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

Answer 1

#color(blue)("volume"=(32pi)/3" units"^3)#

An ellipsoid's volume can be calculated using:

#V=4/3pi*a*b*c#
Where #a, b, c# are the radii of the ellipsoid.

A hemisphere's volume is equal to half that of a sphere:

A sphere's volume is:

#V=4/3pir^3#

Thus, the hemisphere's volume is:

#V=2/3pir^3#

Simply calculate the ellipsoid's volume and deduct the hemisphere's volume to determine the ellipsoid's volume after the hemisphere is removed:

#V=4/3pi*a*b*c-2/3pir^3#
#V=2/3pi(2abc-r^3)#

Changing the values:

#V=2/3pi(2*2*2*3-2^3)#
#V=2/3pi(16)#
#V=(32pi)/3#
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Answer 2

To find the remaining volume of the ellipsoid after removing a hemisphere portion, we need to calculate the volume of the ellipsoid and then subtract the volume of the removed hemisphere.

The formula for the volume of an ellipsoid is given by:

[ V = \frac{4}{3} \pi abc ]

where ( a ), ( b ), and ( c ) are the semi-axes lengths of the ellipsoid.

Given that the semi-axes lengths of the ellipsoid are 2, 2, and 3, respectively, we have:

[ a = 2, \quad b = 2, \quad c = 3 ]

Substitute these values into the formula for the volume of the ellipsoid:

[ V_{\text{ellipsoid}} = \frac{4}{3} \pi (2)(2)(3) ] [ V_{\text{ellipsoid}} = \frac{4}{3} \pi \times 12 ] [ V_{\text{ellipsoid}} = 16 \pi ]

Now, we need to find the volume of the removed hemisphere. The formula for the volume of a hemisphere is:

[ V_{\text{hemisphere}} = \frac{2}{3} \pi r^3 ]

Given that the radius of the hemisphere is 2, we have:

[ r = 2 ]

Substitute this value into the formula for the volume of the hemisphere:

[ V_{\text{hemisphere}} = \frac{2}{3} \pi (2)^3 ] [ V_{\text{hemisphere}} = \frac{2}{3} \pi \times 8 ] [ V_{\text{hemisphere}} = \frac{16}{3} \pi ]

Now, to find the remaining volume of the ellipsoid, we subtract the volume of the removed hemisphere from the volume of the ellipsoid:

[ \text{Remaining Volume} = V_{\text{ellipsoid}} - V_{\text{hemisphere}} ] [ \text{Remaining Volume} = 16 \pi - \frac{16}{3} \pi ] [ \text{Remaining Volume} = \frac{48}{3} \pi - \frac{16}{3} \pi ] [ \text{Remaining Volume} = \frac{32}{3} \pi ]

Therefore, the remaining volume of the ellipsoid after removing the hemisphere portion is ( \frac{32}{3} \pi ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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