To solve this problem, we can apply linear interpolation. This method helps to estimate the volume corresponding to a sounding of 4.13 meters, using the two given data points from the table.

The given points are:

• Sounding 1: $x_1 = 4.10$ meters, $V_1 = 120.91$ cubic meters
• Sounding 2: $x_2 = 4.15$ meters, $V_2 = 124.27$ cubic meters
• Desired sounding: $x = 4.13$ meters

The formula for linear interpolation is:

$V = V_1 + \frac{(V_2 - V_1)(x - x_1)}{(x_2 - x_1)}$

Now, let's plug in the values:

$V = 120.91 + \frac{(124.27 - 120.91)(4.13 - 4.10)}{(4.15 - 4.10)}$

Let me calculate that for you.The volume $V$ corresponding to a sounding of 4.13 meters is approximately 122.93 cubic meters.

Would you like further details or explanations on this?

Here are some related questions to explore:

1. How does linear interpolation work in different contexts?
2. What other methods could estimate volumes between data points?
3. How would this approach differ if the data were non-linear?
4. Can you apply interpolation for more than two data points?
5. How could errors affect interpolation results in real-world scenarios?

Tip: Linear interpolation assumes a straight-line relationship between two points. Always verify if this assumption is valid for your data!