How can I create a triple integration for this question

Question

How can I solve this Problem using triple integration?

Find the volume in the first octant bounded by the planes $x+z=1$, and $y+2z=2$?

I don't want the whole solution I just want the triple integration.

Answer 1

You know that $x\geq0$, $y\geq0$, and $z\geq0$ since the volume is contained in the first octant. Solving for $x$ in your first equation gives $x=1-z$, and similarly solving for $y$ gives $y=2(1-z)$. Since we know that $x\geq0$, we can use this information to find the upper limit for $z$: $$0\leq1-z\Longrightarrow z\leq1$$ So your region is the set $$R=\{(x,y,z)\in\mathbb{R}^3\mid0\leq x\leq1-z,0\leq y\leq2(1-z),0\leq z\leq1\}$$ and the volume of the region is: $$\iiint_R\,dV=\int_0^1\int_0^{2(1-z)}\int_0^{1-z}\,dx\,dy\,dz$$