Hokkaido University 2007: Cross-section and Volume of a Cube

Problem

In coordinate space, consider the solid represented by the following system of inequalities:
$${0 \leqq x \leqq 1,\ 0 \leqq y \leqq 1, 0 \leqq z \leqq 1, \ x^2+y^2+z^2-2xy-1 \geqq 0}$$
Illustrate the cross-section of this solid when cut by the plane $${z=t}$$ on the $${xy}$$ plane, and find the area of this cross-section $${S(t)}$$ and the volume of the solid.

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To calculate the volume of this solid, one can consider the cross-section at $${z=t}$$ and integrate; if the cross-sectional diagram can be drawn, the calculation is relatively simple. If we set $${z=t}$$ in the fourth inequality of the system,
it becomes $${x^2+y^2+t^2-2xy-1 \geqq 0}$$. What kind of figure does this represent? Changing it to an equation, we get $${(x-y)^2+t^2-1 = 0}$$, which becomes $${(x-y)^2-(1-t^2) = 0}$$. Factoring the left side, it will be the area outside the region bounded by two straight lines. (Let's calculate it.)

Opening the link will take you to the following screen.

You can change the value of t by dragging the green point on the slider on the right, and the figure is drawn in the range $${0 \leqq z \leqq t}$$. The top surface represents the cross-section.

Let's try looking at it from various directions using the circular slider on the left. When viewed from directly above, it is clear that it is the area outside the region bounded by two straight lines (line segments).

Once the situation is understood, finding the cross-sectional area $${S(t)=(1-\sqrt{1-t^2})^2}$$ is not difficult, and the integration should not be hard either. The volume is $${5/3 - \pi/2}$$.

*The figures were created using Cinderella (CindyJS).