# Solving Volume of Ellipsoid with Spherical Coordinates

• WayBehind
In summary, the student is having difficulty understanding how to solve a problem in spherical coordinates. The equation they are trying to solve is \rho^2 sin\phi d\rho d\phi d\theta. However, they get lost when trying to integrate sin(\phi). They incorrectly convert to spherical coordinates and end up with \rho=\sqrt{20}. They then try to find \phi, but get stuck. They eventually realize that \rho must be from 0 to \sqrt{20} and solve the equation for \phi.
WayBehind
I know this has been answered in another thread, but it still isn't entirely clear to me. This particular section in class is giving me some major problems and I'm hoping someone can shed some light on things. This is probably one of the easier problems in this assignment and I'm hoping if I can figure it out then the rest will fall into place.

I thought spherical coordinates would be appropriate here, but I can't seem to find a correct answer.

b]1. Homework Statement [/b]
FInd the volume of the ellipsoid $$x^2 + y^2 + 4 z^2 = 100$$

## The Attempt at a Solution

if I convert to spherical coordinates, I can reduce the following:
$$\rho^2sin^2\phi cos^2\theta + \rho^2sin^2\phi sin^2\theta + 4\rho^2cos^2\phi = 100$$
to
$$\rho=\sqrt{20}$$
So I'm trying to set up my equation and get:
$$\int_{0}^{2\pi}\int_{0}^{?}\int_{0}^{\sqrt{20}} d\rho d\phi d\theta$$

I'm fairly certain I'm not really on the right track, if someone could confirm that...
How do I find the value of $$\phi$$?
And I'm not even really sure how to find the equation, although I'm guessing I am just being dense.
Any help would be appreciated.

In spherical coordinates, $\theta$, the "longitude", goes from 0 to $2\pi$, as you have, and $\phi$, the "co-latitude", goes from 0 to $\pi$.

However, the "differential of volume" in spherical coordinates is not "$d\rho d\phi d\theta$". It is $\rho^2 sin(\phi) d\rho d\phi d\theta$.

Thanks. I knew it wasn't simply $d\rho d\phi d\theta$, I just didn't know the equation was $\rho^2 sin\phi$. How do you get that? I'd really like to understand the entire method so that I can apply it to more than just this problem.

How is $\phi$ = 0 to $\pi$? I don't know how to get there, either.

Do I have $\rho$ being from 0 to $\sqrt{20}$ correct?

Okay, I've just gotten back to this problem and I'm now even more confused. Here it is, step by step:

$$\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\sqrt{20}} \rho^2 sin\phi d\rho d\phi d\theta$$
$$\int_{0}^{2\pi}\int_{0}^{\pi} \rho^3/3 sin\phi d\phi d\theta$$
$$= \int_{0}^{2\pi}\int_{0}^{\pi} 20\sqrt{20}/3 sin\phi d\phi d\theta$$
$$\int_{0}^{2\pi} -20\sqrt{20}/3 cos\phi d\theta$$ from 0 to $$\pi$$
$$= \int_{0}^{2\pi} 20\sqrt{20}/3 - 20\sqrt{20}/3 d\phi d\theta$$
$$= \int_{0}^{2\pi} 0$$
= 0

What am I doing wrong?

Thanks

Last edited:
WayBehind said:
Okay, I've just gotten back to this problem and I'm now even more confused. Here it is, step by step:

$$\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\sqrt{20}} \rho^2 sin\phi d\rho d\phi d\theta$$
$$\int_{0}^{2\pi}\int_{0}^{\pi} \rho^3/3 sin\phi d\phi d\theta$$
$$= \int_{0}^{2\pi}\int_{0}^{\pi} 20\sqrt{20}/3 sin\phi d\phi d\theta$$
$$\int_{0}^{2\pi} -20\sqrt{20}/3 cos\phi d\theta$$ from 0 to $$\pi$$
The integral of $sin(\phi)$ is $-cos(\phi)$, not $cos(\phi)$. But since you got "0" that's not your real problem.

$$= \int_{0}^{2\pi} 20\sqrt{20}/3 - 20\sqrt{20}/3 d\phi d\theta$$
Since you don't say how you got this, I can't tell you what you did wrong! Your integrand should be $-20\sqrt{20}/3(cos(\pi)- cos(0))$. What is that? What are $cos(\pi)$ and $cos(0)$?

$$= \int_{0}^{2\pi} 0$$
= 0

What am I doing wrong?

