Answer
We prove: ${I_h} = {I_z} + M{h^2}$
Work Step by Step
Write
${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {x^2} - 2ax + {a^2} + {y^2} - 2by + {b^2}$
$ = {x^2} + {y^2} - 2ax - 2by + {a^2} + {b^2}$
Substituting in ${I_h}$
${I_h} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \left( {{{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2}} \right)\delta \left( {x,y,z} \right){\rm{d}}V$
gives
${I_h} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \left( {{x^2} + {y^2} - 2ax - 2by + {a^2} + {b^2}} \right)\delta \left( {x,y,z} \right){\rm{d}}V$
(1) ${\ \ \ \ }$ ${I_h} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \left( {{x^2} + {y^2}} \right)\delta \left( {x,y,z} \right){\rm{d}}V$
$ - 2a\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} x\delta \left( {x,y,z} \right){\rm{d}}V - 2b\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} y\delta \left( {x,y,z} \right){\rm{d}}V$
$ + \left( {{a^2} + {b^2}} \right)\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \delta \left( {x,y,z} \right){\rm{d}}V$
Using the following facts:
1. By definition:
${I_z} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \left( {{x^2} + {y^2}} \right)\delta \left( {x,y,z} \right){\rm{d}}V$
2. Since the center of mass is at the origin, so $\left( {{x_{CM}},{y_{CM}},{z_{CM}}} \right) = \left( {0,0,0} \right)$.
Thus,
${x_{CM}} = \frac{1}{M}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} x\delta \left( {x,y,z} \right){\rm{d}}V = 0$
${y_{CM}} = \frac{1}{M}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} y\delta \left( {x,y,z} \right){\rm{d}}V = 0$
These implies that
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} x\delta \left( {x,y,z} \right){\rm{d}}V = 0$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} y\delta \left( {x,y,z} \right){\rm{d}}V = 0$
3. $h = \sqrt {{a^2} + {b^2}} $, so ${h^2} = {a^2} + {b^2}$.
4. $M = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \delta \left( {x,y,z} \right){\rm{d}}V$
thus, equation (1) becomes ${I_h} = {I_z} + M{h^2}$.
This proves the Parallel-Axis Theorem.
