Answer
$\int_{-2}^2\int_{x^2}^4\int_0^{2-y/2} f(x,y,z)dzdydx$ and $\int_{0}^{4}\int_{-\sqrt{y}}^{\sqrt{y}}\int_0^{2-y/2} f(x,y,z) dzdxdy$
$\int_{0}^{2} \int_{0}^{4-z} \int_{-\sqrt{y}}^{\sqrt{y}} f(x,y,z)dxdydz$ and $\int_{0}^{4}\int_0^{2-y/2} \int_{-\sqrt{y}}^{\sqrt{y}}f(x,y,z) dxdzdy$
$\int_{-2}^{2} \int_{0}^{2-x^2/2} \int_{x^2}^{4-2z} f(x,y,z)dydzdx$ and $\int_{0}^{2}\int_{-\sqrt{4-2z}}^{\sqrt{4-2z}} \int_{x^2}^{4-2z} f(x,y,z) dydxdz$
Work Step by Step
Case 1: In xy palne: Set $z=0$
Thus, we have $y=x^2, z=4$
Therefore, $\int_{-2}^2\int_{x^2}^4\int_0^{2-y/2} f(x,y,z)dzdydx$ and $\int_{0}^{4}\int_{-\sqrt{y}}^{\sqrt{y}}\int_0^{2-y/2} f(x,y,z) dzdxdy$
Case 2: In yz palne: Set $x=0$
Thus, we have $y=0, z, z=2-\dfrac{y}{2}$
Therefore, $\int_{0}^{2} \int_{0}^{4-z} \int_{-\sqrt{y}}^{\sqrt{y}} f(x,y,z)dxdydz$ and $\int_{0}^{4}\int_0^{2-y/2} \int_{-\sqrt{y}}^{\sqrt{y}}f(x,y,z) dxdzdy$
Case 3: In xz palne: Set $y=0$
Thus, we have $x=0, z=0, z=2$
Therefore, $\int_{-2}^{2} \int_{0}^{2-x^2/2} \int_{x^2}^{4-2z} f(x,y,z)dydzdx$ and $\int_{0}^{2}\int_{-\sqrt{4-2z}}^{\sqrt{4-2z}} \int_{x^2}^{4-2z} f(x,y,z) dydxdz$
