Solid of revolution question: given height of a cylindrical core in a sphere

  • Thread starter zeion
  • Start date
  • #1
zeion
467
1

Homework Statement



A hole is drilled through the center of a ball of radius r, leaving a solid with a hollow cylindrical core of height h. Show that the volume of this solid is independent of the radius of the ball.


Homework Equations



[tex]
V = \int_{a}^{b} 2\pi x (f(x) - g(x)) dx [/tex]

[tex]
V = \int_{a}^{b} \pi ([f(x)]^2 - [g(x)]^2) dx
[/tex]


The Attempt at a Solution



Am I allowed to use the radius of the cylinder? Then I can find the volume of the cylinder and then the radius of the sphere?

I can't really think of a way to do this without referring to any radius..
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,263
619
The 'cylindrical core' isn't really a cylinder. It has rounded ends. Put the core around the x-axis and set up one of those integrals. The second one worked fairly well for me. Now just say what f(x) and g(x) are in terms of r and h.
 
  • #3
zeion
467
1
The 'cylindrical core' isn't really a cylinder. It has rounded ends. Put the core around the x-axis and set up one of those integrals. The second one worked fairly well for me. Now just say what f(x) and g(x) are in terms of r and h.

I thought I had to show that it was independent of the radius of the ball?
 
  • #4
Dick
Science Advisor
Homework Helper
26,263
619
You have to show the result of the integral V is independent of r. The formula for the integral V will contain r. You just have to show that if you do the integral then the result doesn't depend on r. f(x) and/or g(x) will depend on r. That doesn't necessarily mean V does.
 
Last edited:
  • #5
holezch
251
0
hey, I am trying this too.. pretty interesting.

so I should do f(x) = volume of sphere and g(x) = volume of cylinder right ? how should I find the volume of the cylinder without knowing the radius of it... thanks, I need some more clues
 
  • #6
Dick
Science Advisor
Homework Helper
26,263
619
There is really no actual 'cylinder' here. There's a cylindrical cross section of a sphere. f(x) and g(x) aren't volumes. They are the radii of the disks or shells you want to deal with. Just set up the integral. It will magically work out. Trust me.
 
  • #7
holezch
251
0
There is really no actual 'cylinder' here. There's a cylindrical cross section of a sphere. f(x) and g(x) aren't volumes. They are the radii of the disks or shells you want to deal with. Just set up the integral. It will magically work out. Trust me.

oh..right.. I was thinking of taking off the spherical tops and then we will have a cylinder.

Okay, so (sorry, I'm new to this revolution stuff), we have g(x ) that will be shells for the cylindrical part and f(x ) can be shells for the overall sphere part? we would still need a radius for g(x), unless we can express it in a different way :S

edit : Okay, Ididn't know that radii meant radius in plural form.. so.. we have to set up f(x) and g(x) radii for the inner part and the outer part.. :S I have no idea where to start

thanks!
 
Last edited:
  • #8
Dick
Science Advisor
Homework Helper
26,263
619
The length of the cylindrical cross section is h. The radius of the sphere is r. Find the radius of the cylindrical cross section by intersecting it with the sphere. Then find f(x) and g(x). Don't wimp out like zeion. You can probably use either shells or disks. I think I did disks. The integration is not even hard. Show the result is, in fact, independent of r.
 
  • #9
holezch
251
0
okay.. I couldn't find what "Find the radius of the cylindrical cross section by intersecting it with the sphere." is supposed to mean.. could anybody help me out..?
thanks

edit: okay, maybe I don't need to figure out what that was supposed to mean:
if I say that cr is the radius of the cylinder part.. then 2cr is the diameter, h is the length
of the hole, and then we can draw a triangle and get an expression for cr! now it should be easy enough with regular disk integration (I think)
 
Last edited:
  • #10
holezch
251
0
okay, so now should I use the shell method and integrate from the radius of the cylinder to the radius of the sphere?
 
