A cylindrical container, rendered in light transparent blue, is depicted. Inside the cylinder, a sphere is partially visible.  The sphere's top and bottom sections, which are inside the cylinder, are darkly shaded blue.  The visible portions of the sphere are outlined by thin black lines. A dashed black line shows the curve of the sphere intersecting the cylinder's inner surface. A black dot is positioned inside the sphere. A black line, extending from the dot and labeled with the lowercase letter 'r', connects the center to the sphere's circumference. Mrs. Kertye bought a spherical ball for her nephew, but she does not know the radius. She is in search of a cylindrical box that will fit the ball perfectly, that is, it will have the same radius plus a little more so that the ball will fit.

About how much larger is the cylindrical box’s volume compared to the ball’s volume?
How do their surface areas compare to each other? What is the ratio of the sphere’s surface area to the cylinder’s surface area?
Mrs. Kertye wants to pack a cone-shaped toy inside the same box. What is the largest possible cone that could fit inside? Give its volume.

1 answer

To solve the problem, we first need to understand the geometric relationships between the sphere, the cylindrical box, and the cone that Mrs. Kertye wants to pack inside.

  1. Volume of the Sphere: The volume \( V \) of a sphere with radius \( r \) is given by the formula: \[ V_{\text{sphere}} = \frac{4}{3} \pi r^3 \]

  2. Volume of the Cylinder: If we want the cylindrical box to fit the sphere perfectly, the radius \( R \) of the cylinder should be slightly larger than the radius \( r \) of the sphere. We’ll assume \( R = r + \epsilon \) where \( \epsilon \) is a small positive number (adjusted as needed to ensure fit).

    The height \( H \) of the cylinder should be equal to the diameter of the sphere, which is \( 2r \). Thus, the volume \( V \) of the cylinder is: \[ V_{\text{cylinder}} = \pi R^2 H = \pi (r + \epsilon)^2 (2r) = 2\pi (r + \epsilon)^2 r \]

  3. Comparing Volumes: The difference in volume \( V_{\text{cylinder}} - V_{\text{sphere}} \) gives: \[ V_{\text{cylinder}} - V_{\text{sphere}} = 2\pi (r + \epsilon)^2 r - \frac{4}{3} \pi r^3 \] If we neglect \( \epsilon \) for a moment, we would find \( 2\pi r^3 - \frac{4}{3} \pi r^3 = \frac{2\pi r^3}{3} \), which suggests the cylinder's volume can be more than \( \frac{2\pi r^3}{3} \) larger than the sphere's volume depending on how much larger the radius is, making \( \epsilon \) significant.

  4. Surface Areas: The surface area \( A \) of the sphere is given by: \[ A_{\text{sphere}} = 4 \pi r^2 \]

    The surface area of the cylinder, excluding the top and bottom, is the lateral surface area plus the area of the two bases: \[ A_{\text{cylinder}} = 2\pi R H + 2\pi R^2 = 2\pi (r+\epsilon)(2r) + 2\pi (r+\epsilon)^2 \] Ignoring \( \epsilon \): \[ A_{\text{cylinder}} = 4\pi r^2 + 2\pi r^2 = 6\pi r^2 \]

    The ratio of the sphere's surface area to the cylinder's surface area: \[ \text{Ratio} = \frac{A_{\text{sphere}}}{A_{\text{cylinder}}} = \frac{4\pi r^2}{6\pi r^2} = \frac{2}{3} \]

  5. Volume of the Largest Possible Cone: The largest cone that can fit inside would have a base radius equal to \( r \) and height equal to the height of the cylinder \( H = 2r \). The volume \( V \) of the cone is given by: \[ V_{\text{cone}} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^2 (2r) = \frac{2\pi r^3}{3} \]

In summary:

  • The cylinder’s volume is about \( \frac{2\pi r^3}{3} \) larger than the sphere's volume, depending on the size of \( \epsilon \).
  • The ratio of the sphere's surface area to the cylinder's surface area is \( \frac{2}{3} \).
  • The volume of the largest cone that could fit inside is \( \frac{2\pi r^3}{3} \).