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Finding the volume of an ellipsoid by using the volume of a sphere

  • Thread starter Fractal20
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  • #1
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Homework Statement


So I keep coming across problems that suggest finding the Volume of an ellipsoid using the volume of a ball ie:

Find the volume enclosed by the ellipsoid:

(x/a)^2 + (y/b)^2 + (z/c)^2 = 1

by using the fact that the volume of the unit ball in R^3 is 4pi/3


Homework Equations





The Attempt at a Solution



I keep getting stuck on this one. It seems like it will be some change of variables in which abc will eventually just pop out front of the usual expression for the volume of a ball. I keep thinking that I should be able to use spherical coords and a substitution like u = x/a, v = y/b and w = z/c. But then you just get the normal unit ball integral. I feel like it might come out of the limits of the integral in terms of P. That is, the maximum length of p is a function of [itex]\phi[/itex] and [itex]\theta[/itex] in the ellipsoid case. But with the above substitution then P is 1. Or I keep thinking it might come out of dP, but that hasn't been clear either.

I also keep thinking of plugging in for p -> p = [itex]\sqrt{u2 + v2 + w2}[/itex] = [itex]\sqrt{(psin \phi cos \theta)2/a + (psin \phi sin \theta)2/b + (pcos \phi cos \theta)2/c}[/itex] but then the p2 term would cancel out.

Anyway, I am just going in circles. Does anybody have any suggestions?
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement


So I keep coming across problems that suggest finding the Volume of an ellipsoid using the volume of a ball ie:

Find the volume enclosed by the ellipsoid:

(x/a)^2 + (y/b)^2 + (z/c)^2 = 1

by using the fact that the volume of the unit ball in R^3 is 4pi/3


Homework Equations






The Attempt at a Solution



I keep getting stuck on this one. It seems like it will be some change of variables in which abc will eventually just pop out front of the usual expression for the volume of a ball. I keep thinking that I should be able to use spherical coords and a substitution like u = x/a, v = y/b and w = z/c. But then you just get the normal unit ball integral. I feel like it might come out of the limits of the integral in terms of P. That is, the maximum length of p is a function of [itex]\phi[/itex] and [itex]\theta[/itex] in the ellipsoid case. But with the above substitution then P is 1. Or I keep thinking it might come out of dP, but that hasn't been clear either.

I also keep thinking of plugging in for p -> p = [itex]\sqrt{u2 + v2 + w2}[/itex] = [itex]\sqrt{(psin \phi cos \theta)2/a + (psin \phi sin \theta)2/b + (pcos \phi cos \theta)2/c}[/itex] but then the p2 term would cancel out.

Anyway, I am just going in circles. Does anybody have any suggestions?
The volume of the ellipsoid is$$
\iiint_V 1dxdydz$$in the ##xyz## plane. Don't think in terms of spherical coordinates, think more about what the ellipsoid becomes in ##uvw## space under the transformation you are thinking about which I have highlighted. And the Jacobian has something to do with it.
 
  • #3
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The volume of the ellipsoid is$$
\iiint_V 1dxdydz$$in the ##xyz## plane. Don't think in terms of spherical coordinates, think more about what the ellipsoid becomes in ##uvw## space under the transformation you are thinking about which I have highlighted. And the Jacobian has something to do with it.
Is it simply that x/a = u -> dx = adu and etc?

I think I can see it from the Jacobian in Spherical but not Cartesian. Like if u,v,w are defined like normal in spherical coordinates, then making the jacobian makes a factor of a across the first row, b on the second and c on the third which pull out in the determinant.

But the Jacobian in cartesian still seems weird. So:

∫∫∫R(xyz)dV = ∫∫∫R(uvw) J dV


But in cartesian x=ua, y=vb and z=wc only defines the "functions" in terms of one other variable (x = f(u), y=g(v) etc...) then fv(u) is 0 etc. I am pretty new to Jacobians so I am problem just have a misunderstanding with them.
 
  • #4
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Ohhh! I see the Jacobian will just be a diagonal matrix, so it won't be 0. Thanks, sorry I can be a little slow.
 
  • #5
LCKurtz
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Ohhh! I see the Jacobian will just be a diagonal matrix, so it won't be 0. Thanks, sorry I can be a little slow.
So do you see how to finish it?
 
  • #6
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I think so. Just that J then equals abc. Right?
 
  • #7
LCKurtz
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I think so. Just that J then equals abc. Right?
Yes. Assuming you mean ##\left |\frac {\partial(x,y,z)}{\partial(u,v,w)}\right|##. So you get ##abc## times the volume of the unit sphere, which you already know.
 
  • #8
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I'm not too familiar with the notation. So just to be sure. If I saw we have x=f(u), y=g(v), z = h(w) then is what you wrote equivalent to determinate of

fu fv fw
gu gv gw
hu hv hw

? Sorry, I still can't figure out how to write matrices in this forum.
 
  • #9
LCKurtz
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I'm not too familiar with the notation. So just to be sure. If I saw we have x=f(u), y=g(v), z = h(w) then is what you wrote equivalent to determinate of

fu fv fw
gu gv gw
hu hv hw

? Sorry, I still can't figure out how to write matrices in this forum.
Yes. Right click on this to show the tex:$$
J = \left |\begin{array}{ccc}
a & 0 & 0\\
0 & b & 0\\
0 & 0 & c
\end{array}\right| = abc$$
 
  • #10
Curious3141
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Alternatively, you can use a Tex editor. I've found Daum equation editor (a free Chrome app) to be excellent for complicated Tex formatting.
 

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