A spherical ball is measured to have a radius of 10 mm, with a possible measurement error of 0.1 mm. What is the possible change in volume (in mm^3)?
To calculate the possible change in volume due to the measurement error, we'll use the formula for the volume of a sphere:
\[V = \frac{4}{3} \pi r^3\]
Given:
- Initial radius, \(r = 10 \, \text{mm}\)
- Measurement error, \(Δr = 0.1 \, \text{mm}\)
We'll first find the initial volume, then find the volumes with the maximum and minimum possible radii within the given error range, and finally calculate the changes in volume.
1. Initial Volume (\(V_0\)):
\[V_0 = \frac{4}{3} \pi (10 \, \text{mm})^3\]
\[V_0 = \frac{4}{3} \pi (1000 \, \text{mm}^3)\]
\[V_0 = \frac{4000}{3} \pi \, \text{mm}^3\]
2. Volume with Maximum Radius (\(V_{\text{max}}\)):
\[r_{\text{max}} = r + Δr = 10 \, \text{mm} + 0.1 \, \text{mm} = 10.1 \, \text{mm}\]
\[V_{\text{max}} = \frac{4}{3} \pi (10.1 \, \text{mm})^3\]
3. Volume with Minimum Radius (\(V_{\text{min}}\)):
\[r_{\text{min}} = r - Δr = 10 \, \text{mm} - 0.1 \, \text{mm} = 9.9 \, \text{mm}\]
\[V_{\text{min}} = \frac{4}{3} \pi (9.9 \, \text{mm})^3\]
Now, let's calculate these volumes:
\[V_0 \approx \frac{4000}{3} \pi \, \text{mm}^3 \approx 4188.79 \, \text{mm}^3\]
\[V_{\text{max}} \approx \frac{4}{3} \pi (10.1 \, \text{mm})^3 \approx 4241.75 \, \text{mm}^3\]
\[V_{\text{min}} \approx \frac{4}{3} \pi (9.9 \, \text{mm})^3 \approx 4136.82 \, \text{mm}^3\]
Finally, let's find the possible change in volume:
\[ΔV_{\text{max}} = V_{\text{max}} - V_0 \approx 4241.75 \, \text{mm}^3 - 4188.79 \, \text{mm}^3 \approx 52.96 \, \text{mm}^3\]
\[ΔV_{\text{min}} = V_{\text{min}} - V_0 \approx 4136.82 \, \text{mm}^3 - 4188.79 \, \text{mm}^3 \approx -51.97 \, \text{mm}^3\]
The possible change in volume due to the measurement error is approximately \(52.96 \, \text{mm}^3\) (positive or negative).