please solve this calc 3 problem?

Note that z = x^2 - 1 intersects z = 1 - x^2 when x^2 - 1 = 1 - x^2 ==> x = -1, 1.

So, ∫∫∫E (x-y) dV
= ∫(y = 0 to 2) ∫(x = -1 to 1) ∫(z = x^2 - 1 to 1 - x^2) (x - y) dz dx dy
= ∫(y = 0 to 2) ∫(x = -1 to 1) (x - y)z {for z = x^2 - 1 to 1 - x^2} dx dy
= ∫(y = 0 to 2) ∫(x = -1 to 1) (x - y) (2 - 2x^2) dx dy
= ∫(y = 0 to 2) ∫(x = -1 to 1) (0 - y) (2 - 2x^2) dx dy, via x(2 - 2x^2) being odd in x
= ∫(y = 0 to 2) ∫(x = -1 to 1) y(2x^2 - 2) dx dy
= 2 ∫(y = 0 to 2) ∫(x = 0 to 1) y(2x^2 - 2) dx dy, via integrand being even in x
= 2 ∫(y = 0 to 2) y((2/3)x^3 - 2x) {for x = 0 to 1} dy
= (-8/3) ∫(y = 0 to 2) y dy
= (-8/3) * (1/2)y^2 {for y = 0 to 2}
= -16/3.

I hope this helps!