If the radius of a sphere is measured as 5 m with an error of 0.03 m, then find the approximate error in calculating its volume.

Answer

To find the approximate error in calculating the volume of a sphere when the radius is measured with a certain error, we can follow these steps:

### Step 1: Understand the formula for the volume of a sphere
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the sphere.

### Step 2: Identify the given values
We are given:
- The radius \( r = 5 \, \text{m} \)
- The error in the radius \( \Delta r = 0.03 \, \text{m} \)

### Step 3: Differentiate the volume with respect to the radius
To find the approximate change in volume \( \Delta V \) due to a change in radius \( \Delta r \), we need to find the derivative of the volume with respect to the radius:
\[ \frac{dV}{dr} = 4 \pi r^2 \]

### Step 4: Substitute the radius into the derivative
Now, substitute \( r = 5 \, \text{m} \) into the derivative:
\[ \frac{dV}{dr} = 4 \pi (5)^2 = 4 \pi \times 25 = 100 \pi \]

### Step 5: Calculate the approximate change in volume
Using the formula for the approximate change in volume:
\[ \Delta V \approx \frac{dV}{dr} \cdot \Delta r \]
Substituting the values we have:
\[ \Delta V \approx 100 \pi \cdot 0.03 \]
Calculating this gives:
\[ \Delta V \approx 3 \pi \, \text{m}^3 \]

### Final Result
Thus, the approximate error in calculating the volume of the sphere is:
\[ \Delta V \approx 3 \pi \, \text{m}^3 \]

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