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


Homework Equations





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...
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,847
966
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]
 
  • #3
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:
  • #4
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,389
1,044
You are right regarding the Jacobian in your above post.

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

Homework Equations



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 .)
 
  • #5
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.
 
  • #6
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,389
1,044
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 .
 

Related Threads on Triple integral volume

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
959
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
5
Views
965
  • Last Post
Replies
1
Views
736
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
1
Views
7K
Top