# An ellipsoid has radii with lengths of #8 #, #9 #, and #7 #. A portion the size of a hemisphere with a radius of #3 # is removed form the ellipsoid. What is the volume of the remaining ellipsoid?

The remaining volume is

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To find the volume of the remaining ellipsoid after removing a hemisphere, you can use the formula:

[V = \frac{4}{3} \pi abc - \frac{1}{2} \pi r^2 h]

Where:

- (a), (b), and (c) are the semi-axes of the ellipsoid.
- (r) is the radius of the removed hemisphere.
- (h) is the distance between the center of the ellipsoid and the center of the removed hemisphere.

Given:

- (a = 8), (b = 9), (c = 7)
- (r = 3)

First, we need to find (h), which is the distance between the center of the ellipsoid and the center of the removed hemisphere. By the Pythagorean theorem:

[h = \sqrt{c^2 - r^2}]

[h = \sqrt{7^2 - 3^2} = \sqrt{49 - 9} = \sqrt{40} = 2\sqrt{10}]

Now, substitute the values into the formula:

[V = \frac{4}{3} \pi (8)(9)(7) - \frac{1}{2} \pi (3^2)(2\sqrt{10})]

[V = \frac{4}{3} \pi (504) - \frac{1}{2} \pi (9)(2\sqrt{10})]

[V = \frac{4}{3} \pi (504) - 9\pi\sqrt{10}]

[V = 672\pi - 9\pi\sqrt{10}]

So, the volume of the remaining ellipsoid is (672\pi - 9\pi\sqrt{10}).

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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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