It is more like this.

The "square" has a vertical side that is ([y of (2 -x^2)] minus the [y of x^2]).

That is (2 -x^2) -(x^2) = 2 -2x^2 = 2(1 -x^2),

So, the "square" is [2(1 -x^2]^2.

The volume element of the solid in question, dV, has a thickness of dx.

dV = [2(1 -x^2]^2 *dx

The integration is for dx to go from x = -1 to x = 1. ----the intersections of the two parabolas.

So,

V = INT.(-1 --> 1)[4(1 -x^2)^2]dx

V = (4)INT.(-1 --> 1)[1 -2x^2 +x^4]dx

V = 4[x -(2/3)x^3 +(1/5)x^5] | (-1 --> 1)

V = 4{[1 -(2/3)(1) +(1/5)(1)] -[-1 -(2/3)(-1) +(1/5)(-1)]}

V = 4{1 -2/3 +1/5 +1 -2/3 +1/5}

V = 4{2 -4/3 +2/5}

V = 4{(2*3*5 -4*5 +2*3) / 3*5}

V = 64/15 = 4.2667 cu.units ----------------answer.