Answer
$\dfrac{8}{3}$
Work Step by Step
We integrate the integral to calculate the volume as follows:
$V= (2) \int_{-1}^{1} (2) (1-y^2) dy$
Now, $V= 2 [(y-\dfrac{y^3}{3})]_{-1}^{1}$
or, $= 2 [1-\dfrac{1}{3}-(-1) + (\dfrac{-1}{3})]$
or, $=\dfrac{8}{3}$
