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Volume of a Pond (Triple Integral)

  1. Apr 1, 2016 #1
    1. The problem statement, all variables and given/known data
    A circular pond with radius 1 metre and a maximum depth of 1 metre has the shape of a paraboloid, so that its depth z is z = x 2 + y 2 − 1. What is the total volume of the pond? How does this compare with the case where the pond has the same radius but has the shape of a hemisphere?

    2. Relevant equations
    n/a

    3. The attempt at a solution
    \int _{-1}^0\left(\int _{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\left(\int _{x^2+y^2-1}^0\left(\right)dz\right)dy\right)dx

    I was just wondering if this triple integral accurately describes what the question is asking? I searched online but can't find any anything similar.
     
    Last edited: Apr 1, 2016
  2. jcsd
  3. Apr 1, 2016 #2

    BvU

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    Now (with the ##\$\$## around the ##\TeX##) it looks OK. You can see the bounds sitting there properly.
     
  4. Apr 1, 2016 #3

    blue_leaf77

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    Doesn't seem to work. If you want to integrate in Cartesian variables, you should identify the correct limits for each variable. I suggest that you work in cylindrical coordinate instead, because in this case there is only one variable whose integration limits are a function of the other two variables.
     
  5. Apr 1, 2016 #4
    So in cylindrical coordinates if we integrate in the order z,r,θ
    would the limits be r-1 to 0,-r to r and 0 to 2*pi?

    EDIT
    or should it be r-1 to 0,0 to r and 0 to 2*pi?
     
  6. Apr 1, 2016 #5

    BvU

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    Neither. z is not r-1 to 0 (that would be a cone)

    (PS post #2 only said something about the look of things...., not the solubility)
     
  7. Apr 1, 2016 #6
    is it r^2 -1 to 0 ,0 to r and 0 to 2*pi?
     
  8. Apr 1, 2016 #7

    blue_leaf77

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    For your purpose, it will be helpful to think of the pond as being made of a stack of infinitesimally thick disks put on top of another in z direction such that they form a paraboloidal volume. Now consider a disk located at a fixed plane ##z##. The thickness of this disk is ##dz##. For the area of this disk, how will you calculate it using the integral in polar coordinate ##r##, ##\theta##? To do this, first you need to express the radius of this disk as a function of ##z##.
     
  9. Apr 1, 2016 #8
    Thanks guys,
    I realized in my original integral x should have been from -1 to 1 and not 0 to 1. also in polar coordinates if I use is it r^2 -1 to 0 ,0 to 1 and 0 to 2*pi, I get the same answer of (1/2)*pi
     
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