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Homework Help: Finding the volume with Shell method

  1. Sep 12, 2005 #1
    the volume that i get is a negative value, there must be something wrong with my working.
    is shell height = y + 2 - y^2
    or i am wrong?
    tq.
     

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  3. Sep 13, 2005 #2

    Fermat

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    The shell Method:
    When taking a volume of revolution about the y-axis, the formula is,

    [tex]V = \int_{x_0} ^{x_1} xy \ dx[/tex]

    where x is the radius of the shell, dx represents the shell thickness and y is the length, or height, of the shell.

    Since, in this case, you are taking negative y-values, then you will get a negative answer for the volume. Simply change the sign.

    You have two curves to contend with, so you should make up the integral like this,

    [tex]V = \int_0 ^1 xy_1 \ dx + \int_1 ^2 xy_2 \ dx[/tex]

    [tex]\mbox{where}\ y_1\ \mbox{is}\ \sqrt{x}\ \mbox{and}\ y_2\ \mbox{is}\ (x-2).[/tex]
     
  4. Sep 13, 2005 #3
    thank you, but im a bit confused.
    im doing some self-study here, and the formula they wrote in the book is what i wrote in the first line.
    btw, the x-axis is the axis of rotation.

    nvm. i try to figure it out again :)
     
  5. Sep 13, 2005 #4
    ...don't you mean:

    [tex]V=\mathbf{2\pi}\int_{a}^{b}xf(x)dx[/tex]
     
  6. Sep 13, 2005 #5

    Fermat

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    yes,

    thanks for the correction,

    y = f(x)

    and I completely missed out the 2pi !!
     
  7. Sep 13, 2005 #6

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    Sorry about the confusion there. I'm afraid I misread your work. Actually, I'd never heard of the shell method before.

    You working is correct. The shell height/length is (y + 2 - y²). And you will have gotten a volume of -5pi/6 units, yes?
    I got a volume of +5pi/6 units using my volume of revolution method - which I've now found out is called the disk method or washer method.

    You are getting a negative value simply because you are using negative y-values for the radius, but +ve values for the shell height.
     
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