Answer
$\dfrac{2ma^3}{9}$
Work Step by Step
Consider $Volume=\iint_{R} mx dA$
or, $=\int_{-a/3}^{a/3} \int_0^{\sqrt {a^2-9y^2}} mx dx dy$
or, $=\int_{-a/3}^{a/3} [\dfrac{mx^2}{2}]_0^{\sqrt {a^2-9y^2}} dy$
or, $=m \int_0^{a/3} a^2-9y^2 dy$
or, $=m (a^2 y-3y^3)_0^{a/3}$
or, $=\dfrac{2ma^3}{9}$