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Homework Help: Integration By Parts: Volume - help

  1. Feb 11, 2010 #1
    1. The problem statement, all variables and given/known data
    Use the method of cylindrical shells to find the volume generated by rotating the region R bounded by the curves y=e1.6 x, y=e−1.6 x and x=0.6 about the y-axis.


    2. Relevant equations
    V=[tex]$\displaystyle \Large \int _a^c 2pix (yt - yb) dx$[/tex]

    3. The attempt at a solution
    [tex]$\displaystyle \Large 2pi * \int _0^{0.6} x(e^{1.6} - e^{-1.6})dx$[/tex]

    is this right?
    if yes, how do i find the antiderivative of it? i got an answer but its wrong for some reason..
     
  2. jcsd
  3. Feb 11, 2010 #2

    Mark44

    Staff: Mentor

    You're close, but you left off the x parts in your exponential functions. This is what you want:
    [tex] 2\pi * \int _0^{0.6} x(e^{1.6x} - e^{-1.6x})dx[/tex]

    I would split these into two integrals and use integration by parts on each one.
     
  4. Feb 11, 2010 #3

    Mentallic

    User Avatar
    Homework Helper

    Your integral is wrong.
    Watch and learn :smile:

    http://img28.imageshack.us/img28/640/volumeintegration.png [Broken]
    Notice how [itex]y=e^{1.6x}[/itex] and [itex]y=e^{-1.6x}[/itex] are symmetrical about the y-axis. This means we only need to deal with one side of the axis. Let's take the positive side.

    http://img163.imageshack.us/img163/613/cylinder.png [Broken]
    We want to find the volume of the grey glittery bit.

    [tex]r=x[/tex]

    [tex]R=r+\delta x=x+\delta x[/tex]

    [tex]h=y=e^{-1.6x}[/tex]

    Now the volume is
    [tex]\delta V=\pi \left(R^2-r^2\right)h[/tex]

    and substituting into this formula and simplifying we have
    [tex]\delta V=2\pi x e^{-1.6x} \delta x[/tex]

    (we ignore the [itex](\delta x)^2[/itex] since this is a negligible amount)

    So you can now set up your integral accordingly and use integration by parts to solve it. Note that we could have also taken the other side of the x-axis so instead we have [itex]h=e^{1.6x}[/itex] and the limits in our integral will be from -0.6 to 0 instead of 0 to 0.6.
     
    Last edited by a moderator: May 4, 2017
  5. Feb 12, 2010 #4

    Mark44

    Staff: Mentor

    Mentallic,
    I disagree. Slimsta's problem statement says that the region to be rotated is bounded by the two exponential curves and the line x = 0.6. So the region R is above your green-shaded region, and below the red curve, and bounded on the right by the vertical line.
     
  6. Feb 12, 2010 #5

    Mentallic

    User Avatar
    Homework Helper

    Oh my! Yes you're right, I was completely off.

    Now it makes sense as to why the question would bother adding in that other exponential curve :blushing:
     
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