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Volume of Ellipsoid

  • Thread starter WayBehind
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  • #1
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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 [tex]x^2 + y^2 + 4 z^2 = 100[/tex]


Homework Equations





The Attempt at a Solution


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

I'm fairly certain I'm not really on the right track, if someone could confirm that...
How do I find the value of [tex]\phi[/tex]?
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.
 

Answers and Replies

  • #2
HallsofIvy
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In spherical coordinates, [itex]\theta[/itex], the "longitude", goes from 0 to [itex]2\pi[/itex], as you have, and [itex]\phi[/itex], the "co-latitude", goes from 0 to [itex]\pi[/itex].

However, the "differential of volume" in spherical coordinates is not "[itex]d\rho d\phi d\theta[/itex]". It is [itex]\rho^2 sin(\phi) d\rho d\phi d\theta[/itex].
 
  • #3
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Thanks. I knew it wasn't simply [itex]d\rho d\phi d\theta[/itex], I just didn't know the equation was [itex]\rho^2 sin\phi[/itex]. 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 [itex]\phi[/itex] = 0 to [itex]\pi[/itex]? I don't know how to get there, either.

Do I have [itex]\rho[/itex] being from 0 to [itex]\sqrt{20}[/itex] correct?
 
  • #4
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Okay, I've just gotten back to this problem and I'm now even more confused. Here it is, step by step:

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

What am I doing wrong???

Thanks
 
Last edited:
  • #5
HallsofIvy
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Okay, I've just gotten back to this problem and I'm now even more confused. Here it is, step by step:

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

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

[tex]= \int_{0}^{2\pi} 0 [/tex]
= 0

What am I doing wrong???

Thanks
 
  • #6
LCKurtz
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Thanks. I knew it wasn't simply [itex]d\rho d\phi d\theta[/itex], I just didn't know the equation was [itex]\rho^2 sin\phi[/itex]. 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 [itex]x = \rho\sin\phi\cos\theta,\,y = \rho\sin\phi\sin\theta,\,z = \rho\cos\phi[/itex] you get the Jacobian
[tex]
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
[/tex]

so your dV element transfroms from dxdydz to [itex]\rho^2\sin\phi\ d\rho\, d\phi\, d\theta[/itex]

This is also why it is standard to let [itex]\phi[/itex] range from 0 to [itex]\pi[/itex]. Otherwise you would need to deal with absolute value bars on the [itex]\sin\phi[/itex].
 
  • #7
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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 [itex]\rho=\sqrt{20}[/itex]? Something tells me I don't have that right.
I'll continue to work through the problem:
[itex]\int_{0}^{2\pi} -20\sqrt{20}/3 cos\phi d\theta[/itex] from 0 to [itex]\pi [/itex]

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

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

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