Answer
$M= 3 \delta$
$I_x=2 \delta$
$\sqrt {\dfrac{2}{3}}$
Work Step by Step
Consider $M= \delta \int_{0}^{3} \int_{0}^{2x/3} \ dy \ dx= \delta \int_{0}^{3} \dfrac{2x}{3} \ dy \ dx = 3 \delta$
or, $= \delta \int_{0}^{3} \int_{0}^{2x/3} y^2 \ dy \ dx$
or, $= \dfrac{8 \delta}{81} \int_{0}^{3} x^3 \ dx$
or, $=(\dfrac{8 \delta}{81}) (\dfrac{3^4}{4})$
and $I_x=2 \delta$
Thus, $R_x=\sqrt {\dfrac{I_x}{M}}=\sqrt {\dfrac{2}{3}}$
