Answer
$I_x= \dfrac{M(b^2+c^2)}{12}$, $I_y= \dfrac{M(a^2+c^2)}{12}$, $I_z= \dfrac{M(a^2+b^2)}{12}$
Work Step by Step
Let us consider that $I_x=\iiint_{E} (y^2+z^2) \rho(x,y,z) dV=k\int_{-c/2}^{c/2} \int_{-b/2}^{b/2}\int_{-a/2}^{a/2}(y^2+z^2) dx dy dz $
and $k\int_{-c/2}^{c/2} \int_{-b/2}^{b/2}(ay^2+az^2) dy dz =k \int_{-c/2}^{c/2} (\dfrac{ay^3}{3}+az^2y)_{-b/2}^{b/2} dz $
and $k \int_{-c/2}^{c/2} (\dfrac{ab^3}{12}+abz^2) dz =k[\dfrac{ab^3}{12}z+\dfrac{ab}{3}z^3]_{-c/2}^{c/2}$
or, $I_x=k(\dfrac{ab^3z}{12}+\dfrac{abz^3}{3})_{-c/2}^{c/2}= \dfrac{kabc(b^2+c^2)}{12}$
Here, we have $M=kabc$
Therefore, $I_x= \dfrac{M(b^2+c^2)}{12}\\I_y= \dfrac{M(a^2+c^2)}{12}\\I_z= \dfrac{M(a^2+b^2)}{12}$
