1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    x=0
    x=1

    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

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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]

     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook