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

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Answer 1 · 2017-04-12T06:09:00

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

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

$V_{E} = \frac{4}{3} π a b c$ $V_E=4/3piabc$

In the given case:

$V_{E} = \frac{4}{3} π \times 12 \times 11 \times 8$ $V_E=4/3pixx12xx11xx8$

The formula for volume of a hemisphere is:

$V_{H} = \frac{2}{3} π r^{3}$ $V_H=2/3pir^3$

In the given case:

$V_{H} = \frac{2}{3} π \times 9^{3}$ $V_H=2/3pixx9^3$

$V_{H} = \frac{2}{3} π \times 729$ $V_H=2/3pixx729$

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

$V_{E} - V_{H} = (\frac{4}{3} π \times 12 \times 11 \times 8) - (\frac{2}{3} π \times 729)$ $V_E-V_H=(4/3pixx12xx11xx8)-(2/3pixx729)$

Simplify the brackets.

$V_E-V_H=(4/cancel3pixx4cancel12xx11xx8)-(2/cancel3pixx243cancel729)$

$V_E-V_H=(4pixx4xx11xx8)-(2pixx243)$

$V_E-V_H=1408pi-486pi$

$V_E-V_H=(1408-486)pi$

$V_E-V_H=922pi$

Considering $pi$ as $22/7$ , we get:

$V_E-V_H=922xx22/7$

$V_E-V_H=2897.71$

An ellipsoid has radii with lengths of 12 1212 , 11 1111 , and 8 88 . A portion the size of a hemisphere with a radius of 9 99 is removed from the ellipsoid. What is the remaining volume of the ellipsoid?