The base of a solid

The problem states: The base of a solid is the region bounded by the parabolas y=x^2 and y=2-x^2. Find the volume of the solid if the cross-section perpendicular to the x-axis are squares with one side lying along the base.

Here is what I did: x^2=2-x^2 (x-1)(x+2)= 0 x=1 , x=-2
pi *integral from -1 to 1 ((2-x^2)^2 - (x^2)^2 dx
pi*integral from -1 to 1(4-4x^2)dx
pi(4x - 4/3x^3) from limits -1 to 1 volume = 8pi

Is this procedure correct?
Thankx,
Keith:)

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.