We are tasked with estimating the maximum error and relative error in the calculated volume of a spherical ball bearing, using differentials.

Given:

• Diameter $d = 8.4 \, \text{mm}$
• Error in diameter $\Delta d = 0.07 \, \text{mm}$

Step 1: Formula for Volume of a Sphere

The volume $V$ of a sphere is given by:

$V = \frac{4}{3} \pi r^3$

where $r$ is the radius. Since the diameter $d$ is given, we know $r = \frac{d}{2}$.

Step 2: Maximum Error in Volume

We use differentials to estimate the maximum error in volume. The differential of the volume $dV$ with respect to the radius $r$ is:

$dV = \frac{dV}{dr} \cdot dr = 4\pi r^2 \cdot dr$

Here, $dr$ represents the change in radius, which is half the change in diameter, so $dr = \frac{\Delta d}{2}$.

Now, let's calculate the values:

• $r = \frac{8.4}{2} = 4.2 \, \text{mm}$
• $dr = \frac{0.07}{2} = 0.035 \, \text{mm}$

Substitute these values into the differential formula:

$dV = 4\pi (4.2)^2 \cdot 0.035$

Simplifying:

$dV = 4\pi \cdot 17.64 \cdot 0.035$ $dV \approx 7.7335 \, \text{mm}^3$

So, the maximum error in the calculated volume is approximately $7.7335 \, \text{mm}^3$.

Step 3: Relative Error in Volume

The relative error is the ratio of the differential of the volume $dV$ to the actual volume $V$. First, let's calculate the actual volume:

$V = \frac{4}{3} \pi (4.2)^3$ $V = \frac{4}{3} \pi \cdot 74.088$ $V \approx 310.339 \, \text{mm}^3$

The relative error is given by:

$\text{Relative Error} = \frac{dV}{V} = \frac{7.7335}{310.339}$ $\text{Relative Error} \approx 0.0249$

So, the relative error is approximately 0.0249 or 2.49%.

Would you like further details or have any other questions?

Related Questions:

1. How do differentials estimate error in other shapes, such as cubes or cylinders?
2. How would the error change if the diameter's error doubled?
3. How can you use calculus to minimize error in measurements?
4. What is the relationship between absolute and relative error?
5. Can the method of differentials be applied to real-world engineering problems?

Tip: When working with errors, remember that small measurement uncertainties can lead to significant volume discrepancies, especially with geometric shapes like spheres.