An ellipsoid has radii with lengths of #3 #, #4 #, and #2 #. 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?

Question

An ellipsoid has radii with lengths of #3   #, #4  #, and #2  #. 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?

Wyatt Anderson · Accepted Answer

Remaining volume of the ellipsoid is #(80pi)/3# units.

Volume of an ellipsoid with dimensions #a#, #b# and #c# is given by #4/3pixxaxxbxxc#, hence volume of ellipsoid of radii with lengths of #3#, #4# and #2# is #4/3pixx4xx3xx2=32pi#. Now as volume of sphere is given by #4/3pir^3#, volume of hemisphere of radius #2# cut from this is #1/2xx4/3pixx2^3=2/3xx8pi=(16pi)/3# Hence, remaining volume of the ellipsoid is #32pi-(16pi)/3=(80pi)/3#

Allison Allen · Answer

undefined To find the remaining volume of the ellipsoid after removing a hemisphere with a radius of 2, we first calculate the volume of the ellipsoid and then subtract the volume of the hemisphere.

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

$$ V_{\text{ellipsoid}} = \frac{4}{3} \pi a b c $$

where $ a $, $ b $, and $ c $ are the semi-axes of the ellipsoid.

Given that the lengths of the radii of the ellipsoid are 3, 4, and 2, we have:

$ a = 3 $, $ b = 4 $, and $ c = 2 $.

Substitute these values into the formula to find the volume of the ellipsoid:

$$ V_{\text{ellipsoid}} = \frac{4}{3} \pi (3)(4)(2) = \frac{32}{3} \pi $$

Next, we calculate the volume of the hemisphere using the formula:

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

where $ r = 2 $ (radius of the hemisphere).

Substitute $ r = 2 $ into the formula to find the volume of the hemisphere:

$$ V_{\text{hemisphere}} = \frac{2}{3} \pi (2)^3 = \frac{16}{3} \pi $$

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

$$ \text{Remaining volume} = V_{\text{ellipsoid}} - V_{\text{hemisphere}} $$
$$ = \frac{32}{3} \pi - \frac{16}{3} \pi $$
$$ = \frac{16}{3} \pi $$

So, the remaining volume of the ellipsoid after removing a hemisphere with a radius of 2 is $ \frac{16}{3} \pi $ cubic units.