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Volume integral of an ellipsoid with spherical coordinates.

  • Thread starter epiclier
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Homework Statement



By making two successive simple changes of variables, evaluate:

I =[itex]\int\int\int x^{2} dxdydz[/itex]

inside the volume of the ellipsoid:

[itex]\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=R^{2}[/itex]

Homework Equations



dxdydz=r^2 Sin(phi) dphi dtheta dr

The Attempt at a Solution

I will post more here if needed.

So I let x=au y=bv z=cw
then I calculated the jacobian which gave |J|=abc

I then manipulated the integral to get:

[itex]a^{3}bc \int\int\int r^{4} sin^{2}(phi}cos{theta} dphi dtheta dr[/itex]

where r is from 0->R
theta is from 0->2π
phi is from 0->π


Following this through got me the answer of:

V=[itex]\frac{a^{3}bc\pi}{2}[/itex]


Would anybody be able to confirm this answer for me?

Thanks in advance!!!!




Epiclier
 

Answers and Replies

  • #2
Simon Bridge
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First you need a way to check your answer ... consider: what is a volume?
If a, b, and c were in meters - the volume should be in m3 ... is that what you got?

note: what you have for the equation of the ellipse will give a sphere if a=b=c=1 - so maybe you need a term in R there?
 
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  • #3
HallsofIvy
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If the Jacobian is abc, where did that [itex]a^3[/itex] come from?
 
  • #4
Simon Bridge
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Aside:
Lets see if I can get that relation to render...
[itex]a^{3}bc \int\int\int r^{4} sin^{2}(phi}cos{theta} dphi dtheta dr[/itex]
... I see what happened: sin^{2}(phi} won't render because you have closed a ( with a }. It will also render better with extra backslashes: \sin^2(\phi)

[tex]a^{3} bc \int \int \int r^{4} \sin^{2}(\phi) \cos(\theta) d\phi d\theta dr[/tex]... I'm guessing you used the built-in equation editor for part and hand edited a part.
 
  • #5
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Aside:
Lets see if I can get that relation to render...
... I see what happened: sin^{2}(phi} won't render because you have closed a ( with a }. It will also render better with extra backslashes: \sin^2(\phi)

[tex]a^{3} bc \int \int \int r^{4} \sin^{2}(\phi) \cos(\theta) d\phi d\theta dr[/tex]... I'm guessing you used the built-in equation editor for part and hand edited a part.
Yep, thats exactly what happened thanks simon.

First you need a way to check your answer ... consider: what is a volume?
If a, b, and c were in meters - the volume should be in m3 ... is that what you got?

note: what you have for the equation of the ellipse will give a sphere if a=b=c=1 - so maybe you need a term in R there?
Its not actually a volume though, its the integral of a function inside of the volume. I started again anyway because the answer I posted above was way off --> I didnt multiply by the jacobian from cartesian u-v-w to spherical u-v-w properly. So im just going to write the whole lot out again.

So this was the equation for the boundary volume:

[itex]\frac{x^{2}}{b^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = R^{2}[/itex]

Next I substitute to make this all alot easier:
let: x=au
y=bv
z=cw

where u,v,w are variables.

Hence: [itex]u^{2} + v^{2} + w^{2} = R^{2}[/itex]

Now, the Jacobian from x-y-z space to u-v-w space is [itex]abc[/itex]

AND [itex]x^{2} = a^{2} u^{2}[/itex]
AND [itex]u^{2} = r^{2} \sin^{2}(\phi) \cos^{2}(\theta)[/itex]

Therefore, the original integral becomes:

[itex]\int\int\int a b c a^{2} r^{2} \sin^{2}(\phi) \cos^{2}(\theta) r^{2} \sin(\phi) d\theta d\phi dr[/itex]

with the following limits(these will be the same throughout): (0≤r≤R),(0≤[itex]\phi\leq \Pi[/itex]) and (0≤θ≤2 Pi)

I get the following integrals for between stages:

[itex]\int cos^{2}(\theta) d\theta = \frac{1}{2}(\theta + Sin(\theta)Cos(\theta)) [/itex]

[itex]\int Sin^{3}(\phi) d\phi = \frac{1}{12}(Cos(3 \Phi) - 9Cos(\Phi))[/itex]

[itex]\int r^4 dr = r^5/5[/itex]

Then yada yada yada, we solve the integral for said limits and get:

I = [itex]\frac{4 \Pi a^{3} b c R^{5}}{15}[/itex]

Does this seem right to anyone?

Once again, thanks in advance, and sorry for the mess around.


-Epiclier
 
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  • #6
Simon Bridge
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Its not actually a volume though, its the integral of a function inside of the volume.
And what sort of thing does that kind of integral normally give you?

Isn't dxdydz a volume element?
So what sort of thing does the x^2 represent?
(You are right - but that is still the way to think about it.)

At least you've got an R in this one.
What would happen to the units if you'd integrated it inside a rectangular volume?
 
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  • #7
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And what sort of thing does that kind of integral normally give you?

Isn't dxdydz a volume element?
So what sort of thing does the x^2 represent?
(You are right - but that is still the way to think about it.)

At least you've got an R in this one.
What would happen to the units if you'd integrated it inside a rectangular volume?
Erm... No idea, the sum of the function at every point in the volume?

Yes, it is the volume element, so I would guess it represents a scalar field inside of the volume?

Sorry, but I just feel really lost now, I felt good about the last answer I gave, now Im just confused :(


-Epiclier
 
  • #8
Simon Bridge
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If you don't understand what you are doing, you will need outside support to tell you when you have the right answer. I could just tell you - which will do until next time. But what you are training for is to be able to deal with the situation where nobody knows the right answer so you need to wrap your head around self-checking.

This is an integral over a volume.
g(xyz)dxdydz is something multiplied by a small volume ... gdV if you like.
An easy analogy for that in physics is density .. g is a volumetric density function.
It does not have to be a mass-density but that is one that is easier to think about.
It is as if you have a gas which is weirdly compressed along the x direction so that the density varies with the square of the distance.

Then x2dV is the mass, dm, of the volume element.

The integral signs are just telling you to add up all the masses.

That should sort-of help ease some of the "magic box" effect of the problem statement.

If you want to know if the result you got from all that tricksey adding masses makes sense ... try it over a simple volume. Repeat the calculation, only for a box with length a in the x direction, b in y and c in z. Now you don't need the coordinate shift it should be easy.

The final forms should be similar even though the actual details will be different.
If you can account for the differences in terms of differences in the calculations then you have good reason to be confident in your answer.
 

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