To find the approximate change in the volume of the spherical balloon as its radius increases, we use the formula for the volume of a sphere:

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

Step 1: Calculate the volume at $r = 6 \, \text{cm}$

$V_2 = \frac{4}{3} \pi (6)^3 = \frac{4}{3} \pi (216) = 288 \pi \, \text{cm}^3$

Step 2: Calculate the volume at $r = 5.8 \, \text{cm}$

$V_1 = \frac{4}{3} \pi (5.8)^3 = \frac{4}{3} \pi (195.112) \approx 260.149 \pi \, \text{cm}^3$

Step 3: Compute the change in volume

$\Delta V = V_2 - V_1 = 288 \pi - 260.149 \pi \approx 27.851 \pi \, \text{cm}^3$

Step 4: Approximate the numerical value

Using $\pi \approx 3.1416$: $\Delta V \approx 27.851 \times 3.1416 \approx 87.5 \, \text{cm}^3$

Final Answer:

The approximate change in volume is:

$\boxed{87.5 \, \text{cm}^3}$

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Would you like further details or another example? Here are 5 related questions to explore:

1. How does the rate of change of volume depend on the radius of a sphere?
2. What happens to the volume change if the radius increases by a smaller increment, say from $5.8$ cm to $5.9$ cm?
3. Can we use calculus to estimate the volume change for small radius increments?
4. How does the surface area of a sphere change with a similar radius increase?
5. If the balloon is inflated to a pressure, how does that relate to the change in volume?

Tip: When estimating changes, using the derivative can simplify calculations for small increments.