If the diameter of a softball is 3.5 in.

To find the difference in volumes between a softball and a baseball, we can use the formula for the volume of a sphere:

\[ V = \frac{4}{3} \pi r^3 \]

First, we need to calculate the radius for both the softball and the baseball.

Step 1: Calculate the radii

• Radius of the softball: \[ r_s = \frac{3.5 , \text{in}}{2} = 1.75 , \text{in} \]
• Radius of the baseball: \[ r_b = \frac{2.9 , \text{in}}{2} = 1.45 , \text{in} \]

Step 2: Calculate the volumes

• Volume of the softball: \[ V_s = \frac{4}{3} \pi r_s^3 = \frac{4}{3} \times 3.14 \times (1.75)^3 \] Calculating \( (1.75)^3 \): \[ (1.75)^3 = 5.359375 \] Now substituting back: \[ V_s = \frac{4}{3} \times 3.14 \times 5.359375 \approx \frac{4 \times 3.14 \times 5.359375}{3} \] Calculating \( 4 \times 3.14 \): \[ 4 \times 3.14 = 12.56 \] Now multiply: \[ 12.56 \times 5.359375 \approx 67.3 \] Then divide by 3: \[ V_s \approx \frac{67.3}{3} \approx 22.43 , \text{in}^3 \]

Now rounding: \[ V_s \approx 22.4 , \text{in}^3 \]

• Volume of the baseball: \[ V_b = \frac{4}{3} \pi r_b^3 = \frac{4}{3} \times 3.14 \times (1.45)^3 \] Calculating \( (1.45)^3 \): \[ (1.45)^3 = 3.049625 \] Now substituting back: \[ V_b = \frac{4}{3} \times 3.14 \times 3.049625 \approx \frac{4 \times 3.14 \times 3.049625}{3} \] Calculating: \[ 4 \times 3.14 = 12.56 \] Now multiply: \[ 12.56 \times 3.049625 \approx 38.34 \] Then divide by 3: \[ V_b \approx \frac{38.34}{3} \approx 12.78 , \text{in}^3 \]

Step 3: Calculate the difference

Now, find the difference in volumes: \[ \text{Difference} = V_s - V_b \approx 22.4 - 12.8 \approx 9.6 , \text{in}^3 \]

Final Answer

Thus, the approximate difference in their volumes is:

9.6 in³