Understanding the Problem: Sphere Volume and Surface Area

This problem involves the concepts of volume and surface area of a sphere. When a solid material like an iron ball is melted and recast into smaller shapes, the total volume of the material remains constant. We are given a large solid iron ball which is melted to form 64 smaller solid balls of equal size. We need to find the ratio of the surface area of the original larger ball to the sum of the surface areas of all 64 smaller balls.

Step-by-Step Solution to the Sphere Problem

Let's denote the properties of the larger iron ball and the smaller iron balls:

• Let the radius of the larger ball be \(R\).
• Let the radius of each smaller ball be \(r\).

Volume Relationship

The volume of a sphere with radius \(x\) is given by the formula \(V = \frac{4}{3}\pi x^3\).

The volume of the large ball is \(V_{large} = \frac{4}{3}\pi R^3\).

The volume of one small ball is \(V_{small} = \frac{4}{3}\pi r^3\).

Since the large ball is melted and its entire volume is used to make 64 smaller balls, the total volume remains the same:

\(V_{large} = 64 \times V_{small}\)

\(\frac{4}{3}\pi R^3 = 64 \times \frac{4}{3}\pi r^3\)

We can cancel out the common terms \(\frac{4}{3}\pi\) from both sides:

\(R^3 = 64 r^3\)

To find the relationship between \(R\) and \(r\), we take the cube root of both sides:

\(\sqrt[3]{R^3} = \sqrt[3]{64 r^3}\)

\(R = \sqrt[3]{64} \times \sqrt[3]{r^3}\)

\(R = 4r\)

So, the radius of the large ball is 4 times the radius of each smaller ball.

Surface Area Calculation

The surface area of a sphere with radius \(x\) is given by the formula \(A = 4\pi x^2\).

The surface area of the large ball is \(A_{large} = 4\pi R^2\).

Substitute \(R = 4r\) into the formula for the large ball's surface area:

\(A_{large} = 4\pi (4r)^2\)

\(A_{large} = 4\pi (16r^2)\)

\(A_{large} = 64\pi r^2\)

The surface area of one small ball is \(A_{small} = 4\pi r^2\).

We need the sum of the surface areas of all 64 smaller balls. This is \(64 \times A_{small}\).

Sum of surface areas of small balls = \(64 \times (4\pi r^2)\)

Sum of surface areas of small balls = \(256\pi r^2\)

Finding the Ratio of Surface Areas

We need to find the ratio of the surface area of the larger ball to the sum of the surface areas of all the smaller balls:

Ratio = \(\frac{A_{large}}{\text{Sum of surface areas of small balls}}\)

Ratio = \(\frac{64\pi r^2}{256\pi r^2}\)

Cancel out the common term \(\pi r^2\):

Ratio = \(\frac{64}{256}\)

Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 64:

Ratio = \(\frac{64 \div 64}{256 \div 64} = \frac{1}{4}\)

The ratio as a decimal is \(\frac{1}{4} = 0.25\).

Thus, the ratio of the surface area of the larger ball to the sum of the surface areas of all the smaller balls is 0.25.

Revision Table: Sphere Formulas

Additional Information: Scaling of Volume and Surface Area

When a sphere's radius changes by a factor \(k\), its volume changes by a factor of \(k^3\), and its surface area changes by a factor of \(k^2\).

In this problem, we found that the large radius \(R\) is 4 times the small radius \(r\), so \(R = 4r\). The scaling factor for the radius from small to large is \(k=4\).

• Ratio of radii: \(\frac{R}{r} = 4\)
• Ratio of volumes: \(\frac{V_{large}}{V_{small}} = \frac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi r^3} = \frac{R^3}{r^3} = \left(\frac{R}{r}\right)^3 = 4^3 = 64\). This confirms our starting point that the large ball's volume equals 64 small balls' volumes.
• Ratio of surface areas (single large vs single small): \(\frac{A_{large}}{A_{small}} = \frac{4\pi R^2}{4\pi r^2} = \frac{R^2}{r^2} = \left(\frac{R}{r}\right)^2 = 4^2 = 16\). The large ball's surface area is 16 times that of a single small ball.

However, the question asks for the ratio of the large ball's surface area to the *sum* of the surface areas of *all 64* small balls.

Sum of surface areas of small balls = \(64 \times A_{small}\)

The required ratio is \(\frac{A_{large}}{64 \times A_{small}}\). We know \(A_{large} = 16 \times A_{small}\), so:

Ratio = \(\frac{16 \times A_{small}}{64 \times A_{small}}\) = \(\frac{16}{64}\) = \(\frac{1}{4}\) = \(0.25\).

This confirms the result obtained earlier. The total surface area increases significantly when a large ball is broken down into many smaller balls, even though the volume remains constant.