {V}_{h}$

${V}_{e} = \frac{4}{3} \pi a b c$

${V}_{h} = \frac{2}{3} \pi {r}^{3}$

$a , b , c = 5 , 5 , 8 \text{respectively}$

$r = 3$

$\pi = 3.142$

Hence substituting the values into the formula;

${R}_{v e} = {V}_{e} - {V}_{h}$

${R}_{v e} = \frac{4}{3} \pi \left(5 \times 5 \times 8\right) - \frac{2}{3} \pi {\left(3\right)}^{3}$

${R}_{v e} = \frac{4}{3} \pi \left(200\right) - \frac{2}{3} \pi \left(27\right)$

${R}_{v e} = \frac{4 \left(200\right)}{3} \pi - \frac{2 \left(27\right)}{3} \pi$

${R}_{v e} = \frac{800}{3} \pi - \frac{54}{3} \pi$

${R}_{v e} = \frac{800 - 54}{3} \pi$

${R}_{v e} = \frac{746}{3} \pi$

${R}_{v e} = 248.67 \pi$

${R}_{v e} = 248.67 \times 3.142$

${R}_{v e} = 781.31 c {m}^{3}$

Therefore the remaining volume of the ellipsoid is $781.31 c {m}^{3}$