Setting up a triple integral to find volume of a region

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Homework Help Overview

The discussion revolves around setting up a triple integral to find the volume of a region bounded by a sphere defined by the equation x²+y²+z²=a² and an ellipsoid given by (x²/4a²)+(4y²/a²)+(9z²/a²)=1. Participants are exploring the appropriate coordinate systems and intervals for integration.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of spherical coordinates for the integration, with some expressing uncertainty about the intervals to use. There are attempts to derive the limits for both the sphere and the ellipsoid, with questions about the setup and specific expressions for the integrals.

Discussion Status

Some participants have provided guidance on the intervals for the sphere and ellipsoid, while others are still seeking clarity on the setup of the triple integral. There is an ongoing exploration of the correct approach without a clear consensus on the final form of the integrals.

Contextual Notes

Participants mention confusion regarding the variable 'a' and its implications on the intervals. There is also a request for examples from textbooks, indicating a lack of similar problems for reference.

yolanda
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Homework Statement



I need to set up the triple integral to find the volume of the region bounded by the sphere: x^2+y^2+z^2=a^2 and the ellipsoid: (x^2/4a^2)+(4y^2/a^2)+(9z^2/a^2)=1.

Homework Equations



above

The Attempt at a Solution



graph.jpg


I'm not sure which interval I should be using here. I made a 3D graph of the region, but the a variable is really throwing me off. Can anyone point me in the right direction here? Thanks in advance!
 
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In what type of coordinates? Spherical coordinates looks good.
 
I was considering spherical as well. That would work. I'm not picky on which coordinate system we use.
 
To calculate the volume of the sphere which is x2 + y2 + z2 = a2 you should use the following for the cartesian coordinate.

-a [tex]\leq[/tex] z [tex]\leq[/tex] a

-[tex]\sqrt{a^2 - z^2}[/tex] [tex]\leq[/tex] y [tex]\leq[/tex] [tex]\sqrt{a^2 - z^2}[/tex]

-[tex]\sqrt{a^2 - y^2 - z^2}[/tex] [tex]\leq[/tex] x [tex]\leq[/tex] [tex]\sqrt{a^2 - y^2 - z^2}[/tex]

And you can do it the same for the ellipsoid!
 
So, for the ellipsoid, should it be the following?:

-a<=z<=a
-sqrt[-x^2-4(9*z^2-a^2)]/4 <=y<= sqrt[-x^2-4(9*z^2-a^2)]/4 (shouldn't have x values...hmmm)

I'm a bit lost here :confused: Would someone mind setting up the triple integral for me? I think having a look at the final product would open up some doors in my brain. No need to evaluate. Thanks again, people!
 
Vsphere = [tex]\int[/tex][tex]^{a}_{-a}[/tex][tex]\int[/tex][tex]^{\sqrt{a^2 - z^2}}_{-\sqrt{a^2 - z^2}}[/tex][tex]\int[/tex][tex]^{\sqrt{a^2 - y^2 - z^2}}_{-\sqrt{a^2 - y^2 - z^2}}[/tex]dx dy dz

and for the ellipsoid [tex]\frac{-a}{3}[/tex] [tex]\leq[/tex] z [tex]\leq[/tex] [tex]\frac{a}{3}[/tex] and ...
 
Oh ok, I think I get that part now.

-a/3 <= z <= a/3

-sqrt[(a^2-9z^2)/4] <= y <= sqrt[(a^2-9z^2)/4]

-sqrt[4a^2-16y^2-36z^2] <= x <= sqrt[4a^2-16y^2-36z^2]

So, now that I have my intervals for both shapes, what comes next? I need the volume of the ellipsoid, but only the part that's within the sphere. Sorry to ask so many questions, but my book has no similar examples that I can work with. I really appreciate your help.
 
I solved it in spherical coordination and I think it is correct, if it is not, somebody please tell me why?

V= [tex]\int[/tex] [tex]\int[/tex] [tex]\int[/tex] r2sin[tex]\phi[/tex] dr d[tex]\theta[/tex] d[tex]\phi[/tex]

0 [tex]\leq[/tex] r [tex]\leq[/tex] [tex]\frac{4 \sqrt{2}a}{sin\phi \sqrt{12 cos^2(\theta) +20}}[/tex]

0 [tex]\leq[/tex] [tex]\theta[/tex] [tex]\leq[/tex] 2 [tex]\pi[/tex]

0 [tex]\leq[/tex] [tex]\phi[/tex] [tex]\leq[/tex] [tex]\pi[/tex]
 
Last edited:

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