# An ellipsoid has radii with lengths of #4 #, #5 #, and #7 #. A portion the size of a hemisphere with a radius of #5 # is removed form the ellipsoid. What is the remaining volume of the ellipsoid?

The volume of the ellipsoid is

The hemisphere's volume is

The amount of volume left is

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The volume of an ellipsoid with radii (a), (b), and (c) is given by the formula:

[ V_{\text{ellipsoid}} = \frac{4}{3}\pi abc ]

For the given ellipsoid with radii (a = 4), (b = 5), and (c = 7), the volume is:

[ V_{\text{ellipsoid}} = \frac{4}{3}\pi (4)(5)(7) = \frac{4}{3}\pi (140) = \frac{560}{3}\pi ]

The volume of a sphere is given by:

[ V_{\text{sphere}} = \frac{4}{3}\pi r^3 ]

The volume of a hemisphere (half of a sphere) with radius (r = 5) is half of the sphere's volume, so:

[ V_{\text{hemisphere}} = \frac{1}{2} \left( \frac{4}{3}\pi (5)^3 \right) = \frac{1}{2} \left( \frac{4}{3}\pi (125) \right) = \frac{250}{3}\pi ]

To find the remaining volume of the ellipsoid after the hemisphere has been removed, subtract the volume of the hemisphere from the volume of the ellipsoid:

[ V_{\text{remaining}} = V_{\text{ellipsoid}} - V_{\text{hemisphere}} = \frac{560}{3}\pi - \frac{250}{3}\pi = \frac{310}{3}\pi ]

Therefore, the remaining volume of the ellipsoid, after removing a portion the size of a hemisphere with radius 5, is ( \frac{310}{3}\pi ) cubic units.

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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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