Sphere-Cylinder Volume: Using Cylindrical Shells Method for Solid Calculation

  • Thread starter Thread starter Justabeginner
  • Start date Start date
  • Tags Tags
    Volume
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
6 replies · 3K views
Justabeginner
Messages
309
Reaction score
1

Homework Statement


A cylindrical hole of radius R is drilled through the center of a sphere of radius 3. Use the method of cylindrical shells to find the volume of the remaining solid, given that the solid is 6 cm. high.


Homework Equations


V= ∫2* ∏rh dr (shell method)


The Attempt at a Solution



I think this is the form to be used for this problem, but I am probably mistaken.
2 * ∫2* ∏ * x (R^2 - x^2) dx with a= r and b= R
4∏ ∫x (R^2 - x^2) dx

Then I'm stumped on how to do this :/ Thank you for your help.
 
Physics news on Phys.org
Justabeginner said:

Homework Statement


A cylindrical hole of radius R is drilled through the center of a sphere of radius 3. Use the method of cylindrical shells to find the volume of the remaining solid, given that the solid is 6 cm. high.


Homework Equations


V= ∫2* ∏rh dr (shell method)


The Attempt at a Solution



I think this is the form to be used for this problem, but I am probably mistaken.
2 * ∫2* ∏ * x (R^2 - x^2) dx with a= r and b= R
4∏ ∫x (R^2 - x^2) dx

Then I'm stumped on how to do this :/ Thank you for your help.

You have a = r but there is no r in the problem. Do you mean a = 3? Otherwise your integral looks OK, so just integrate it.

[Edit] That should be ##\sqrt{R^2-x^2}## in that integrand.
 
Last edited:
Oh wow, I see what I did wrong. So my final answer turns out to be

4/3 * pi (R^2 - 9) ^(3/2)

Is this correct? Thank you.
 
Justabeginner said:
Oh wow, I see what I did wrong. So my final answer turns out to be

4/3 * pi (R^2 - 9) ^(3/2)

Is this correct? Thank you.

I didn't notice you left off the square root over the ##R^2-x^2## in your integral formula, but apparently you had it correct on your paper since that looks like the right answer.
 
Great, thank you so much! I feel more confident about how to do these problems now :)
 
Thank you for your insight on that. I will keep that in mind if I ever come across such a problem again.