Answer
(a) the mass density: $\delta = \frac{M}{A} = \frac{M}{{4ab}}$
(b) ${I_x} = \frac{M}{3}{b^2}$ and ${I_0} = \frac{M}{3}\left( {{a^2} + {b^2}} \right)$
(c) the radius of gyration about the $x$-axis:
${r_g} = \frac{b}{{\sqrt 3 }}$
Work Step by Step
We have the region ${\cal R}$, a rectangle $\left[ { - a,a} \right] \times \left[ {b, - b} \right]$.
The area of ${\cal R}$ is $A = \left( {2a} \right)\cdot\left( {2b} \right) = 4ab$.
(a) Since the density is uniform, the mass density is
$\delta = \frac{M}{A} = \frac{M}{{4ab}}$
(b) Evaluate the moment of inertia relative to the $x$-axis:
${I_x} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {y^2}\delta \left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = - a}^a \mathop \smallint \limits_{y = - b}^b {y^2}\delta \left( {x,y} \right){\rm{d}}y{\rm{d}}x$
$ = \frac{M}{{4ab}}\mathop \smallint \limits_{x = - a}^a \mathop \smallint \limits_{y = - b}^b {y^2}{\rm{d}}y{\rm{d}}x$
$ = \frac{M}{{4ab}}\left( {\mathop \smallint \limits_{x = - a}^a {\rm{d}}x} \right)\left( {\mathop \smallint \limits_{y = - b}^b {y^2}{\rm{d}}y} \right)$
$ = \frac{M}{{12ab}}\left( {x|_{ - a}^a} \right)\left( {{y^3}|_{ - b}^b} \right) = \frac{M}{{12ab}}\left( {2a} \right)\left( {2{b^3}} \right) = \frac{M}{3}{b^2}$
Evaluate the moment of inertia relative to the $y$-axis:
${I_y} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {x^2}\delta \left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = - a}^a \mathop \smallint \limits_{y = - b}^b {x^2}\delta \left( {x,y} \right){\rm{d}}y{\rm{d}}x$
$ = \frac{M}{{4ab}}\mathop \smallint \limits_{x = - a}^a \mathop \smallint \limits_{y = - b}^b {x^2}{\rm{d}}y{\rm{d}}x$
$ = \frac{M}{{4ab}}\left( {\mathop \smallint \limits_{x = - a}^a {x^2}{\rm{d}}x} \right)\left( {\mathop \smallint \limits_{y = - b}^b {\rm{d}}y} \right)$
$ = \frac{M}{{12ab}}\left( {{x^3}|_{ - a}^a} \right)\left( {y|_{ - b}^b} \right) = \frac{M}{{12ab}}\left( {2{a^3}} \right)\left( {2b} \right) = \frac{M}{3}{a^2}$
The polar moment of inertia, ${I_0}$
${I_0} = {I_x} + {I_y} = \frac{M}{3}\left( {{a^2} + {b^2}} \right)$
(c) From part (b), we obtain ${I_x} = \frac{M}{3}{b^2}$.
By definition, the radius of gyration about the $x$-axis:
${r_g} = {\left( {\frac{{{I_x}}}{M}} \right)^{1/2}} = {\left( {\frac{{{b^2}}}{3}} \right)^{1/2}} = \frac{b}{{\sqrt 3 }}$
