How can we find the volume by integration for $x>0, y>0, z>0$ and $z^2<x+y<2z$?

Question

Intersection of formulas

I am trying to find the volume but I can't use the methods in most videos where I should let every 2 variables be zero and solve for the third since there are no values to start with.

Answer 1

All points of this solid $S$ are lying over points in the first quadrant of the $(x,y)$-plane. For each such point $(x,y)$ we have $$(x,y,z)\in S\quad\Leftrightarrow\quad {x+y\over2}<z<\sqrt{x+y}\ .$$ It turns out that we need $x+y<2\sqrt{x+y}$, which then implies $x+y<4$. It follows that our domain of integration $S'$ is given by $$S'=\bigl\{(x,y)\bigm| x>0, \ y>0, \ x+y<4\bigr\}\ ,$$ and we have $${\rm vol}(S)=\int_{S'}f(x,y)\>{\rm d}(x,y)=\int_0^4\int_0^{4-x}f(x,y)\>dy\>dx\ ,$$ whereby $$f(x,y):=\sqrt{x+y}-{x+y\over2}\ .$$