An ellipsoid has radii with lengths of #8 #, #6 #, and #5 #. A portion the size of a hemisphere with a radius of #2 # is removed form the ellipsoid. What is the volume of the remaining ellipsoid?

Question

An ellipsoid has radii with lengths of #8   #, #6  #, and #5  #. A portion the size of a hemisphere with a radius of #2  # is removed form the ellipsoid. What is the volume of the remaining ellipsoid?

James Atkins · Accepted Answer

#=988.54#

The Volume of an Ellipsoid with radii #=8,6 and 5#

#=pi/6times#(major-axis)#times#(minor axis)#times#(vertcal-axis)

#=pi/6(2times8)(2times5)(2times6)#

#=pi/6(16times10times12)#

#=320pi#

#=1005.3#

Volume of an Hemisphere#=2/3(pir^3)# where #r=2# is the radius #=2/3pi(2)^3#

#=2/3pitimes8#

#=16/3pi#

#=16.76#

Thus, the Ellipsoid's residual volume

#=1005.3-16.76#

#=988.54#

Caroline Aucoin · Answer

undefined To find the volume of the remaining ellipsoid after removing a portion in the shape of a hemisphere, you can calculate the volume of the ellipsoid and then subtract the volume of the removed hemisphere.

The formula for the volume of an ellipsoid is:

$$ V = \frac{4}{3} \pi abc $$

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

Given that the semi-axes lengths of the ellipsoid are $ a = 8, b = 6, $ and $ c = 5 $, we can calculate its volume using the formula:

$$ V_{\text{ellipsoid}} = \frac{4}{3} \pi (8)(6)(5) $$

$$ V_{\text{ellipsoid}} = \frac{4}{3} \pi (240) $$

$$ V_{\text{ellipsoid}} = 320 \pi $$

The volume of a hemisphere can be calculated using the formula:

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

Given that the radius of the removed hemisphere is $ r = 2 $, we can calculate its volume:

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

$$ V_{\text{hemisphere}} = \frac{16}{3} \pi $$

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

$$ V_{\text{remaining}} = V_{\text{ellipsoid}} - V_{\text{hemisphere}} $$

$$ V_{\text{remaining}} = 320 \pi - \frac{16}{3} \pi $$

$$ V_{\text{remaining}} = \frac{960}{3} \pi - \frac{16}{3} \pi $$

$$ V_{\text{remaining}} = \frac{944}{3} \pi $$

Therefore, the volume of the remaining ellipsoid after removing the portion in the shape of a hemisphere is $ \frac{944}{3} \pi $.