  • #11
Dick
Science Advisor
Homework Helper
26,263
619
okay, so now should I use the shell method and integrate from the radius of the cylinder to the radius of the sphere?

That should work.
 
  • #12
holezch
251
0
That should work.

thanks, so I'll something like:
r - radius of cylinder ( sqrt(4R^2 - h^2) )
R - radius of sphere
[tex]2 \pi \int \sqrt{R^{2} - x^{2} [/tex]
integral goes from r to R

I think I'll have to use a substitution x = Rcosu or something of the like, but I foresee a problem.. when I have to evaluate this integral from sqrt(4R^2 - h^2) to R^2.. it looks like this may get a little messy. Am I on the right track doing this substitution? thanks..
 
  • #13
Dick
Science Advisor
Homework Helper
26,263
619
sqrt(R^2-x^2) is half the length of your shell, right? Where's the radius of the shell? You integration expression is incomplete. And I don't quite agree with your expression for the radius of the cylinder either, I think you've got the diameter there.
 
  • #14
holezch
251
0
sqrt(R^2-x^2) is half the length of your shell, right? Where's the radius of the shell? You integration expression is incomplete. And I don't quite agree with your expression for the radius of the cylinder either, I think you've got the diameter there.

Right, I have to multiply whatever I get by 2 after. and you're right, I should divide that thing I got by 2, I had a diameter.

so it should be

[tex] 4 \pi \int x \sqrt{R^{2} - x^{2} [/tex]

from r to R , where r is [tex] \sqrt {4R^{2} - h^{2}}/2 [/tex]

Is that alright? do you think that I might have some trouble calculating this, since I'll have to evaluate at [tex] \sqrt {4R^{2} - h^{2}}/2 [/tex] .. and should I use x = Rcosu?
thanks a lot

so, doing that:

[tex] 4 \pi \int x \sqrt{R^{2} - x^{2} [/tex]

let x = Rcosu
then
[tex] - \int R^{2}cosu \sqrt{R^{2}(1-cos^{2}u)} sinu du = - \int R^{3}sin^{2}(u)cos(u)du [/tex]
, then
[tex]
y = sinu,
- R^{3} \int y^2 dy
[/tex]
the limits are all r to R, but for my last sub.. I have to change the limits to sin(r) and sin(R). Dealing with sin(r) and sin(R) is my problem.. should I not have done trigonometric substitutions?
thanks
 
Last edited:
  • #15
Dick
Science Advisor
Homework Helper
26,263
619
You didn't need to do trig substitutions u=R^2-x^2 will do the job.
 
  • #16
holezch
251
0
You didn't need to do trig substitutions u=R^2-x^2 will do the job.

thanks, I thought about that.. but I don't think that it's valid.. when you solve for x, the range of x of your new integrand is only positive, in our original integral, x can be negative or positive. I guess I'm wrong though? What have I mistaken?
 
  • #17
Dick
Science Advisor
Homework Helper
26,263
619
thanks, I thought about that.. but I don't think that it's valid.. when you solve for x, the range of x of your new integrand is only positive, in our original integral, x can be negative or positive. I guess I'm wrong though? What have I mistaken?

You are rotating the positive region around the y-axis. That's the full solid of revolution. I don't see any reason to worry about negative x values.
 
  • #18
holezch
251
0
You are rotating the positive region around the y-axis. That's the full solid of revolution. I don't see any reason to worry about negative x values.

okay, so I got it down to this with u = R^2 - x^2

[tex] -4 \pi \int \sqrt{u} du[/tex]

after subbing and algebra. . at this point, I lost track of those constants :S

anyway, am I on the right track?

then

[tex] -4 \pi * 2(R^2 - x^2)^{3/2} / 3 [/tex] evaluated at r to R

[tex] -4 \pi * 2(R^{2} - (4R^{2} - h^{2})/2)^{3/2} / 3 [/tex]

and I got lost here, am I right so far? thanks
 
Last edited:
  • #19
Dick
Science Advisor
Homework Helper
26,263
619
You agreed that the limit (the radius of the cylinder) was sqrt(4R^2-h^2)/2, right? Where did the 2 go? You should check the other constants as well. They don't all look right. But the limit is the most important.
 
