An ellipsoid has radii with lengths of $3$ $3$ , $6$ $6$ , and $5$ $5$ . A portion the size of a hemisphere with a radius of $3$ $3$ is removed form the ellipsoid. What is the volume of the remaining ellipsoid?

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Answer 1 · 2018-08-07T08:50:13

The volume remaining is $= 320.4 u^{3}$ $=320.4u^3$

The volume of an ellipsoid is

$V_{e} = \frac{4}{3} π a b c$ $V_e=4/3piabc$

The volume of a hemisphere is

$V_{h} = \frac{1}{2} \cdot \frac{4}{3} π r^{3} = \frac{2}{3} π r^{3}$ $V_h=1/2*4/3pir^3=2/3pir^3$

Here,

$a = 3$ $a=3$

$b = 6$ $b=6$

$c = 5$ $c=5$

$r = 3$ $r=3$

The remaining volume is

$V_{r} = V_{e} - V_{h}$ $V_r=V_e-V_h$

$= \frac{4}{3} π a b c - \frac{2}{3} π r^{3}$ $=4/3piabc-2/3pir^3$

$= \frac{2}{3} π (2 a b c - r^{3})$ $=2/3pi(2abc-r^3)$

$= \frac{2}{3} π (2 \cdot 3 \cdot 6 \cdot 5 - 3^{3})$ $=2/3pi(2*3*6*5-3^3)$

$= \frac{2}{3} π \cdot (180 - 27)$ $=2/3pi*(180-27)$

$= \frac{2 \cdot 153}{3} π$ $=(2*153)/3pi$

$= 320.4 u^{3}$ $=320.4u^3$

An ellipsoid has radii with lengths of 33 , 66 , and 55 . A portion the size of a hemisphere with a radius of 33 is removed form the ellipsoid. What is the volume of the remaining ellipsoid?