Determination of a volume. Consider, in the Cartesian plane, the square Q having vertices in the points $(-1, 0), (1, 0), (0, -1)$, and $(0, 1)$.
The sections of a solid with planes orthogonal to $y=0$ are squares having two opposite vertices situated on Q sides. Determine the volume of the solid.
THIS IS MY SOLUTION BUT THE RESULT IS WRONG:
The equation of AB side is: $y = -x + 1$
The equation of BC side is: $y = x - 1$
$$PP' = 2(-x + 1) = -2x + 2$$
is the diagonal of the little squares (the book speaks about OPPOSITE VERTICES)
The side of the little squares measures: ${-2x+2\over \sqrt 2} = -\sqrt2 x + \sqrt 2$
Area = $(- \sqrt 2 x + \sqrt 2)^2 = 2(x^2) - 4x + 2$
$$V = \int_{-1}^1 2x^2-4x+2\,dx = 16/3$$
... but the books says 4/3
Where did I wrong??
 A: Recall that the volume is the integral of the cross sectional areas. The two graphs that make the base square are
$$\begin{cases} y_1 = 1-|x| & -1\le x\le 1 \\ y_2=|x|-1 & -1\le x\le 1\end{cases}$$
So that the area of the square orthogonal to it is just the square of the side-length, but all you have is the diagonal length as $y_1-y_2$. However we know that $s\sqrt{2}=d$ is the relation between sides and diagonals, so the cross sectional area is just i.e. $\left({1\over\sqrt{2}}(y_1-y_2)\right)^2$ integrated across $-1\le x\le 1$, which gives us the formula for volume:

$$V={1\over 2}\int_{-1}^12^2(1-|x|)^2\,dx$$

Doing the algebra out this gives:
$$V=2\left(\int_{-1}^11-2|x|+|x|^2\,dx\right)$$
$$=2\left(\int_{-1}^11\,dx-2\int_{-1}^1 |x|\,dx +\int_{-1}^1 x^2\,dx\right)$$
Now the first integral is easily seen to be $2$, the second is twice the area under $|x|$ which by basic geometry is $2$ because each of the two triangles has area ${1\over 2}$, so the first and second integral cancel one another out, and we are left with
$$\int_{-1}^1 x^2\,dx = {2\over 3}$$
which leaves us with a grand total of $V={4\over 3}$.
