Find the volume of the region by integration

Question

I have to find the volume of the region bounded by $x + z = 1$;$ y + 2z = 2$;$ x = 0$;$ y = 0 $;$ z = 0$;

I tried to sketch the graph separately in the $y-z$ plane and then in $x-z$ plane. But I am stuck on the $xy$ plane. I know that the limits for $z$ for integration are from $0$ to $1$ and I don't know how to find the $x$ and $y$ limits. How should I proceed further?

Answer 1

let $f_1(x,y)= 1-x$ and $f_2(x,y)= 1-\frac y 2 $

$f_1 =0 $ on the line $x=1$ and $f_2 =0 $ on the line $y=2$ so the region of integration is $x \in (0,1)\; ; \;y \in (0,2) $

$f_1 = f_2 \Rightarrow y=2x$

$f_1 > f_2$ when $y > 2x$ and $f_2 > f_1$ when $y < 2x$

$$ V = \int _0^2 \int _0^\frac y 2 (1-\frac y2) dx dy + \int _0^2 \int _\frac y2^1 (1-x) dx dy $$

Answer 2

Hint: You have $z<1-x$ AND $z<1-y/2$. Separate the two cases where $1-x<1-y/2$ OR $1-x>1-y/2$

$z$ will have to be bounded by the smaller of the two bounds. So you separate the xy axis by $1-x<1-y/2$ and integrate them separately