
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=2x^2. Find the volume of the solid if the crosssection perpendicular to the xaxis are squares with one side lying along the base.
Here is what I did: x^2=2x^2 (x1)(x+2)= 0 x=1 , x=2
pi *integral from 1 to 1 ((2x^2)^2  (x^2)^2 dx
pi*integral from 1 to 1(44x^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.