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Homework Help: Triple integral for cone in cylindrical coordinates.

  1. Jun 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Find limits of integration for volume of upside down cone with vertex on origin and base at z=1/sqrt(2). Angle at vertex is pi/2. Do this in cylindrical coordinates.

    2. Relevant equations

    3. The attempt at a solution
    My inner integral conflicts with the books solution. So in my triple integral, the outsides are right: 0<theta<2pi ; 0<r<1/sqrt(2)

    But my inner integral is r<z<1/sqrt(2) and the book says its r<z<1.

    Where does this 1 come from? I thought the max height was 1/sqrt(2)?

    Sorry if my formats messed up. Typing this on my phone.
  2. jcsd
  3. Jun 22, 2012 #2
  4. Jun 22, 2012 #3


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    It looks to me like you did the limits of integration correctly.
  5. Jun 22, 2012 #4
    Yea. After looking at that link, I think the books solution is wrong. Thanks. I spent like an hour trying to figure out what I did wrong. Ill ask my professor to make sure.
  6. Jun 22, 2012 #5
    I've followed my link above and I think your limits for z are r < z < 1/√2

    My triple integral is

    int (0 to 1/√2) r dr int (0 to 2∏) dθ int (r to 1/√2) dz
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