Answer
$I_x= \dfrac{M(b^2+c^2)}{12}$, $I_y= \dfrac{M(a^2+c^2)}{12}$, $I_z= \dfrac{M(a^2+b^2)}{12}$
$M=kabc$
Work Step by Step
We have $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 $
$=k \int_{-c/2}^{c/2} (\dfrac{ay^3}{3}+az^2y)_{-b/2}^{b/2} dz $
$=k \int_{-c/2}^{c/2} (\dfrac{a(b/2+b/2)^3}{3}+az^2(b/2+b/2)dz $
$=(\dfrac{ab^3zk}{12}+\dfrac{abz^3k}{3})_{-c/2}^{c/2}$
$=(\dfrac{ab^3(c/2+c/2)k}{12}+\dfrac{ab(c/2+c/2)^3k}{3})$
$= \dfrac{kabc(b^2+c^2)}{12}$
Suppose $M=kabc$
Then , we get $I_x= \dfrac{M(b^2+c^2)}{12}$
and $I_y= \dfrac{M(a^2+c^2)}{12}$ and $I_z= \dfrac{M(a^2+b^2)}{12}$
