Calculating the volume with double integral

Question

Hello i am trying to calculate the volume for a double integral but i am having problem with define the integral because it is not given in a pure form. I have $z = xy$, $x+y+z=1$ $z=0$ my approach is to set the function for a integral to be $$\int_Dxy$$ and to find the $limits for$ $dy$ i set $z$ to be zero it is also given by definition and i get $y = 1-x$ after that i set both $z$ and $y$ to zero and i get $x = 1$ so i have the following limits $$\int_0^1 \int_0^{x-1}xy$$ but i am not getting the right answer after evaluating the integral. What confusing me here is that the integral is not given by default here also the other thing that confuses me is i have the same problem but to be solved with triple integral. I am thinking maybe for the volume i just need $dydx$ without a function but i am not sure. Thank you for any help in advance.

Answer 1

Note that
$$1-x-y\geq xy\geq 0\quad \text{in $D_1=\left\{y\in \left[0,\frac{1-x}{1+x}\right],x\in[0,1]\right\}$}$$
and $$0\leq 1-x-y\leq xy\quad \text{in $D_2=\left\{y\in \left[\frac{1-x}{1+x},1-x\right],x\in[0,1]\right\}$}.$$ Therefore, the volume is given by $$\begin{align}V&=\iint_{D_1}xy dxdy+\iint_{D_2}(1-x-y) dxdy\\ &=\int_0^1\int_0^{\frac{1-x}{1+x}}xy dydx+ \int_0^1\int_{\frac{1-x}{1+x}}^{1-x}(1-x-y) dydx. \end{align}.$$ The final result should be $17/12-2\ln(2)$.

Answer 2

The limits of integration in your $$\int_0^1 \int_0^{x-1}xy$$ are not correct.

Note that you have to intersect $$x+y+z=1$$ with $$z=xy$$ which gives you $$ y=\frac {1-x}{1+x}$$

Thus the limits of integration are $$\int_0^1 \int_0^{\frac {1-x}{1+x}}xy dydx$$