# An ellipsoid has radii with lengths of 12 , 11 , and 8 . A portion the size of a hemisphere with a radius of 9  is removed from the ellipsoid. What is the remaining volume of the ellipsoid?

Apr 12, 2017

The remaining volume is $922 \pi$ or $2897.71$.

#### Explanation:

The formula for the volume of an ellipsoid where the three radii are represented by $a$, $b$, and $c$, is:

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

In the given case:

${V}_{E} = \frac{4}{3} \pi \times 12 \times 11 \times 8$

The formula for volume of a hemisphere is:

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

In the given case:

${V}_{H} = \frac{2}{3} \pi \times {9}^{3}$

${V}_{H} = \frac{2}{3} \pi \times 729$

We need to determine the volume of the ellipsoid when the hemisphere is removed from it, which is;

${V}_{E} - {V}_{H} = \left(\frac{4}{3} \pi \times 12 \times 11 \times 8\right) - \left(\frac{2}{3} \pi \times 729\right)$

Simplify the brackets.

${V}_{E} - {V}_{H} = \left(\frac{4}{\cancel{3}} \pi \times 4 \cancel{12} \times 11 \times 8\right) - \left(\frac{2}{\cancel{3}} \pi \times 243 \cancel{729}\right)$

${V}_{E} - {V}_{H} = \left(4 \pi \times 4 \times 11 \times 8\right) - \left(2 \pi \times 243\right)$

${V}_{E} - {V}_{H} = 1408 \pi - 486 \pi$

${V}_{E} - {V}_{H} = \left(1408 - 486\right) \pi$

${V}_{E} - {V}_{H} = 922 \pi$

Considering $\pi$ as $\frac{22}{7}$, we get:

${V}_{E} - {V}_{H} = 922 \times \frac{22}{7}$

${V}_{E} - {V}_{H} = 2897.71$