How Do You Calculate the Volume of This Complex Solid in 3D Space?

[twoflower]

Hi,

I have to

Compute volume of the solid

<br /> 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\}<br />

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

<br /> 1-x-y = xy<br />

<br /> x(y+1)=1-y<br />

<br /> x = \frac{1-y}{1+y}<br />

So for fixed y, if

<br /> x \leq \frac{1-y}{1+y}<br />

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

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

<br /> 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<br />

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?

 

[Dale]

I think your upper limit on x for the second integral should be 1-y instead of 1
<br /> \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<br />
-Dale