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Shell Method

  1. Sep 7, 2014 #1

    squelch

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    This is from a physics course, but felt more appropriate to post here.

    I just want some sanity checking on my procedure, since I'm not this far in my calculus course yet but am having to work through it anyway for physics.

    I have no idea how to approach part c, not even an inkling of where to begin. If all you give me are Google search terms, then I'll be happy.


    1. The problem statement, all variables and given/known data

    A cylinder of radius R and length L is given.

    a) Use the shell method to write the infinitesimal volume DV
    b) Integrate dv to obtain the volume of the cylinder.
    c) The density of the cylinder is given by [itex]\rho = {\rho _0}(1 - \frac{r}{R})[/itex] where [itex]{\rho _0}[/itex] is constant.

    2. Relevant equations

    NA.

    3. The attempt at a solution

    a) A cylinder can be divided into infinitesimal shells of height L and width dr. Therefore, the infinitesimal volume is given by:
    [tex]dv = 2\pi LRdr[/tex]

    b) This infinitesimal volume can then be integrated as:
    [tex]V = \int_0^R {2\pi LRdr = 2\pi LR\int_0^R {dr = 2\pi LR(\left. r \right|_0^R) = 2\pi L{R^2}} } [/tex]

    c) No idea.
     
  2. jcsd
  3. Sep 7, 2014 #2

    vela

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    Not quite. The radius of the shell is ##r##, not ##R##, so ##dv = 2 \pi r L\,dr##.

    Recall that density is mass per unit volume.
     
  4. Sep 7, 2014 #3
    The idea is to add the infinitesimal volumes of all the shells, right? But doesn't every shell have a different radius? In the equation you wrote in a) you used capital R indicating that all shells have equal radius. If you correct this, you will easily get the correct answer in b). Additionally, by giving you the density it probably wants you to find the mass of the cylinder. Doesn't the problem statement ask for this explicitly?
     
  5. Sep 7, 2014 #4

    squelch

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    So, by way of correction,

    a) A cylinder can be divided into infinitesimal shells of height L and width dr a distance r from the center axis. Therefore, the infinitesimal volume is given by:
    [tex]dv = 2\pi Lrdr[/tex]

    b) This infinitesimal volume can then be integrated as:
    [tex]V = \int_0^R {2\pi Lrdr = 2\pi L\int_0^R {rdr = 2\pi L(\frac{1}{2}\left. {{r^2}} \right|_0^R) = \pi L{R^2}} } [/tex]

    Might have caught that if I looked at the given formula for the volume of a cylinder a bit more carefully, but what can you do.
     
  6. Sep 7, 2014 #5

    squelch

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    I just noticed that I didn't include that c) was asking for the mass of the cylinder. My mistake.

    Attempting to solve for the mass of the cylinder:

    [tex]\begin{array}{l}
    \rho = \frac{{{m_c}}}{{\pi L{R^2}}} = {\rho _0}\left( {1 - \frac{r}{R}} \right)\\
    {m_c} = {\rho _0} \cdot (\pi L{R^2}) \cdot \left( {1 - \frac{r}{R}} \right)\\
    {m_c} = {\rho _0} \cdot (\pi LR) \cdot \left( {R - r} \right)
    \end{array}[/tex]
     
  7. Sep 7, 2014 #6

    vela

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    What is ##r## supposed to represent in your calculation of the mass?
     
  8. Sep 7, 2014 #7

    squelch

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    It was given as part of the density equation in the problem, so I wasn't sure how to work it away.
     
  9. Sep 7, 2014 #8

    vela

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    You're being told that the density varies with the distance from the axis. You can't use ##\rho = m/V## for the cylinder taken as a whole because that only holds for constant density. For an infinitesimal shell, however…
     
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