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The volume of a sphere can be found using the formula \( V = \frac{4}{3} \pi r^3 \). In this formula, \( V \) stands for the volume of the sphere, \( \pi \) is a constant approximately equal to 3.14159, and \( r \) is the radius of the sphere.
The formula helps us determine how much space the sphere occupies.
To use the formula effectively, it is crucial to know the radius of the sphere. The volume calculated using this formula is stated in cubic units, reflecting the three-dimensional space the sphere occupies.
Understanding this formula is pivotal for solving real-world problems related to spheres, whether they involve balls, planets, or even tiny beads. Using this formula, you can compute how much liquid a spherical container might hold, for instance.

In any calculation involving the volume of a sphere, finding the radius is an essential step. Since the problem gives the diameter of the sphere as 12 millimeters, the radius is half of that.
To find the radius, simply divide the diameter by 2:

• Diameter = 12 mm
• Radius \( r = \frac{12}{2} = 6 \text{ mm} \)

Knowing the radius allows us to move forward with the volume formula and correctly calculate the space inside the sphere.
Occasionally, problems only provide the diameter, so dividing by two is a key step to proceeding with the volume calculation.

To find the precise volume of the sphere, once the radius is known, substitute it into the volume formula.
With a radius of 6 mm, the calculation becomes:

• First, calculate \( 6^3 \), which equals 216
• Then, multiply 216 by \( \frac{4}{3} \pi \)
• Resulting in \( V = \frac{4}{3} \pi \times 216 = 288\pi \) cubic millimeters

This is known as the exact volume since it involves the use of \( \pi \) in its symbolic form, retaining complete precision.
Exact volume is useful in situations where the most accurate measurement is required without numerical rounding or approximations.

Often, we need a numerical approximation of volume for practical purposes. This involves using an estimated value for \( \pi \). In this case, we use \( \pi \approx 3.14159 \), which is a commonly used approximation. To find the approximate volume:

• Multiply the exact volume expression \( 288\pi \) by \( 3.14159 \)
• This yields an approximate volume of \( 904.78 \) cubic millimeters

Approximations are helpful in everyday scenarios where exact values are not necessary, and calculations need to be concise and easily manageable.
This approach allows scientists and mathematicians to communicate results effectively when exact symbolic representation is not feasible or required.