Thanks

WayBehind said:
Thanks. I knew it wasn't simply $d\rho d\phi d\theta$, I just didn't know the equation was $\rho^2 sin\phi$. How do you get that? I'd really like to understand the entire method so that I can apply it to more than just this problem.

When you make the change of variables $x = \rho\sin\phi\cos\theta,\,y = \rho\sin\phi\sin\theta,\,z = \rho\cos\phi$ you get the Jacobian
$$J = \left| \begin{array}{ccc} x_\rho&y_\rho&z_\rho\\ x_\phi&y_\phi&z_\phi\\ x_\theta&y_\theta&z_\theta \end{array}\right | = \rho^2\sin\phi$$

so your dV element transfroms from dxdydz to $\rho^2\sin\phi\ d\rho\, d\phi\, d\theta$

This is also why it is standard to let $\phi$ range from 0 to $\pi$. Otherwise you would need to deal with absolute value bars on the $\sin\phi$.

Thanks to both of you, that helps quite a bit.

I see where I went wrong with coming up with zero. The signs were messing me up.

So now I'm on to my next stumbling block, and I'm quite certain I'm just being dense.
Is $\rho=\sqrt{20}$? Something tells me I don't have that right.
I'll continue to work through the problem:
$\int_{0}^{2\pi} -20\sqrt{20}/3 cos\phi d\theta$ from 0 to $\pi$

$= \int_{0}^{2\pi} 40\sqrt{20}/3 d\theta$
Correct, so far?
$= 40\sqrt{20}\theta/3$ from 0 to $2\pi$

Which gives me
$80\sqrt{20}\pi/3$
And that isn't right. I'll have to try and work this out again this evening.

## 1. How do you calculate the volume of an ellipsoid using spherical coordinates?

To calculate the volume of an ellipsoid using spherical coordinates, you can use the formula V = (4/3)πabc, where a, b, and c are the semi-axes of the ellipsoid.

## 2. What are the advantages of using spherical coordinates to solve for the volume of an ellipsoid?

Using spherical coordinates can simplify the calculation of the volume of an ellipsoid, as it involves integrating over a single variable instead of three. It also allows for easier visualization of the shape.

## 3. Is there a specific method for converting from Cartesian coordinates to spherical coordinates to solve for the volume of an ellipsoid?

Yes, there is a specific method for converting from Cartesian coordinates to spherical coordinates. It involves using the equations x = ρsinφcosθ, y = ρsinφsinθ, and z = ρcosφ, where ρ, φ, and θ represent the radial distance, polar angle, and azimuthal angle, respectively.

## 4. Can you use spherical coordinates to solve for the volume of any ellipsoid?

Yes, spherical coordinates can be used to solve for the volume of any ellipsoid, regardless of its orientation or eccentricity.

## 5. Are there any limitations to using spherical coordinates to solve for the volume of an ellipsoid?

One limitation of using spherical coordinates is that it may not be the most efficient method for solving for the volume of an ellipsoid with a high eccentricity. In these cases, it may be more efficient to use a different coordinate system such as ellipsoidal coordinates.

• Calculus and Beyond Homework Help
Replies
3
Views
752
• Calculus and Beyond Homework Help
Replies
4
Views
2K
• Calculus and Beyond Homework Help
Replies
7
Views
1K
• Differential Geometry
Replies
0
Views
286
• Calculus and Beyond Homework Help
Replies
7
Views
2K
• Calculus and Beyond Homework Help
Replies
34
Views
2K
• Calculus and Beyond Homework Help
Replies
9
Views
8K
• Electromagnetism
Replies
3
Views
679
• Calculus and Beyond Homework Help
Replies
4
Views
999
• Calculus and Beyond Homework Help
Replies
7
Views
783