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