Answer
$\dfrac{-16}{3}$
Work Step by Step
Consider $I=\iiint_E (x-y) dV$
$I= \int_{-1}^1 \int_{0}^{2} \int_{1-x^2}^{x^2-1} (x-y) dz dy dx= \int_{-1}^1 \int_{0}^{2} (x-y)[z]_{1-x^2}^{x^2-1} dy dx$
or, $=\int_{-1}^1 \int_{0}^{2}(2) (x-y)(x^2-1) dy dx$
or, $=(-2) \int_{-1}^1 [x(1-x^2)y-(1-x^2)\dfrac{y^2}{2}]_0^2dx$
or, $= (-4) \int^{-1}_{1}[x-x^3-1+x^{2}] dx$
or, $= (-4)[\dfrac{x^2}{2}-\dfrac{x^2}{4}-x+\dfrac{x^3}{3}]^{-1}_{1}$
or, $=\dfrac{-16}{3}$
