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Finding the Volume

  1. Oct 27, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.

    y=http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmsy10/alpha/144/char70.png [Broken]x−1[PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmmi10/alpha/144/char3B.png [Broken] [Broken] y=0[PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmmi10/alpha/144/char3B.png [Broken] [Broken] x=5;

    about the line y = 8.
    2. Relevant equations


    3. The attempt at a solution

    I tried using the washer disk method and got the integral of:

    (pi)(8-sqrt(x-1))^2dx and integrating it from x = 0-5.

    Doesn't work.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Oct 27, 2014 #2

    mfb

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    What do you mean with "doesn't work"? That integral is certainly calculable.
     
  4. Oct 27, 2014 #3

    Dick

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    sqrt(x-1) isn't even defined at x=0. Draw a sketch of the function and boundaries and rethink this. They really are washers, not disks.
     
  5. Oct 27, 2014 #4
    I'm confused. Don't know what you mean.
     
  6. Oct 27, 2014 #5

    HallsofIvy

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    What, exactly, do you not understand? What is [tex]\sqrt{x- 1}[/tex] when x= 0 or 1/2? What does the graph of [tex]y= \sqrt{x-1}[/tex] look like?
     
  7. Oct 27, 2014 #6
    I got it with this:

    (pi)(8)^2-(pi)(8-sqrt(x-1))^2dx
     
  8. Oct 28, 2014 #7

    HallsofIvy

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    You still have completely missed the point! But I am hoping that your integration "from x= 0- 5" was a simple misprint. If x< 1, then x-1 < 0 so [itex]\sqrt{x- 1}[/itex] is not a real number. If [itex]y= \sqrt{x- 1}[/itex] then [itex]x= y^2- 1[/itex]. That is a parabola, opening to the right with vertex at (1, 0). Your integration must be from x= 1 to 5, not 0 to 5.
     
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