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