An ellipsoid has radii with lengths of #6 #, #9 #, and #4 #. 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?

Answer 1

The remaining volume is: #V =614/3pi#

I'll provide this link for an ellipsoid's volume.

#V = 4/3piabc#
We are given #a = 6#, #b = 9#, and #c = 4#:
#V = 288pi#

I'll provide this source regarding a sphere's volume.

#V = 4/3pir^3#

Hemisphere requires a division of 2:

#V = 4/6pir^3#
We are given #r = 5#:
#V = 250/3pi#

The difference is the remaining volume:

#V =614/3pi#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To find the remaining volume of the ellipsoid after removing a hemisphere with a radius of 5, we can use the formula for the volume of an ellipsoid and subtract the volume of the removed hemisphere.

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

[ V = \frac{4}{3} \pi a b c ]

Where ( a ), ( b ), and ( c ) are the semi-axes of the ellipsoid.

Given that the semi-axes of the ellipsoid are 6, 9, and 4, respectively, the volume of the ellipsoid before removing the hemisphere is:

[ V_{\text{ellipsoid}} = \frac{4}{3} \pi \times 6 \times 9 \times 4 ]

To find the volume of the hemisphere, we use the formula for the volume of a sphere:

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

Where ( r ) is the radius of the hemisphere, given as 5.

So, the volume of the hemisphere to be removed is:

[ V_{\text{hemisphere}} = \frac{2}{3} \pi \times 5^3 ]

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}} ]

[ \text{Remaining volume} = \frac{4}{3} \pi \times 6 \times 9 \times 4 - \frac{2}{3} \pi \times 5^3 ]

[ \text{Remaining volume} = \frac{4}{3} \pi \times 6 \times 9 \times 4 - \frac{2}{3} \pi \times 125 ]

[ \text{Remaining volume} = \frac{4}{3} \pi \times 216 - \frac{2}{3} \pi \times 125 ]

[ \text{Remaining volume} = \frac{864}{3} \pi - \frac{250}{3} \pi ]

[ \text{Remaining volume} = \frac{614}{3} \pi ]

So, the remaining volume of the ellipsoid after removing the hemisphere is ( \frac{614}{3} \pi ) cubic units.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

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.

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.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7