Answer
${I_0} = 8000$ $kg\cdot{m^2}$
${I_x} = 4000$ $kg\cdot{m^2}$
Work Step by Step
We have the region ${\cal D}$ defined by the disk ${x^2} + {y^2} \le 16$. Since the disk has radius $4$ meters, the area is $A = 16\pi $ ${m^2}$.
The total mass $M$ is $1000$ kg. Since the mass density is uniform, so the mass density is given by
$\delta = \frac{M}{A} = \frac{{1000}}{{16\pi }} = \frac{{125}}{{2\pi }}$ $kg/{m^2}$
We choose to compute in polar coordinates, so the description of ${\cal D}$:
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le 4,0 \le \theta \le 2\pi } \right\}$
The polar moment of inertia, ${I_0}$ is given by
${I_0} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {{x^2} + {y^2}} \right)\delta \left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{r = 0}^4 \mathop \smallint \limits_{\theta = 0}^{2\pi } {r^2}\delta \left( {x,y} \right)r{\rm{d}}r{\rm{d}}\theta $
$ = \frac{{125}}{{2\pi }}\mathop \smallint \limits_{r = 0}^4 \mathop \smallint \limits_{\theta = 0}^{2\pi } {r^3}{\rm{d}}r{\rm{d}}\theta $
$ = \frac{{125}}{{2\pi }}\left( {\mathop \smallint \limits_{r = 0}^4 {r^3}{\rm{d}}r} \right)\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta } \right)$
$ = \frac{{125}}{{2\pi }}\left( {\frac{1}{4}{r^4}|_0^4} \right)\left( {2\pi } \right) = 8000$
So, ${I_0} = 8000$ $kg\cdot{m^2}$.
We observe that ${I_0} = 2{I_x}$. We verify this statement by writing ${I_0}$ in rectangular coordinates:
${I_0} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {{x^2} + {y^2}} \right)\delta \left( {x,y} \right){\rm{d}}A = \frac{{125}}{{2\pi }}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {{x^2} + {y^2}} \right){\rm{d}}y{\rm{d}}x$
${I_0} = \frac{{125}}{{2\pi }}\left( {\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {x^2}{\rm{d}}y{\rm{d}}x + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {y^2}{\rm{d}}y{\rm{d}}x} \right)$
Since ${\cal D}$ is both vertically and horizontally simple region, we can switch the order of integration in the second integral above, such that
${I_0} = \frac{{125}}{{2\pi }}\left( {\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {x^2}{\rm{d}}y{\rm{d}}x + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {y^2}{\rm{d}}x{\rm{d}}y} \right)$
Since $x$ and $y$ are dummy variables, we have
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {x^2}{\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {y^2}{\rm{d}}x{\rm{d}}y$
Therefore,
${I_0} = \frac{{125}}{{2\pi }}\left( {2\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {x^2}{\rm{d}}y{\rm{d}}x} \right) = 2\left( {\frac{{125}}{{2\pi }}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {x^2}{\rm{d}}y{\rm{d}}x} \right)$
Since ${I_x} = \frac{{125}}{{2\pi }}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {x^2}{\rm{d}}y{\rm{d}}x$, we conclude that ${I_0} = 2{I_x}$.
Thus, ${I_x} = 4000$ $kg\cdot{m^2}$.
