An ellipsoid has radii with lengths of #2 #, #1 #, and #6 #. 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?

To find the remaining volume of the ellipsoid after removing a hemisphere portion, we need to calculate the volume of the original 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, 1, and 6, respectively, we substitute these values into the formula:

[ V_{\text{ellipsoid}} = \frac{4}{3} \pi \times 2 \times 1 \times 6 = 16 \pi ]

The formula for the volume of a hemisphere is:

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

where ( r ) is the radius of the hemisphere.

Given that the radius of the hemisphere removed from the ellipsoid is 2, we substitute this value into the formula:

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

To find the remaining volume of the ellipsoid, we subtract the volume of the hemisphere from the volume of the ellipsoid:

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

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