Volume of generated solid by rotation

Click For Summary
SUMMARY

The volume of the solid generated by rotating the curve defined by y = 1/x from x = 1 to x = 5 about the x-axis can be calculated using the formula V = π ∫ (f(x))^2 dx. The relevant integral for this problem is π ∫ (1/x)^2 dx, which simplifies to π [ -1/x ] from 1 to 5. This approach successfully yields the correct volume, confirming the effectiveness of the method discussed.

PREREQUISITES
  • Understanding of integral calculus
  • Familiarity with the concept of volume of revolution
  • Knowledge of the function y = 1/x
  • Ability to perform definite integrals
NEXT STEPS
  • Study the method of calculating volumes of solids of revolution using the disk method
  • Learn about the washer method for calculating volumes of solids with holes
  • Explore applications of integration in physics, particularly in calculating moments of inertia
  • Review the use of integral calculus in real-world engineering problems
USEFUL FOR

Students studying calculus, particularly those focusing on integral calculus and applications in geometry and physics. This discussion is also beneficial for educators seeking to clarify methods for teaching volume calculations of solids of revolution.

ombudsmansect
Messages
29
Reaction score
0

Homework Statement



Find the volume of the solid generated where y = 1/x for 1<=x<=5 is rotated about the x axis.

Homework Equations



I = r^2 dm, sqrt(1 + (dy/dx)^2) dx,

The Attempt at a Solution



So I have found the length of the curve and can henceforth find surface area and so on but I cannot figure out the step to take to get to volume, I understand that it probably isn't too difficult after the first part, so does anyone knw what the relevant formula is to solve for this, any help much appreciated :) btw solid will b the shape of a spinning top in a way ty :)
 
Physics news on Phys.org
ombudsmansect said:

Homework Statement



Find the volume of the solid generated where y = 1/x for 1<=x<=5 is rotated about the x axis.

Homework Equations



I = r^2 dm, sqrt(1 + (dy/dx)^2) dx,

The Attempt at a Solution



So I have found the length of the curve and can henceforth find surface area and so on but I cannot figure out the step to take to get to volume, I understand that it probably isn't too difficult after the first part, so does anyone knw what the relevant formula is to solve for this, any help much appreciated :) btw solid will b the shape of a spinning top in a way ty :)

Hi again Ombudsmand,

The first thing that I notice here is

I = \int r^2 dm which is the definition of the moment of ineteria..


Look this up in the Calculus bible and you will find what you are looking for..
 
Last edited:
hey :) Is ther really a calculus bible online?? lol but i can't find it in my textbook and all ones i find online are specific to certain shapes but I am sure that like the third integral or something like tht hsould do it. I can find the area under the curve then i know i have 2pi degrees of rotation but how to put that into volumetric terms within the given bounds? thanks again for helpin me out
 
ombudsmansect said:
hey :) Is ther really a calculus bible online?? lol but i can't find it in my textbook and all ones i find online are specific to certain shapes but I am sure that like the third integral or something like tht hsould do it. I can find the area under the curve then i know i have 2pi degrees of rotation but how to put that into volumetric terms within the given bounds? thanks again for helpin me out

Edwards and Penney...

Roting a solid about a fixed axis..

Should be a bell Ring :)
 
hmmm perhaps all i need is (pi) integral of (f(x))^2 dx, ill give it a shot and see what happens
 
lol! de ja vu frm reading that lol, well i used (pi) integral of (f(x))^2 dx and got the correct answer so all looks well. Thanks heaps again u r really good at helping ppl saved me twice tonite already! have a good one suz :D
 
ombudsmansect said:
lol! de ja vu frm reading that lol, well i used (pi) integral of (f(x))^2 dx and got the correct answer so all looks well. Thanks heaps again u r really good at helping ppl saved me twice tonite already! have a good one suz :D

You are welcome :)
 

Similar threads

Replies
2
Views
2K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K