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