Answer
$$\dfrac{8}{3}$$
Work Step by Step
We need to integrate the integral to compute the volume.
$$ Volume = (2) \int_{-1}^{1} (2) \times (1-y^2) dy \\ =[2y-\dfrac{2y^3}{3}]_{-1}^{1} \\= 2 [1-\dfrac{1}{3}-(-1) + (\dfrac{-1}{3})] \\=\dfrac{8}{3}$$