If the radius of a sphere is measured as 9cm with an error of 0.03 cm, 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: Differentiate the volume with respect to the radius
To find the approximate error in volume, we need to differentiate the volume with respect to the radius \( r \):
\[ \frac{dV}{dr} = 4 \pi r^2 \]

### Step 3: Relate the change in volume to the change in radius
The change in volume \( dV \) can be expressed in terms of the change in radius \( dr \):
\[ dV = \frac{dV}{dr} \cdot dr \]
Substituting the derivative we found:
\[ dV = 4 \pi r^2 \cdot dr \]

### Step 4: Substitute the values
We know:
- The radius \( r = 9 \) cm
- The error in the radius \( dr = 0.03 \) cm

Now substituting these values into the equation:
\[ dV = 4 \pi (9)^2 \cdot 0.03 \]

### Step 5: Calculate \( dV \)
Calculating \( 9^2 \):
\[ 9^2 = 81 \]
Now substituting back:
\[ dV = 4 \pi \cdot 81 \cdot 0.03 \]
Calculating \( 4 \cdot 81 \):
\[ 4 \cdot 81 = 324 \]
So we have:
\[ dV = 324 \pi \cdot 0.03 \]
Calculating \( 324 \cdot 0.03 \):
\[ 324 \cdot 0.03 = 9.72 \]
Thus, we find:
\[ dV = 9.72 \pi \]

### Step 6: Final answer
The approximate error in calculating the volume is:
\[ dV \approx 9.72 \pi \text{ cm}^3 \]
If we approximate \( \pi \) as \( 3.14 \):
\[ dV \approx 9.72 \times 3.14 \approx 30.54 \text{ cm}^3 \]

### Conclusion
The approximate error in the volume of the sphere is \( 9.72 \pi \text{ cm}^3 \) or approximately \( 30.54 \text{ cm}^3 \).

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