Answer
$$\dfrac{9}{2}$$
Work Step by Step
$$I=\int_{0}^{3/2} \int_{-\sqrt {9-4t^2}}^{\sqrt {9-4t^2}} t \ ds \ dt \\=\int_{0}^{3/2} 2t \sqrt {9-4t^2} \ dt \\=[\dfrac{-(9-4t^2)^{3/2}}{6}]_{0}^{3/2} \\=-(\dfrac{1}{6}) (0-9^{3/2})] \\=\dfrac{27}{6} \\=\dfrac{9}{2}$$
