I am currently confused in how to find volume enclosed between two surfaces $z_1=f_1(x,y),z_2(x,y)=f_2(x,y)$
After going through some online resources, I thought the general way of doing this is:
$\iint_{D} (z_2-z_1) dxdy$ where it will be $(z_2-z_1)$ or $(z_1-z_2)$ depending on the surfaces.
$D=\{(x,y)\in \mathbb{R}^2: f_1(x,y)=f_2(x,y)$ (eliminating $z$ from the two equation)$\}$
However I got stuck while solving the problem .
Find the volume between the surfaces $x+y+2z=2,2x+y+z=4$ in the first octent.
I could figure out that it would be of the form
$$\iint _D \left\{(4-(2x+y))-\frac{(2-(x+y))}{2}\right\}\,dxdy$$
However how do I figure out $D$?
Should I take the projection of the two surfaces on the $xy$ plane and then find out the limits as we did in double integration?
Or should I equate the two surfaces and then find out the limit?
The first octant cut by the plane $x+y+2z=2$ is the tetrahedron $$T_1=\{(x,y,z):x+y+2z\leq 2, x\geq 0, y\geq0, z\geq 0\}.$$ The first octant cut by the plane $2x+y+z=4$ is the tetrahedron $$T_2=\{(x,y,z):2x+y+z\leq 4, x\geq 0, y\geq0, z\geq 0\}.$$ Notice that $T_1\subset T_2$: if $x+y+2z\leq 2$ and $x,y,z\geq 0$ then $$2x+y+z\leq 2(2-y-2z)+y+z=4-y-3z\leq 4.$$
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Hence the desired volume is $$\begin{align}\text{vol}(T_2)-\text{vol}(T_1)&= \iint_{D_2}\Big(4-2x-y\Big)dxdy -\iint_{D_1}\Big(1-\frac{x+y}{2}\Big)dxdy\\ &=\frac{16}{3}-\frac{2}{3}=\frac{14}{3} \end{align} $$ where $$D_1=\{(x,y):x+y\leq 2, x\geq 0, y\geq0\}\;\text{and}\; D_2=\{(x,y):2x+y\leq 4, x\geq 0, y\geq0\}.$$