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Cal ii volumes

  1. Aug 28, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider the solid obtained by rotating the region bounded by the given curves about y=10

    Find the volume V of this solid using DISK METHOD

    2. Relevant equations

    y=5e^-x, y=5, x=4

    3. The attempt at a solution

    setting up the picture easy. but i'm having trouble knowing HOW to find the radius. any help would be cool. thank you.
  2. jcsd
  3. Aug 28, 2010 #2


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    Homework Helper

    draw a horizontal line from y=10, then draw a vertical line down & see where it intersects the region, the radius is the vertical disatnce from y=10 to the point in your region

    from that you should find an innner (ri) & outer (r0) radius, that may be both be functions of x.

    then set up the integral over x
  4. Aug 28, 2010 #3


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    corrected above after drawing it
  5. Aug 28, 2010 #4
    Okay, I think I know what you are saying. so far I got pitimes the integral from 0 to 4 of [(10-5e^(-x)]^2 -(5)^2 dx

    what do you think
  6. Aug 29, 2010 #5
    Yep, that's it!
  7. Aug 29, 2010 #6
    Dude man, I keep getting the wrong answer, or at least www.webassign.net tells me so. SO...the problem must lie within my algebra.

    I freaking hate webassign.net. You do all this work by setting up the problem, doing the calculus, etc, but it aint worth crap because you get the final answer incorrect. B.S. man, you know what I mean?

    By the way, www.webassign.net is my MANDATORY online homework I have to do. No TEXT BOOK problems were assigned this semester. They're all on webassign. Argh.
  8. Aug 29, 2010 #7
    [tex] V = \pi \int_0^4 \left[ (100 - 100e^{-x} + 25e^{-2x}) - 25 \right] dx [/tex]

    [tex] = \pi \int_0^4 25e^{-x}(e^{-x} - 4)dx \ + \ 75\pi x|_0^4 [/tex]

    Let [tex] u = e^{-x} - 4 \ \Rightarrow \ du = -e^{-x}dx. [/tex]

    So [tex] V = 300\pi - 25 \pi \int u du [/tex]

    [tex] = 300\pi - 25 \pi \left( \frac{(e^{-x} - 4)^2}{2} \right) \right|_0^4. [/tex]

    Does that help?
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