Answer
The six different integrals are: $\int_{-2}^2\int_{x^2}^4\int_0^{2-y/2} f(x,y,z)dzdydx$ 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
a) In xy palne: put $z=0$
This implies that $y=x^2, z=4$
Thus, $\int_{-2}^2\int_{x^2}^4\int_0^{2-y/2} f(x,y,z)dzdydx$ and $\int_{0}^{4}\int_{-\sqrt{y}}^{\sqrt{y}}\int_0^{2-y/2} f(x,y,z) dzdxdy$
b) In yz palne: put $x=0$
This implies that $y=0, z, z=2-\dfrac{y}{2}$
Thus, $\int_{0}^{2} \int_{0}^{4-z} \int_{-\sqrt{y}}^{\sqrt{y}} f(x,y,z)dxdydz$ and $\int_{0}^{4}\int_0^{2-y/2} \int_{-\sqrt{y}}^{\sqrt{y}}f(x,y,z) dxdzdy$
c) In xz palne: put $y=0$
This implies that $x=0, z=0, z=2$
Thus, $\int_{-2}^{2} \int_{0}^{2-x^2/2} \int_{x^2}^{4-2z} f(x,y,z)dydzdx$ and $\int_{0}^{2}\int_{-\sqrt{4-2z}}^{\sqrt{4-2z}} \int_{x^2}^{4-2z} f(x,y,z) dydxdz$
Hence, The six different integrals are: $\int_{-2}^2\int_{x^2}^4\int_0^{2-y/2} f(x,y,z)dzdydx$ 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 $\int_{0}^{2}\int_{-\sqrt{4-2z}}^{\sqrt{4-2z}} \int_{x^2}^{4-2z} f(x,y,z) dydxdz$