  • #20
holezch
251
0
You agreed that the limit (the radius of the cylinder) was sqrt(4R^2-h^2)/2, right? Where did the 2 go? You should check the other constants as well. They don't all look right. But the limit is the most important.

right, I had a /2 in there, forgot to put it in with latex.. , and right, my constants were wrong, I had 2 * 2 * 2 and I put 6 :redface: anyway, I was more focused on getting rid of that R term, will it work out like this? I edited LATEX now, should that negative sign be there? I got it from du/-2x = dx
I can't get rid of that R^2 term when I expand that expression.. this is such a setback
thanks
 
Last edited:
  • #21
Dick
Science Advisor
Homework Helper
26,263
619
(sqrt(4R^2-h^2)/2)^2 is not (4R^2-h^2)/2. The R DOES cancel.
 
  • #22
Dick
Science Advisor
Homework Helper
26,263
619
Look. (sqrt(4R^2-h^2)/2)^2=(4R^2-h^2)/2^2=(4R^2-h^2)/4=R^2-h^2/4.
 
  • #23
holezch
251
0
Right, I realized that so I deleted my post... ahah thanks, I guess it all worked out in the end.. I should be neater and more careful.. does that negative sign belong there?
I get [tex] -4 h^{3}/3 [/tex] in the end
Thanks for staying in touch :)
 
  • #24
holezch
251
0
hold on, the R's did cancel, but I just realized that I never changed my limits.. this changes a lot.. after the substitution, my limits should be from sqrt(R^2 - r) to sqrt(R^2 - R) (I think..)
 
  • #25
Dick
Science Advisor
Homework Helper
26,263
619
hold on, the R's did cancel, but I just realized that I never changed my limits.. this changes a lot.. after the substitution, my limits should be from sqrt(R^2 - r) to sqrt(R^2 - R) (I think..)

I'd think that one over again. You did a substitution u=R^2-x^2 to do the integral but then you changed the variable back to x. You don't need to change the limits. They are still r to R.
 
  • #26
holezch
251
0
I'd think that one over again. You did a substitution u=R^2-x^2 to do the integral but then you changed the variable back to x. You don't need to change the limits. They are still r to R.

hm, if I understand how this works, I am changing my x to a function now: [tex] \sqrt{R^{2} - u } [/tex]

and from the sub formula:

[tex] \int\stackrel{g(b)}{g(a)} f(u) du = \int \stackrel{b}{a} f(g(x)) g'(x) dx [/tex]

so, from the original integral, I had f(u) (or x), and now I've expanded that to a function g to include g(x)g'(x) ... so I think I need to change my limits?

thanks and by the way, could you tell me how to include the limits of integration with LaTeX? thanks
 
  • #27
Dick
Science Advisor
Homework Helper
26,263
619
hm, if I understand how this works, I am changing my x to a function now: [tex] \sqrt{R^{2} - u } [/tex]

and from the sub formula:

[tex] \int\stackrel{g(b)}{g(a)} f(u) du = \int \stackrel{b}{a} f(g(x)) g'(x) dx [/tex]

so, from the original integral, I had f(u) (or x), and now I've expanded that to a function g to include g(x)g'(x) ... so I think I need to change my limits?

thanks and by the way, could you tell me how to include the limits of integration with LaTeX? thanks

You already basically solved the problem and now you are overshooting the finish line. You gave the answer as a constant times (R^2-x^2)^(3/2). That's in terms of x, NOT u. There is no reason on God's green earth to change the limits to u limits. Stop that. TeX for the limits on an integral is like \int^b_a. I think.
 

Suggested for: Solid of revolution question: given height of a cylindrical core in a sphere

Replies
0
Views
5K
  • Last Post
Replies
8
Views
841
Replies
5
Views
1K
  • Last Post
Replies
4
Views
11K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
7
Views
803
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
7
Views
4K
Top