1 Attachment(s)
Volume from revolved area?
Hi,
Hope I can explain this well. I have a function that gives me the deflection of a circular/disk membrane as a function of the radius. From this I can get the cross sectional area/profile of the disks displacement in terms of the radius. I would now like to revolve this area to create a volume.
My problem is that when ever I look this up on the internet I find equations for the volume if I revolve the area aournd the x-axis whilst I want to revolve it around the y-axis.
My inital attempt was to invert the function tbut this has proven impossible :(
My questions are:
1) Would the volume be same reguardless of which axis I choose to revolve around?
2) Is there an better way of doing this?
Attachment 22711
Re: Volume from revolved area?
Quote:
Originally Posted by
Bwts 
Hi,
Hope I can explain this well. I have a function that gives me the deflection of a circular/disk membrane as a function of the radius. From this I can get the cross sectional area/profile of the disks displacement in terms of the radius. I would now like to revolve this area to create a volume.
My problem is that when ever I look this up on the internet I find equations for the volume if I revolve the area aournd the x-axis whilst I want to revolve it around the y-axis.
My inital attempt was to invert the function tbut this has proven impossible :(
My questions are:
1) Would the volume be same reguardless of which axis I choose to revolve around?
2) Is there an better way of doing this?
Attachment 22711
If you want to revolve the region around the y axis, first write the function as a function of y, i.e. x = f(y).
Then, imagine the area being approximated using horizontal rectangular strips.
Each rectangular strip has length 
and width 
, (some small change in y).
When you rotate the strips, they create cylinders, each of radius and depth 
.
So the volume of each strip is 
, and therefore the volume of the region is approximated by
![\displaystyle \begin{align*} V &\approx \sum{\pi x^2 \Delta y} \\ V &\approx \sum{\pi\left[f(y)\right]^2\Delta y} \end{align*}](http://latex.codecogs.com/png.latex?\displaystyle \begin{align*} V &\approx \sum{\pi x^2 \Delta y} \\ V &\approx \sum{\pi\left[f(y)\right]^2\Delta y} \end{align*})
As you increase the number of subdivisions, i.e. make 
and 
, this sum converges to an integral, and the approximation becomes exact.
So ![\displaystyle V = \int_a^b{\pi \left[f(y)\right]^2\,dy}](http://latex.codecogs.com/png.latex?\displaystyle V = \int_a^b{\pi \left[f(y)\right]^2\,dy})
, where 
are your y endpoints.
1 Attachment(s)
Re: Volume from revolved area?
Hi,
Thanks for your speedy response. My problem is getting x=f(y) see attched image.
evrything is constant except x. I keep getting a nasty looking polynomial in x which I cant reduce further
Attachment 22712
Re: Volume from revolved area?
Quote:
Originally Posted by
Bwts 1) Would the volume be same reguardless of which axis I choose to revolve around?
No. Try to imagine the two.
Quote:
Originally Posted by
Bwts 2) Is there a better way of doing this?
Yes, integrate along the x axis and rotate the vertical strips around the y axis to make hollow cylinders.
The formula is then
 dx)
... as explained admirably at Solid of revolution - Wikipedia, the free encyclopedia. I thought that page had a typo, but I was misreading 'revolving an area between curve and x-axis' as 'revolving around the x axis'. Maybe you did, too...
Quote:
Originally Posted by
Bwts My problem is that when ever I look this up on the internet I find equations for the volume if I revolve the area aournd the x-axis
Re: Volume from revolved area?
Hi Tom,
Just so I'm clear. I should intergrate my function along x then multiply this intergrated function by the original function and intergrate the product between my x limits?
Re: Volume from revolved area?
Integrate with respect to x, yes, and between your x limits, yes... not sure I follow your multiplying proposal... what you should be planning, anyway, is to integrate the product which is the result of multiplying your function (the height of each hollow cyclinder) by 2 pi x (the circumference of the same cylinder).
Re: Volume from revolved area?
Ahh yes I get you now...thanks Tom
