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Homework Help: Can't get this one

  1. Oct 10, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the exact volume of the region bounded by:

    y = e^3x
    y = e^x

    Revolved around x axis

    2. Relevant equations

    3. The attempt at a solution

    The partitions will be disks of radius e^3x - e^x and height of delta x.

    Therefore, the volume of a partition is

    pi(e^3x - e^x)^2 delta x

    Integrating this from 0 to 1 gives me a final integral of:

    [itex] \pi \int^{1}_{0} e^{6x} - e^{2x} dx[/itex]

    Which evaluated for exact gives me:

    [itex]\pi(\frac{e^{6} - 1}{6} - \frac{e^{2} + 1}{2})[/itex]

    But I am incorrect. Why?
  2. jcsd
  3. Oct 10, 2011 #2


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    They're (vertical) washers, with outer radius e3x and inner radius ex.

    As it turns out, that's what you have in your integral expression. (But two wrongs don't make a right.)

    Check your signs when you substitute the limits of integration into your anti-derivative.
  4. Oct 10, 2011 #3


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    No, they won't. In fact there are NO "disks". The region between the two graphs form a "washer". One way to do this is to find the volume wnem the disk formed by [itex]e^{3x}[/itex] is rotated around the x-axis, find the volume when the disk formed by [itex]e^x[/itex] is rotated around the x-axis, and then subtracting

    No. [itex]\pi\int e^{3x}dx- \pi\int e^x dx= \pi \int (e^3x- e^x) dx[/itex]

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