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

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

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Answer 1 · 2018-02-22T09:47:19

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;

$Volume of an ellipsoid \Rightarrow V_{e} = \frac{4}{3} π a b c$ $"Volume of an ellipsoid" rArr V_e = 4/3piabc$

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

$Volume of a hemisphere \Rightarrow V_{h} = \frac{2}{3} π r^{3}$ $"Volume of a hemisphere" rArr V_h = 2/3pir^3$

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;

$Remaining volume of an ellipsoid x R_{v e} = V_{e} - V_{h}$ $"Remaining volume of an ellipsoid" color(white)x R_(ve) = V_e - V_h$

$R_{v e} = V_{e} - V_{h}$ $R_(ve) = V_e - V_h$

Where;

$R_{v e} = Remaining volume of an ellipsoid$ $R_(ve) = "Remaining volume of an ellipsoid"$

$V_{e} = Volume of an ellipsoid$ $V_e = "Volume of an ellipsoid"$

$V_{h} = Volume of a hemisphere$ $V_h = "Volume of a hemisphere"$

$a, b, c = lengths$ $a, b, c = "lengths"$

$r = radius$ $r = "radius"$

$R_{v e} = V_{e} - V_{h}$ $R_(ve) = V_e - V_h$

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

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

$a, b, c = 5, 5, 8 respectively$ $a, b, c = 5, 5, 8 "respectively"$

$r = 3$ $r = 3$

$π = 3.142$ $pi = 3.142$

Hence substituting the values into the formula;

$R_{v e} = V_{e} - V_{h}$ $R_(ve) = V_e - V_h$

$R_{v e} = \frac{4}{3} π (5 \times 5 \times 8) - \frac{2}{3} π {(3)}^{3}$ $R_(ve) = 4/3pi(5 xx 5 xx 8)- 2/3pi(3)^3$

$R_{v e} = \frac{4}{3} π (200) - \frac{2}{3} π (27)$ $R_(ve) = 4/3pi(200)- 2/3pi(27)$

$R_{v e} = \frac{4 (200)}{3} π - \frac{2 (27)}{3} π$ $R_(ve) = (4(200))/3 pi - (2(27))/3 pi$

$R_{v e} = \frac{800}{3} π - \frac{54}{3} π$ $R_(ve) = 800/3 pi - 54/3 pi$

$R_{v e} = \frac{800 - 54}{3} π$ $R_(ve) = (800 - 54)/3 pi$

$R_{v e} = \frac{746}{3} π$ $R_(ve) = 746/3 pi$

$R_{v e} = 248.67 π$ $R_(ve) = 248.67 pi$

$R_{v e} = 248.67 \times 3.142$ $R_(ve) = 248.67 xx 3.142$

$R_{v e} = 781.31 c m^{3}$ $R_(ve) = 781.31cm^3$

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

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

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Explanation:

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