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Volume by cyndrical shell

  1. Feb 10, 2007 #1

    tony873004

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    I think I've got it set up correctly, but I'm stuck on the integration.

    Use the method of cylindrical shells to find the volume generated by rotating the region bounded by y=e^(-x^2), x=0, x=1, rotated about the y axis.

    [tex]
    \begin{array}{l}
    \int\limits_{\rm{0}}^{\rm{1}} {\pi r^2 h\,\,dx} = \int\limits_{\rm{0}}^{\rm{1}} {\pi x^2 e^{ - x^2 } \,\,dx} = \\
    \\
    \pi \int\limits_{\rm{0}}^{\rm{1}} {x^2 e^{ - x^2 } \,\,dx} \\
    \end{array}
    [/tex]

    I just don't know how to anti-deriv xe^(-x^2). I've tried substitution with u= -x2 with no joy. I'll bet this is easy, but it's got me stumped.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 10, 2007 #2
    So if I understand correctly you rotate a Gaussian around the y-axis and want to know the volume enclosefd up to a distance 1 from the y-axis?

    If that is the case I would reconsider the first integral you wrote down. keep in mind that you want to 'add' a lot of infinetesimally high cilinders (volume pi x^2 dy) and use the relation you have between x and y to change variables.

    Good luck!

    PS:
    how about -(1/2)e^(-x^2)...
     
  4. Feb 10, 2007 #3

    HallsofIvy

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    First, [itex]xe^{-x^2}[/itex] is simple to integrate, as Wiemster says, but your problem is integrating [itex]x^2e^{-x^2}[/itex] which is whole different matter.

    Your real problem is that your formula, [itex]\pi r^2h dx[/itex], is wrong. It can't possibly be a differential of volume- it has units of distance4! A cylindrical shell has volume [itex]2\pi r[/itex], the circumference around the shell, times h, times the thickness,dy= dx. In this case, the thickness is dr, the height is the y value, [itex]h= e^{-x^2}[/itex], and the radius is r= x.
    That does, in fact, reduce to
    [tex]\pi \int_0^1 xe^{-x^2}dx[/tex]
    and the substitution [itex]u= x^2[/itex] works nicely.
     
  5. Feb 10, 2007 #4

    tony873004

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    Sorry, I forgot to square x. I meant the anti-deriv of x^2*e^(-x^2).
     
  6. Feb 10, 2007 #5
    radius = function; in this case e^(-x^2). But is it x*e^(-x^2) or e^(-x^2)?
    h = dy or dx

    If it's x*e^(-x^2), simple u substitution if it's the other it's just it's own derivative however you get ugly numbers.
     
  7. Feb 11, 2007 #6

    tony873004

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    Thanks, everyone. I got the answer (confirmed by the back of the book). But I had to use u=-x^2 instead of x^2, or I would have had to anti-deriv e^-u, which I don't know how to do. Can that be done?

    My mistake, as Halls of Ivy pointed out was using pi r squared for circumference instead of 2pi r. oops... Last chapter we were using areas instead of shells. I guess I just got too used to using pi r squared.

    [tex]
    \begin{array}{l}
    \int\limits_{\rm{0}}^{\rm{1}} {\left[ {{\rm{circumference}}} \right]\left[ {{\rm{height}}} \right]\left[ {{\rm{thickness}}} \right]} \\
    \\
    {\rm{circumference}} = 2\pi r,\,\,\,r = x \\
    {\rm{height}} = e^{ - x^2 } \\
    {\rm{thickness}} = \Delta x \\
    \\
    \int\limits_{\rm{0}}^{\rm{1}} {2\pi rh\,\,dx} = \int\limits_{\rm{0}}^{\rm{1}} {2\pi xe^{ - x^2 } \,\,dx} = \\
    \\
    2\pi \int\limits_{\rm{0}}^{\rm{1}} {xe^{ - x^2 } \,\,dx} ,\,\,\,\,u = - x^2 ,\,\,\frac{{du}}{{dx}} = - 2x,\,\,dx = \frac{{du}}{{ - 2x}} \\
    \\
    2\pi \int\limits_{\rm{0}}^{\rm{1}} {xe^{ - x^2 } \,\,dx} = 2\pi \int\limits_{\rm{0}}^{\rm{1}} {xe^u \,\,\frac{{du}}{{ - 2x}}} = \frac{{2\pi }}{{ - 2}}\int\limits_{\rm{0}}^{\rm{1}} {e^u \,\,du} = - \pi \int\limits_{\rm{0}}^{\rm{1}} {e^u \,\,du} = \\
    \\
    \left. { - \pi e^u } \right]_0^1 = \left. { - \pi e^{ - x^2 } } \right]_0^1 = \left( { - \pi e^{ - 1^2 } } \right) - \left( { - \pi e^{ - 0^2 } } \right) = - \pi \left( {e^{ - 1} - 1} \right) = \pi \left( { - e^{ - 1} + 1} \right) =\pi \left( {1 - 1/e} \right)
    \\
    \end{array}
    [/tex]

    Now although I got the right answer, I think there's still an intermediate mistake I'm making. Once I switch from dx to du, I have to switch the limits of the integration, don't I? I've been getting lazy since I know I'm going to switch back to x before solving. How do I do that? Since my u=-x^2 in this do I just change my 0 and 1 to 0 and -1 wherever I'm using du? And if I do that, should I even bother switching back to x before solving, or just solve with u and my newly-computed limits, which gives me the same answer? I've seen it done both ways, and I'm not sure if there is an advantage of one over the other.

    Also, what's the difference between
    [tex]\begin{array}{l}
    \int_0^1 {} \\
    and \\
    \int\limits_0^1 {} \\
    \end{array}[/tex]
     
    Last edited: Feb 11, 2007
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