Volume of generated solid by rotation

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Homework Help Overview

The problem involves finding the volume of the solid generated by rotating the curve defined by y = 1/x over the interval [1, 5] about the x-axis. This falls within the subject area of calculus, specifically dealing with solids of revolution.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the moment of inertia and the volume of the solid, with one suggesting the use of an integral formula. There are questions about how to apply the area under the curve to find the volume, and some participants express uncertainty about the relevant formulas.

Discussion Status

There is an ongoing exploration of different approaches to the problem, with some participants suggesting specific integral forms. One participant indicates they have found a formula that works for them, while others are still seeking clarity on the steps involved.

Contextual Notes

Participants mention difficulty in locating comprehensive resources or textbooks that cover the necessary formulas for solids of revolution, indicating a potential gap in available materials for reference.

ombudsmansect
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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 :)
 
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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 :)
 

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