Understanding the Percentage Error in the Volume of a Sphere due to Radius Measurement Error

The accuracy of measurements is crucial in many scientific and engineering applications. In the context of calculating the volume of a sphere, an error in measuring its radius can significantly impact the calculated volume. This article explores how to determine the percentage error in the volume of a sphere when there is a known error in measuring the radius. We will use basic principles of calculus and error propagation to derive the solution.

1. The Volume of a Sphere: A Fundamental Formula

The volume V of a sphere is given by the formula:

V frac{4}{3} pi r^3

where r is the radius of the sphere. Understanding how this formula behaves under small changes in the radius is necessary to determine the percentage error in volume.

2. Determining the Relationship Between Percentage Errors

When the radius r has a small percentage error, the volume V will also have an error proportional to the exponent of r. This concept can be derived from the differential approximation of the error propagation formula:

For the volume formula V k r^n, the percentage error in V can be given by:

text{Percentage Error in } V n times text{Percentage Error in } r

In the case of the volume of a sphere, n 3 because the volume is proportional to r^3.

3. Applying the Formula to the Situation at Hand

Given that the percentage error in the radius is 2%, we can apply the formula to find the percentage error in the volume:

text{Percentage Error in } V 3 times text{Percentage Error in } r 3 times 2 6%

Therefore, if there is a 2% error in measuring the radius, the percentage error in the volume of the sphere is 6%.

4. Error Propagation and Absolute Error

It is also valuable to consider the absolute error introduced by the radius measurement. If the radius measurement has an error of ±3% (meaning the radius could be 0.97r or 1.03r), the volume of the sphere can be affected by the cube of this error:

text{New Volume} frac{4}{3} pi (1.03r)^3 frac{4}{3} pi 1.092727r^3

By subtracting the original volume from the new volume:

frac{4}{3} pi 1.092727r^3 - frac{4}{3} pi r^3 frac{4}{3} pi r^3 (1.092727 - 1) frac{4}{3} pi r^3 0.092727

The absolute error in the volume is 0.092727 times frac{4}{3} pi r^3, which represents a 9.2727% change in the volume due to a 3% error in the radius.

5. Summary and Conclusion

Accurately measuring the radius is essential for determining the volume of a sphere. A 2% error in the radius measurement results in a 6% error in the volume, while a ±3% error in the radius can cause an absolute volume error of 9.2727%. Understanding this relationship helps in estimating the effects of measurement errors and ensures more reliable calculations in related fields.

Key takeaways:

The volume of a sphere depends on the cube of the radius. A small percentage error in the radius can result in a larger percentage error in the volume. Propagation of errors in measurements is crucial for accurate calculations.

Keywords: volume of a sphere, percentage error, radius measurement error