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