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?

Question

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?

Evelyn Arboleda · Accepted Answer

The remaining volume is #=2054.6u^3# Volume of ellipsoid is #=4/3piabc# Volume of henisphere is #=2/3pir^3# Remaining volume #=4/3piabc-2/3pir^3# #=2/3pi(2abc-r^3)# #=2/3pi(2*8*9*7-3^3)# #=2/3pi(1008-27)# #=2/3pi*981# #=2054.6#

Autumn Armstrong · Answer

undefined 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 $672\pi - 9\pi\sqrt{10}$.