Answer
$\int_{0}^{1}\int_{\sqrt x}^{1} \int_0^{1-y} f(x,y,z)dzdydx\\ \int_{0}^{1}\int_{0}^{y^2} \int_0^{1-y} f(x,y,z)dzdxdy\\\int_{0}^{1}\int_{0}^{1-\sqrt x} \int_{\sqrt x}^{1-z} f(x,y,z)dydzdx\\\int_{0}^{1}\int_{0}^{(1-z)^2} \int_{\sqrt x}^{1-z} f(x,y,z)dydxdz\\\int_{0}^{1}\int_{0}^{1-z} \int_{0}^{y^2} f(x,y,z)dxdydz\\\int_{0}^{1}\int_{0}^{1-y} \int_{0}^{y^2} f(x,y,z)dxdzdy$
Work Step by Step
The limits for $x$ is from $0$ to $1$ ; $y$ from $\sqrt x$ to $1$
This gives us: $\int_{0}^{1}\int_{\sqrt x}^{1} \int_0^{1-y} f(x,y,z)dzdydx\\ \int_{0}^{1}\int_{0}^{y^2} \int_0^{1-y} f(x,y,z)dzdxdy$
When the limits for $x$ from $0$ to $1$ ; $y$ from $1-z$ to $\sqrt x$ and $z$ from $0$ to $1-\sqrt x$
This gives us: $\int_{0}^{1}\int_{0}^{1-\sqrt x} \int_{\sqrt x}^{1-z} f(x,y,z)dydzdx\\\int_{0}^{1}\int_{0}^{(1-z)^2} \int_{\sqrt x}^{1-z} f(x,y,z)dydxdz$
When the limits for $x$ is from $0$ to $1$ and for $y$ it is from $0$ to $1-z$ and for $z$ from $0$ to $y^2$
This gives us: $\int_{0}^{1}\int_{0}^{1-z} \int_{0}^{y^2} f(x,y,z)dxdydz\\\int_{0}^{1}\int_{0}^{1-y} \int_{0}^{y^2} f(x,y,z)dxdzdy$
