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Math Help - Volume from revolved area?

  1. #1
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    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?

    Volume from revolved area?-arearevolve.png
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  2. #2
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    Re: Volume from revolved area?

    Quote Originally Posted by Bwts View Post
    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?

    Click image for larger version. 

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    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 \displaystyle x and width \displaystyle \Delta y , (some small change in y).

    When you rotate the strips, they create cylinders, each of radius \displaystyle x and depth \displaystyle \Delta y.

    So the volume of each strip is \displaystyle \pi x^2 \Delta y, 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*}

    As you increase the number of subdivisions, i.e. make \displaystyle n \to \infty and \displaystyle \Delta y \to 0, this sum converges to an integral, and the approximation becomes exact.

    So \displaystyle V = \int_a^b{\pi \left[f(y)\right]^2\,dy}, where \displaystyle a, b are your y endpoints.
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  3. #3
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    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


    Volume from revolved area?-area_eq.png
    Last edited by Bwts; November 11th 2011 at 07:35 AM.
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    Re: Volume from revolved area?

    Quote Originally Posted by Bwts View Post
    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 View Post
    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

    \displaystyle V = \int_a^b{2\pi x f(x) 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 View Post
    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
    Last edited by tom@ballooncalculus; November 11th 2011 at 08:09 AM.
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  5. #5
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    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?
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  6. #6
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    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).
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  7. #7
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    Re: Volume from revolved area?

    Ahh yes I get you now...thanks Tom
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