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Volume using the disk method

  1. Mar 4, 2010 #1
    1. The problem statement, all variables and given/known data
    Use the functions:
    y=x+1, y=0, x=0, x=2

    Find the volume when rotated around y=3
    Find the volume when rotated around y=-1

    2. Relevant equations
    PI[tex]\int[/tex] R2dx or perhaps PI[tex]\int[/tex] R2 - r2dx

    3. The attempt at a solution

    i don't think that this is a particularly hard question, but I just can't seem to get the right answer. (perhaps I'm making a silly mistake)

    So first I graphed the equations, all in the first quadrant, vertical line at x=2. bounded region looks like a trapezoid.

    attempt:
    PI[tex]\int[/tex] (from 0-2) of [3-(x+1)]2 dx

    however the answer is 46pi/3 and I get 8pi/3 <-- very obviously wrong.
    is this a washer method problem?

    for the next, PI[tex]\int[/tex] (from 0-2) of [1+(x+1)]2 dx

    i get 32pi/3 and the answer is 50pi/3


    thanks :)
     
  2. jcsd
  3. Mar 5, 2010 #2

    Mark44

    Staff: Mentor

    It's the second one.
    The volume of your typical volume element is [itex]\Delta V = \pi (R^2 - r^2) \Delta x[/itex]. It's not [itex]\Delta V = \pi (R - r)^2 \Delta x[/itex]. Do you see what you're doing wrong?
     
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