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Triple integral volume

1. Homework Statement
Find the volume of the object, defined by these inequalities(?): x^2+y^2+z^2≤4, (x-1)^2+y^2≥1, (x+1)^2+y^2≥1


2. Homework Equations



3. The Attempt at a Solution

First we draw the object, and realize that it's a sphere with 2 circles in it with radius 1 at (-1,0) and (1,0). Our object is defined outside these circles, so I imagine we could consider the square with 2 cylinders drilled out. If we can find the volume of one of these cylinders from 0 to z we are done. (Vsphere-4*Vcylinder)
Is this the way to go or do we have other solutions? This one doesnt give me right answer...
 

HallsofIvy

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I suspect that you have calculated the volume of the cylinders incorrectly. The top and bottom are pieces of the sphere not planes. Yes, this is a sphere of radius 2 and so its volume, ignoring the cylinders for the moment, is [itex](16/3)\pi[/itex].

By symmetry, it is sufficient to find volume of one cylinder, then subtract twice that from the volume of the sphere. Also, I think I would be inclined to shift the coordinate system so that the center of one cylinder is at the origin: if we let x'= x- 1, y'= y, z'= z, then one cylinder has equation [itex]x'^2+ y'^2= 1[/itex] and the sphere has equation [itex](x'+1)^2+ y'^2+ z'^2= x'^2+ 2x'+ 1+ y'^2+ z'^2= 4[/itex], [itex](x'^2+ y'^2)+ 2x'+ z'^2= 3[/itex].

Now, convert to cylindrical coordinates: [itex]x'= rcos(\theta)[/itex], [itex]y'= r sin(\theta)[/itex], [itex]z'= z[/itex] so the equation of the sphere becomes [itex]r^2+ 2r cos(\theta)+ z^2= 3[/itex] so that [itex]z= \pm (3- r^2- 2r cos(\theta))^{1/2}[/itex]. Then the volume of the cylinder is
[tex]\int_{\theta= 0}^{2\pi}\int_{r= 0}^1\int_{z= -(3- r^2- 2rcos(\theta))^{1/2}}^{(3- r^2- 2rcos(\theta))^{1/2}} r cos(\theta)dzdrd\theta[/tex]
 
Hi, thx for answering. This was kinda the vision I had, tho i was unfamiliar with cylindrical coordinates. I dont understand where cos(θ) comes from in the integrand? The jacobian from the substitution would just be r, no? Or have I missed something fundamental here? The double integral gets kinda messy after we've integrated with respect to z, in my mind.

And I dont seem to get the correct answer, which should be 128/9.
 
Last edited:

SammyS

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You are right regarding the Jacobian in your above post.

1. Homework Statement
Find the volume of the object, defined by these inequalities(?): x^2+y^2+z^2≤4, (x-1)^2+y^2≥1, (x+1)^2+y^2≥1

2. Homework Equations

3. The Attempt at a Solution

First we draw the object, and realize that it's a sphere with 2 circles in it with radius 1 at (-1,0) and (1,0). Our object is defined outside these circles, so I imagine we could consider the square with 2 cylinders drilled out. If we can find the volume of one of these cylinders from 0 to z we are done. (Vsphere-4*Vcylinder)
Is this the way to go or do we have other solutions? This one doesnt give me right answer...
Yes.

You are to find the volume enclosed by the sphere, but external to the two cylinders.

So, find the portion of one cylinder enclosed by the sphere and above z = 0. Then use your expression, (Vsphere-4*Vcylinder) .

What is the equation of the circle, (x-1)2+y2 = 1, in polar coordinates?

(I haven't worked the problem, but I suspect the answer is more like 128π/9 .)
 
In polar coordinates it would be r^2-2rcosθ=0 ⇔ r=2cosθ. So it will go from 2cosθ and follow the sphere upwards... nah Im having a hard time visualizing it :(

Yes, the answer is probably a small typo then.
 

SammyS

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In polar coordinates it would be r^2-2rcosθ=0 ⇔ r=2cosθ. So it will go from 2cosθ and follow the sphere upwards... nah I'm having a hard time visualizing it :(

Yes, the answer is probably a small typo then.
I would integrate over z first: from 0 to √(4-r2)

Then r goes from 0 to 2cos(θ) .

Then θ : from -π/2 to π/2 .
 

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