Rotational volume of an ellipse.

## Your answer

 Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: To avoid this verification in future, please log in or register.

## 1 Answer

To compute the volume we only need to consider the first quadrant of the ellipse. Then we simply double this volume.

We make up the quadrant volume by adding together the volumes of infinitesimally thin discs of height y and thickness dx. The volume of each disc is πy²dx. The total volume is the integral of this. The range for x is 0 to a, the length of the semimajor axis. y²=b²(1-x²/a²) so we need to integrate πb²(1-x²/a²)dx.

πb²(x-x³/(3a²)) is the result so we now put x=a the upper limit: πb²(a-a/3)=2πab²/3. (The lower limit where x=0 gives us zero.) Double this to get the volume of the ellipsoid: 4πab²/3.

by Top Rated User (763k points)

0 answers
1 answer
1 answer
0 answers
1 answer
1 answer
0 answers
1 answer
1 answer
1 answer