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Volume: is this ok?

  1. Jan 15, 2006 #1
    Hi,

    I have to

    Compute volume of the solid

    [tex]
    A = \left\{ [x,y,z] \in \mathbb{R}^3, 0 \leq z, x+ y+z \leq 1, z \leq xy, x \geq 0, y \geq 0\right\}
    [/tex]

    I draw it..and the important step is to find out, where is [itex]z[/itex] bounded with either of [itex]x+ y+z \leq 1[/itex] or [itex]z \leq xy[/itex]. To find out the dividing points, in which the second of those inequalities gets the rule, I did as follows:

    [tex]
    1-x-y = xy
    [/tex]

    [tex]
    x(y+1)=1-y
    [/tex]

    [tex]
    x = \frac{1-y}{1+y}
    [/tex]

    So for fixed [itex]y[/itex], if

    [tex]
    x \leq \frac{1-y}{1+y}
    [/tex]

    then [itex]z[/itex] is bounded by [itex]z \leq xy[/itex]. In the remaining area, [itex]x+ y+z \leq 1[/itex] sets the upper bound for [itex]z[/itex].

    So it gives me two integrals, the sum of which will be the volume I am supposed to get:

    [tex]
    I = \iiint_{A}1 \ dx\ dy\ dz = I_1 + I_2 = \int_{0}^{1}\int_{0}^{\frac{1-y}{1+y}}\int_{0}^{xy}1\ dz\ dx\ dy\ +\ \int_{0}^{1}\int_{\frac{1-y}{1+y}}^{1}\int_{0}^{1-x-y} 1\ dz\ dx\ dy
    [/tex]

    If this is correct approach, then in the official solution on web there is a mistake, since these two integrals I can already check in Maple and I computed them right.

    Is this ok?
     
    Last edited: Jan 15, 2006
  2. jcsd
  3. Jan 16, 2006 #2

    Dale

    Staff: Mentor

    I think your upper limit on x for the second integral should be 1-y instead of 1
    [tex]
    \int_{0}^{1}\int_{0}^{\frac{1-y}{1+y}}\int_{0}^{xy}1\ dz\ dx\ dy\ +\ \int_{0}^{1}\int_{\frac{1-y}{1+y}}^{1-y}\int_{0}^{1-x-y} 1\ dz\ dx\ dy
    [/tex]
    -Dale
     
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