Rewrite order of $\int_0^1\int_0^{\sqrt{y}} \int_y^1 \, dz \, dx \, dy$ to $dx\,dy\,dz$ and $dy\,dz\,dx$

82 Views Asked by At

I need to change the order of $$\int_0^1\int_0^{\sqrt{y}}\int_y^1\,dz\,dx\,dy$$ to $dx\,dy\,dz$ and $dy\,dz\,dx.$

I can extract the inequalities to get $1≤z≤y$, $0≤x≤\sqrt y$, $0≤y≤1$, but I get confused when it comes to rewriting the bounds.

I know that I start with the outermost bound, so take $dx\,dy\,dz$, I think $0≤z≤1$. I'm not sure if that's right, but that's what I think I get when I take the smallest possible lowest bound the the largest possible upperbound. Then I don't know how to continue