# Volume: is this ok?

1. Jan 15, 2006

### twoflower

Hi,

I have to

Compute volume of the solid

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

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:

$$1-x-y = xy$$

$$x(y+1)=1-y$$

$$x = \frac{1-y}{1+y}$$

So for fixed $y$, if

$$x \leq \frac{1-y}{1+y}$$

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:

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

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. Jan 16, 2006

### Staff: Mentor

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