Calculating Volume of an Ellipsoid Using Spherical Coordinates

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In summary, the problem is to find the volume of the ellipsoid x^2+y^2+5z^2=16. The solution involves changing to spherical coordinates, with the upper limit for p depending on whether x and y are zero or not. Alternatively, one can use cylindrical coordinates or a convenient parametrization and calculate the Jacobian for the triple integral.
  • #1
PsychonautQQ
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Homework Statement


find the volume of the ellipsoid x^2+y^2+5z^2=16



The Attempt at a Solution


so I assume I must first change it to spherical coordinates. I figure I evaluate θ from 0 to 2∏ abd ∅ from 0 to ∏, but the p is giving me a challenge. if z is zero then p goes from 0 to 4, but if x and y are zero then p goes from 0 to (16/5)^(1/2)? am I on the right track? What do I do now??
 
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  • #2
Hi PsychonautQQ! :smile:
PsychonautQQ said:
so I assume I must first change it to spherical coordinates. …

(If you must integrate) wouldn't ordinary x,y,z integration be easier? :wink:
 
  • #3
Cylindrical coordinates would seem to be a better choice than spherical.

In either case, you want to express the outer boundary in terms of the coordinates you decide to work with. In the case for spherical coordinates, solve for ##\rho## as a function of ##\phi## to get the upper limit for ##\rho##.
 
  • #4
@PsychonautQQ: Have you studied change of variables and Jacobians yet?
 
  • #5
As with the analogous problem of the area of an ellipse, also here I'd use the most convenient parametrization possible. I'd suggest
[tex]\vec{r}(\lambda,\theta,\phi)=\lambda \begin{pmatrix}
a \cos \phi \sin \theta \\
b \sin \phi \sin \theta \\
c \cos \theta
\end{pmatrix}, \quad \lambda \in [0,1], \quad \phi \in [0, 2\pi], \quad \theta \in [0,\pi].[/tex]
Calculate the Jacobian and do the triple integral. it's not too difficult.
 

What is the formula for finding the volume of an elipsoid?

The formula for finding the volume of an elipsoid is V = (4/3)πabc, where a, b, and c are the semi-principal axes of the elipsoid and π is the mathematical constant pi.

How do you measure the semi-principal axes of an elipsoid?

The semi-principal axes of an elipsoid can be measured by finding the longest, middle, and shortest distances from the center of the elipsoid to its surface. These distances are represented by a, b, and c in the volume formula.

Can the volume of an elipsoid be negative?

No, the volume of an elipsoid cannot be negative. The volume of any three-dimensional object represents the amount of space it occupies and therefore cannot be a negative value.

What units should be used for the semi-principal axes and volume?

The semi-principal axes and volume should have consistent units of measurement, such as centimeters or inches. It is important to make sure all measurements are in the same units before calculating the volume.

Is there a simpler way to find the volume of an elipsoid?

Yes, there is a simpler way to find the volume of an elipsoid. Some calculators and online tools have a built-in function for calculating the volume of an elipsoid. All you need to do is input the values for the semi-principal axes and the calculation will be done for you.

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