The radius of a sphere is 4 cm with a possible error of 0.01 cm . Then absolute error in volume is

Options

A. `64 cm^(3)`
B. `pm 0 . 64 pi cm^(3)`
C. ` 6 . 4 pi cm^(3)`
D. `pm 0.0075 cm^(3)`

Answer

To find the absolute error in the volume of a sphere given the radius and its possible 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:
\[ V = \frac{4}{3} \pi r^3 \]

### Step 2: Differentiate the volume with respect to the radius
To find the absolute error in volume, we need to differentiate the volume formula with respect to the radius \( r \). The differential of volume \( dV \) can be expressed as:
\[ dV = \frac{dV}{dr} \cdot dr \]
Calculating the derivative:
\[ \frac{dV}{dr} = 4 \pi r^2 \]
Thus, we have:
\[ dV = 4 \pi r^2 \cdot dr \]

### Step 3: Substitute the known values
We know:
- The radius \( r = 4 \) cm
- The possible error in radius \( dr = 0.01 \) cm

Now substituting these values into the equation:
\[ dV = 4 \pi (4^2) \cdot 0.01 \]
Calculating \( 4^2 \):
\[ 4^2 = 16 \]
So,
\[ dV = 4 \pi (16) \cdot 0.01 \]
This simplifies to:
\[ dV = 64 \pi \cdot 0.01 \]
Calculating further:
\[ dV = 0.64 \pi \]

### Step 4: Calculate the numerical value
Using the approximate value of \( \pi \approx 3.14 \):
\[ dV \approx 0.64 \times 3.14 \approx 2.0096 \text{ cm}^3 \]

### Final Result
The absolute error in volume is:
\[ dV \approx 0.64 \pi \text{ cm}^3 \text{ or approximately } 2.01 \text{ cm}^3 \]