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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?

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# Homework Help: Volume: is this ok?

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