A sphere has a diameter of 18 centimeters. What is the volume, in cubic centimeters, of the sphere?

1. TRANSLATE the problem information

• Given information:
  • Sphere diameter = 18 cm
  • Need to find volume

• What this tells us: Since radius is needed for the volume formula, \(\mathrm{r = 18/2 = 9\,cm}\)

2. INFER the approach

• We need the volume formula for a sphere: \(\mathrm{V = \frac{4}{3}\pi r^3}\)
• We'll substitute our radius value and calculate

3. SIMPLIFY the calculation

• \(\mathrm{V = \frac{4}{3}\pi(9)^3}\)
• Calculate \(\mathrm{9^3 = 729}\)
• \(\mathrm{V = \frac{4}{3}\pi(729)}\)
• \(\mathrm{V = (4 \times 729 \div 3)\pi}\)
• Calculate: \(\mathrm{4 \times 729 = 2916}\), then \(\mathrm{2916 \div 3 = 972}\)
• \(\mathrm{V = 972\pi}\)

Answer: C (\(\mathrm{972\pi}\))

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students use the diameter directly in the volume formula instead of converting to radius first.

They substitute \(\mathrm{d = 18}\) into \(\mathrm{V = \frac{4}{3}\pi r^3}\), treating the 18 as if it were the radius:
\(\mathrm{V = \frac{4}{3}\pi(18)^3 = \frac{4}{3}\pi(5832) = 7776\pi}\)

This may lead them to select Choice D (\(\mathrm{7776\pi}\))

Second Most Common Error:

Poor SIMPLIFY execution: Students make arithmetic errors when calculating \(\mathrm{(4 \times 729)/3}\), especially if they try to work with fractions incorrectly.

Some might incorrectly calculate \(\mathrm{729 \div 3 = 243}\) first, then multiply by 4 to get 972, but mix up the order of operations and get confused about where \(\mathrm{\pi}\) belongs in the expression.

This leads to confusion and guessing among the remaining choices.

The Bottom Line:

This problem tests whether students remember that volume formulas use radius, not diameter, and whether they can handle multi-step arithmetic with \(\mathrm{\pi}\) expressions. The key insight is recognizing that diameter must be halved before using any radius-based formula.