Generalized Chinese Puzzle

We can write the volume of the intersection body as 4 times the volume of the missing part from the solid in the figure. Cosider the height of the body as $H$ . ${{V}_{\operatorname{int}}}=4\int\limits_{V}{1dV}=4\int\limits_{0}^{H}{\int\limits_{0}^{f(z)}{\int\limits_{0}^{f(z)}{1dx}}}dydz$ The final form we will use later is: ${{V}_{\operatorname{int}}}=4\int\limits_{0}^{H}{{{f(z)}^{2}}}dz$ The volume of revolution of the half cross section is equal to the area of half cross section times the length of revolution of the center of the mass. Let's note the distance from the center of mass to $z$ -axis with $D$ . ${{V}_{rev}}=A2\pi D$ Where $A$ is the area under the function. We know the formula for the center of mass: $D=\frac{\int\limits_{A}{kdA}}{A}=\frac{\int\limits_{0}^{H}{\left( \int\limits_{0}^{f(z)}{kdk} \right)dz}}{A}$ Where $k$ is the axis perpendicular to $z$ -axis ( the coordinate of element $dA$ ). Substituting and simplifying the area we get: ${{V}_{rev}}=2\pi \int\limits_{0}^{H}{\left( \left. \frac{{{k}^{2}}}{2} \right|_{0}^{f(z)} \right)dz}=\pi \int\limits_{0}^{H}{f{{(z)}^{2}}dz}$ We see now that the ratio is: $\frac{{{V}_{\operatorname{int}}}}{{{V}_{rev}}}=\frac{4}{\pi }\approx1.2732$

Let $y = f(z)$ be a point on the graph. Then for this value of $z$ , the cross-section of the first solid is a square with side length $2y$ , so its area is $4y^2$ . The cross-section of the second solid is a circle with radius $y$ , so its area is $\pi y^2$ . The ratio of these areas is $\frac{4}{\pi}$ , so the ratio of the volumes is also $\frac{4}{\pi}$ .