The radius of a sphere is 8 cm and 0.02 cm is the error in its measurement. Find the approximate error in its volume.

Answer

To find the approximate error in the volume of a sphere given the radius and the error in its measurement, we can follow these steps:

### Step 1: Understand the problem
We are given:
- Radius of the sphere, \( r = 8 \) cm
- Error in measurement of radius, \( dr = 0.02 \) cm

We need to find the approximate error in the volume, denoted as \( dV \).

### Step 2: Write the formula for the volume of a sphere
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]

### Step 3: Differentiate the volume with respect to the radius
To find the approximate error in volume, we differentiate the volume with respect to the radius:
\[ dV = \frac{dV}{dr} \cdot dr \]
Using the formula for volume:
\[ \frac{dV}{dr} = 4 \pi r^2 \]
Thus, we can write:
\[ dV = 4 \pi r^2 \cdot dr \]

### Step 4: Substitute the values into the differentiated equation
Now, we substitute \( r = 8 \) cm and \( dr = 0.02 \) cm into the equation:
\[ dV = 4 \pi (8^2) \cdot 0.02 \]

### Step 5: Calculate \( 8^2 \)
Calculating \( 8^2 \):
\[ 8^2 = 64 \]

### Step 6: Substitute \( 64 \) back into the equation
Now we substitute \( 64 \) into the equation:
\[ dV = 4 \pi (64) \cdot 0.02 \]

### Step 7: Calculate \( 4 \cdot 64 \)
Calculating \( 4 \cdot 64 \):
\[ 4 \cdot 64 = 256 \]

### Step 8: Substitute \( 256 \) into the equation
Now we substitute \( 256 \) back into the equation:
\[ dV = 256 \pi \cdot 0.02 \]

### Step 9: Calculate \( 256 \cdot 0.02 \)
Calculating \( 256 \cdot 0.02 \):
\[ 256 \cdot 0.02 = 5.12 \]

### Step 10: Write the final expression for \( dV \)
Thus, we have:
\[ dV = 5.12 \pi \text{ cm}^3 \]

### Step 11: Approximate the value of \( \pi \)
Using \( \pi \approx 3.14 \):
\[ dV \approx 5.12 \cdot 3.14 \approx 16.0948 \text{ cm}^3 \]

### Step 12: Final result
The approximate error in the volume of the sphere is:
\[ dV \approx 16.1 \text{ cm}^3 \]