Setting up a triple integral to find volume of a region

In summary, 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, you can use spherical coordinates and set up the triple integral as follows:Vsphere = \int^{a}_{-a}\int^{\sqrt{a^2 - z^2}}_{-\sqrt{a^2 - z^2}}\int^{\sqrt{a^2 - y^2 - z^2}}_{-\sqrt{a^2 - y^2 - z^2}}dx dy dzVellipsoid
  • #1
yolanda
8
0

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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  • #2
In what type of coordinates? Spherical coordinates looks good.
 
  • #3
I was considering spherical as well. That would work. I'm not picky on which coordinate system we use.
 
  • #4
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!
 
  • #5
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!
 
  • #6
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 ...
 
  • #7
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.
 
  • #8
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:

What is a triple integral?

A triple integral is an extension of a regular integral in calculus, where instead of integrating over a two-dimensional region, we integrate over a three-dimensional region. It is used to find the volume of a three-dimensional object or region.

What is the process for setting up a triple integral?

The process for setting up a triple integral involves breaking the region into small, simple shapes such as cubes, rectangular prisms, or cylinders. Each shape is then represented by its own triple integral, with the limits of integration determined by the boundaries of the shape.

How do you determine the limits of integration for a triple integral?

The limits of integration for a triple integral are determined by the boundaries of the region being integrated. This can be done by visualizing the region in three dimensions and identifying the points where the boundaries intersect the x, y, and z-axes. These points will serve as the limits for each variable in the triple integral.

What are the different types of triple integrals?

There are three types of triple integrals: type I, type II, and type III. Type I triple integrals are used when the region is bounded by planes parallel to the yz-plane, type II when the region is bounded by planes parallel to the xz-plane, and type III when the region is bounded by planes parallel to the xy-plane.

Can you use a triple integral to find the volume of an irregularly shaped region?

Yes, a triple integral can be used to find the volume of an irregularly shaped region. This involves breaking the region into smaller, simpler shapes and setting up multiple triple integrals to cover the entire region. The sum of these integrals will give the volume of the irregularly shaped region.

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