Answer
${I_x} = 0$ and ${I_0} = 0$
Work Step by Step
Using the information in Exercise 29, we have the region ${\cal R}$, a rectangle $\left[ { - a,a} \right] \times \left[ {b, - b} \right]$.
We have the mass density of $\delta \left( {x,y} \right) = x$.
Evaluate the moment of inertia relative to the $x$-axis:
${I_x} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {y^2}\delta \left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = - a}^a \mathop \smallint \limits_{y = - b}^b {y^2}\delta \left( {x,y} \right){\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = - a}^a \mathop \smallint \limits_{y = - b}^b x{y^2}{\rm{d}}y{\rm{d}}x$
$ = \left( {\mathop \smallint \limits_{x = - a}^a x{\rm{d}}x} \right)\left( {\mathop \smallint \limits_{y = - b}^b {y^2}{\rm{d}}y} \right)$
$ = \frac{1}{6}\left( {{x^2}|_{ - a}^a} \right)\left( {{y^3}|_{ - b}^b} \right) = \frac{1}{6}\left( 0 \right)\left( {2{b^3}} \right) = 0$
Evaluate the moment of inertia relative to the $y$-axis:
${I_y} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {x^2}\delta \left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = - a}^a \mathop \smallint \limits_{y = - b}^b {x^2}\delta \left( {x,y} \right){\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = - a}^a \mathop \smallint \limits_{y = - b}^b {x^3}{\rm{d}}y{\rm{d}}x$
$ = \left( {\mathop \smallint \limits_{x = - a}^a {x^3}{\rm{d}}x} \right)\left( {\mathop \smallint \limits_{y = - b}^b {\rm{d}}y} \right)$
$ = \frac{1}{4}\left( {{x^4}|_{ - a}^a} \right)\left( {y|_{ - b}^b} \right)$
Since $\mathop \smallint \limits_{x = - a}^a {x^3}{\rm{d}}x = \frac{1}{4}\left( {{x^4}|_{ - a}^a} \right) = 0$, so, ${I_y} = 0$.
The polar moment of inertia, ${I_0}$
${I_0} = {I_x} + {I_y} = 0$
