The volume of a sphere is given by
`V=4/3 piR^(3)`
where `R` is the radius of the radius of the sphere. Find the change in volume of the sphere as the radius is increased from `10.0 cm` to `10.1 cm`. Assume that the rate does not appreciably change between `R=10.0 cm` to `R=10.1 cm`

Options

A. `10 pi cm^(3)`
B. `20 pi cm^(3)`
C. `30 pi cm^(3)`
D. `40 pi cm^(3)`

Explanation

To find the change in volume of a sphere as the radius increases from 10.0 cm to 10.1 cm, we can follow these steps:

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

### Step 2: Differentiate the volume with respect to the radius.
To find the change in volume with respect to a small change in radius, we differentiate the volume formula:
\[ \frac{dV}{dR} = 4 \pi R^2 \]

### Step 3: Calculate the change in radius.
The change in radius \( dR \) is given by:
\[ dR = R_{final} - R_{initial} = 10.1 \, \text{cm} - 10.0 \, \text{cm} = 0.1 \, \text{cm} \]

### Step 4: Substitute the radius and change in radius into the derivative.
Now, substitute \( R = 10.0 \, \text{cm} \) and \( dR = 0.1 \, \text{cm} \) into the differentiated volume formula:
\[ dV = \frac{dV}{dR} \cdot dR = (4 \pi (10.0)^2) \cdot (0.1) \]

### Step 5: Calculate the change in volume.
Now, calculate \( dV \):
\[ dV = 4 \pi (100) \cdot (0.1) = 400 \pi \cdot 0.1 = 40 \pi \, \text{cm}^3 \]

### Step 6: Approximate the value of \( dV \).
Using \( \pi \approx 3.14 \):
\[ dV \approx 40 \cdot 3.14 = 125.6 \, \text{cm}^3 \]

### Final Answer:
The change in volume of the sphere as the radius increases from 10.0 cm to 10.1 cm is approximately \( 125.6 \, \text{cm}^3 \).

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