Let's analyze the problem step by step.

Given:

• The diameter of a spherical ball bearing is $7.8 \, \text{mm}$.
• Possible error in measurement $\Delta d = 0.1 \, \text{mm}$.

The goal is to estimate:

• (a) The maximum error in the volume calculation (already solved as $9.5581 \, \text{mm}^3$).
• (b) The relative error in the calculated volume using differentials.

Part (b): Estimating Relative Error in Volume

1. The formula for the volume of a sphere is: $V = \frac{4}{3} \pi r^3$ where $r$ is the radius. The diameter $d = 7.8 \, \text{mm}$, so the radius is: $r = \frac{d}{2} = \frac{7.8}{2} = 3.9 \, \text{mm}.$

2. Differentiating $V$ with respect to $r$ gives the differential of the volume: $dV = 4 \pi r^2 \, dr.$ The maximum error in $r$ is half the error in the diameter, so: $\Delta r = \frac{\Delta d}{2} = \frac{0.1}{2} = 0.05 \, \text{mm}.$

3. Now, relative error $\frac{dV}{V}$ is given by: $\frac{dV}{V} = \frac{3 \, dr}{r}.$ Substituting $dr = 0.05 \, \text{mm}$ and $r = 3.9 \, \text{mm}$: $\frac{dV}{V} = \frac{3 \times 0.05}{3.9} \approx 0.0385.$

Thus, the relative error in the volume is approximately 0.0385, or 3.85%.

Would you like further details or have any questions?

Here are some related questions to consider:

1. How does the volume error scale with the diameter's error?
2. What would the error be if the ball's diameter was measured at 8.0 mm instead?
3. How is the differential formula used to approximate errors in other shapes, like cylinders?
4. Can you compute the absolute error if the diameter's error increased to 0.2 mm?
5. What happens to the volume error if the radius is halved?

Tip: When estimating errors using differentials, the relative error in volume is often a multiple of the relative error in the radius, especially for powers like cubic terms.