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.

Question

A cylinder, rendered in a light translucent blue, is shown containing a sphere, also in a light translucent blue. 

The cylinder has thick black outlines, and the sphere is defined by two dashed black lines indicating the top and bottom circular sections of the sphere touching the cylinder. These sections are slightly smaller in diameter than the cylinder's opening.

Inside the cylinder, a sphere of the same translucent blue is visible, with its upper and lower sections inside the cylinder's boundaries. 

A thin, solid black line represents the radius (r) of the sphere, extending from the center of the sphere to the edge of the sphere's circumference at the level within the cylinder, marked with a small black circle to denote the sphere's center.
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?

Answer 1

To compare the surface areas of the sphere and the cylinder, let's start by determining their respective surface areas.

1. Surface area of the sphere: The surface area \( A_s \) of a sphere with radius \( r \) is given by the formula: \[ A_s = 4\pi r^2. \]

2. Surface area of the cylinder: A cylinder has a surface area that consists of the lateral area plus the area of the two circular bases. For a cylinder with radius \( R \) and height \( h \), the surface area \( A_c \) is given by: \[ A_c = 2\pi R h + 2\pi R^2, \] where \( 2\pi R h \) is the lateral surface area, and \( 2\pi R^2 \) is the area of the two circular bases.

Since the sphere fits perfectly inside the cylinder, the radius of the cylinder \( R \) must be larger than the sphere's radius \( r \) (let’s say \( R = r + \epsilon \), where \( \epsilon \) is a small positive value for the extra space). The height \( h \) of the cylinder will be equal to the diameter of the sphere: \[ h = 2r. \]

Substituting \( R \) and \( h \) into the formula for the surface area of the cylinder:

1. Replace \( R \) with \( r + \epsilon \) in the lateral surface area and base areas: \[ A_c = 2\pi (r + \epsilon)(2r) + 2\pi (r + \epsilon)^2 \]

2. This simplifies to: \[ A_c = 4\pi (r + \epsilon)r + 2\pi (r + \epsilon)^2. \] Further simplification would yield: \[ A_c = 4\pi r^2 + 4\pi r\epsilon + 2\pi (r^2 + 2r\epsilon + \epsilon^2). \]

3. Aggregating the terms, we obtain: \[ A_c = 4\pi r^2 + 4\pi r\epsilon + 2\pi r^2 + 4\pi r\epsilon + 2\pi \epsilon^2 = 6\pi r^2 + 8\pi r\epsilon + 2\pi \epsilon^2. \]

4. Therefore, we have the surface areas calculated as:

   • Sphere: \( A_s = 4\pi r^2 \)
   • Cylinder: \( A_c = 6\pi r^2 + 8\pi r\epsilon + 2\pi \epsilon^2 \)

5. Ratio of surface areas: To find the ratio of the sphere's surface area to the cylinder's surface area: \[ \text{Ratio} = \frac{A_s}{A_c} = \frac{4\pi r^2}{6\pi r^2 + 8\pi r\epsilon + 2\pi \epsilon^2}. \] After cancelling \( \pi \), we can simplify this to: \[ \text{Ratio} = \frac{4 r^2}{6 r^2 + 8 r\epsilon + 2 \epsilon^2}. \]

As \( \epsilon \) becomes very small (meaning \( R \) is only slightly larger than \( r \)), this ratio approaches: \[ \text{Ratio} \approx \frac{4r^2}{6r^2} = \frac{4}{6} = \frac{2}{3}. \]

Thus, the ratio of the sphere’s surface area to the cylinder’s surface area is approximately \( \frac{2}{3} \) when \( \epsilon \) is very small.