Volume of generated solid by rotation

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..
 
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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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