An Ancient Chinese Puzzle 2

Let the cylinders have the equations $x^2 + z^2 \leq R^2$ and $y^2 + y^2 \leq R^2$ , respectively. The surface consists of four equal sections.

To find the area of one of these sections, parametrize the surface of the first cylinder as $x = R \sin\theta, z = R \cos\theta$ . The intersection with the other cylinder is given by $\left.\begin{array}{l} x^2 + z^2 = R^2 \\ y^2 + z^2 \leq R^2\end{array}\right\}\ \ \Longrightarrow\ \ -R\sin\theta = -x < y < x = R\sin\theta$ . Integrating, we have $A = \int_0^\pi \int_{-R\sin\theta}^{R\sin\theta} R d\theta\ dy$ $= \int_0^\pi (2R^2\sin\theta) = \left.-2R^2\cos\theta\right|_0^\pi = 4R^2.$ Thus the total area is $4\cdot 4R^2 = 16R^2 = \boxed{1600\ \text{cm}^2}$ .

nice problem

Given a cartesian system {x,y,z}, the intersection of the two cylinders of axis (Ox) and (Oz) and radius R is a 3D curve given by the following set of equations:

$x² + y ² = R ²$

$y² + z² = R²$

Using the cylindrical coordinates $(r, theta)$ of the (Ox) cylinder

The system becomes:

$r ² = R ²$

$z² = R²cos²(theta)$

The curve is therefore given by the two parametered curves:

$z1 = R|cos(theta)|$

$z2 = -R|cos(theta)|$

Surface lying on the (Ox) cylinder is deduced by integrating $|z1-z2| \times R \times d(theta)$ over 360°, which gives $8R²$ . Total surface of the object is twice this value (surface on the (Oz) cylinder is egal for symmetryconsiderations).

Hence $1600$ .