Answer
$\int_{0}^{1}\int_{\sqrt x}^{1} \int_0^{1-y} f(x,y,z)dzdydx$ and $\int_{0}^{1}\int_{0}^{y^2} \int_0^{1-y} f(x,y,z)dzdxdy$
$\int_{0}^{1}\int_{0}^{1-\sqrt x} \int_{\sqrt x}^{1-z} f(x,y,z)dydzdx$ and $\int_{0}^{1}\int_{0}^{(1-z)^2} \int_{\sqrt x}^{1-z} f(x,y,z)dydxdz$
$\int_{0}^{1}\int_{0}^{1-z} \int_{0}^{y^2} f(x,y,z)dxdydz$ and $\int_{0}^{1}\int_{0}^{1-y} \int_{0}^{y^2} f(x,y,z)dxdzdy$
Work Step by Step
Given: $z=1-y; y=\sqrt x$
Case 1: Here, we have the bounds for $x$ is from $0$ to $1$ and for $y$ it is from $\sqrt x$ to $1$
Therefore, $\int_{0}^{1}\int_{\sqrt x}^{1} \int_0^{1-y} f(x,y,z)dzdydx$ and $\int_{0}^{1}\int_{0}^{y^2} \int_0^{1-y} f(x,y,z)dzdxdy$
Case 2: Here, we have the bounds for $x$ is from $0$ to $1$ and for $y$ it is from $1-z$ to $\sqrt x$ and for $z$ from $0$ to $1-\sqrt x$
Therefore, $\int_{0}^{1}\int_{0}^{1-\sqrt x} \int_{\sqrt x}^{1-z} f(x,y,z)dydzdx$ and $\int_{0}^{1}\int_{0}^{(1-z)^2} \int_{\sqrt x}^{1-z} f(x,y,z)dydxdz$
Case 3: Here, we have the bounds for $x$ is from $0$ to $1$ and for $y$ it is from $0$ to $1-z$ and for $z$ from $0$ to $y^2$
Therefore, $\int_{0}^{1}\int_{0}^{1-z} \int_{0}^{y^2} f(x,y,z)dxdydz$ and $\int_{0}^{1}\int_{0}^{1-y} \int_{0}^{y^2} f(x,y,z)dxdzdy$
