Problem 900

Solution. Let the radius of the cylinder be $r$, and the common center of the sphere and the cylinder be $O$. A plane passing through the axis of the cylinder cuts a rectangle from the cylinder and a circle from the sphere.

The volume of the sphere: $\frac{4 R^{3} \pi}{3}$. The volume of the cylinder: $r^{2} \pi m$, where $m=\frac{4}{3} R$.

The radius of the base circle of the cylinder can be determined using the Pythagorean theorem in the right triangle shown in the diagram:

$$r^{2}=R^{2}-\left(\frac{2}{3} R\right)^{2}=\frac{5}{9} R^{2}$$

Substituting into the volume of the cylinder:

$$V_{\mathrm{h}}=\frac{5}{9} R^{2} \pi \cdot \frac{4}{3} R=\frac{20}{27} R^{3} \pi$$

The ratio of the two volumes:

$$\frac{V_{\mathrm{h}}}{V_{\mathrm{g}}}=\frac{\frac{20}{27}}{\frac{4}{3}}=\frac{5}{9}$$

The volume of the cylinder is $\frac{5}{9}$ of the volume of the sphere.

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Generalization. Let the height of the cylinder be $x$ times the radius of the sphere: $m=R x(0<x<2)$. Then, from the plane section of the cylinder,

$$R^{2}=\frac{R^{2}}{4} x^{2}+r^{2}$$

From this, $4 r^{2}=R^{2}\left(4-x^{2}\right)$

$$r^{2}=\frac{R^{2}\left(4-x^{2}\right)}{4}$$

The ratio of the volumes:

$$\frac{V_{\mathrm{h}}}{V_{\mathrm{g}}}=\frac{\frac{1}{4} R^{2}\left(4-x^{2}\right) \pi R x}{\frac{4 R^{3} \pi}{3}}=\frac{3}{16}\left(4 x-x^{3}\right)$$

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We plot the function $y=4 x-x^{3}$ in the coordinate system. The derivative of the function is:

$$y^{\prime}=4-3 x^{2}=0, \quad \text { if } \quad x= \pm \sqrt{\frac{4}{3}} \approx 1.1547$$

The function can have an extremum where its derivative is 0. Indeed, at $x=\sqrt{\frac{4}{3}} \approx 1.1547$, the function has a maximum, which in our case means that if the height of the cylinder is approximately 1.1547 times the radius of the sphere, then the ratio of the volumes is approximately 0.57735.

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