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