An ellipsoid has radii with lengths of $3$ $3$ , $6$ $6$ , and $1$ $1$ . A portion the size of a hemisphere with a radius of $2$ $2$ is removed form the ellipsoid. What is the volume of the remaining ellipsoid?

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An ellipsoid has radii with lengths of $3$ $3$ , $6$ $6$ , and $1$ $1$ . A portion the size of a hemisphere with a radius of $2$ $2$ is removed form the ellipsoid. What is the volume of the remaining ellipsoid?

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Answer 1 · 2016-07-23T02:44:56

$\frac{56}{3} π$ $56/3 pi$

Explanation:

The volume of the original ellipsoid was $\frac{4}{3} π \times 3 \times 6 \times 1 = 24 π$ $4/3 pi xx 3 xx 6 xx 1 =24 pi$
The volume taken away is $\frac{1}{2} \times \frac{4}{3} π \times 2^{3} = \frac{16}{3} π$ $1/2 xx 4/3 pi xx 2^3 = 16/3 pi$
Thus the volume remaining is $24 π - \frac{16}{3} π = \frac{56}{3} π$ $24 pi - 16/3 pi = 56/3 pi$

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An ellipsoid has radii with lengths of $3$ $3$ , $6$ $6$ , and $1$ $1$ . A portion the size of a hemisphere with a radius of $2$ $2$ is removed form the ellipsoid. What is the volume of the remaining ellipsoid?

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An ellipsoid has radii with lengths of 33 , 66 , and 11 . A portion the size of a hemisphere with a radius of 22 is removed form the ellipsoid. What is the volume of the remaining ellipsoid?

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An ellipsoid has radii with lengths of $3$ $3$ , $6$ $6$ , and $1$ $1$ . A portion the size of a hemisphere with a radius of $2$ $2$ is removed form the ellipsoid. What is the volume of the remaining ellipsoid?