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
We are given that $z=1-y; y=\sqrt x$
a) when$x$ is from $0$ to $1$ and for $y$ it is from $\sqrt x$ to $1$
Thus, $\int_{0}^{1}\int_{\sqrt x}^{1} \int_0^{1-y} f(x,y,z)dzdydx$ and $\int_{0}^{1}\int_{0}^{y^2} \int_0^{1-y} f(x,y,z)dzdxdy$
b) when $x$ is from $0$ to $1$ and for $y$ it is from $1-z$ to $\sqrt x$ and for $z$ from $0$ to $1-\sqrt x$
Thus, $\int_{0}^{1}\int_{0}^{1-\sqrt x} \int_{\sqrt x}^{1-z} f(x,y,z)dydzdx$ and $\int_{0}^{1}\int_{0}^{(1-z)^2} \int_{\sqrt x}^{1-z} f(x,y,z)dydxdz$
c) when $x$ is from $0$ to $1$ and for $y$ it is from $0$ to $1-z$ and for $z$ from $0$ to $y^2$
Thus, $\int_{0}^{1}\int_{0}^{1-z} \int_{0}^{y^2} f(x,y,z)dxdydz$ and $\int_{0}^{1}\int_{0}^{1-y} \int_{0}^{y^2} f(x,y,z)dxdzdy$
Hence, The six different integrals are:
$\int_{0}^{1}\int_{\sqrt x}^{1} \int_0^{1-y} f(x,y,z)dzdydx\\ \int_{0}^{1}\int_{0}^{y^2} \int_0^{1-y} f(x,y,z)dzdxdy\\
\int_{0}^{1}\int_{0}^{1-\sqrt x} \int_{\sqrt x}^{1-z} f(x,y,z)dydzdx \\ \int_{0}^{1}\int_{0}^{(1-z)^2} \int_{\sqrt x}^{1-z} f(x,y,z)dydxdz\\\int_{0}^{1}\int_{0}^{1-z} \int_{0}^{y^2} f(x,y,z)dxdydz\\\int_{0}^{1}\int_{0}^{1-y} \int_{0}^{y^2} f(x,y,z)dxdzdy$
