Question

If the diameter of a softball is 3.5 in.
and the diameter of a baseball is 2.9 in.
, what is the approximate difference in their volumes?

Use 3.14 for π
and round your answer to the nearest tenth.

Answer 1

To calculate the difference in the volumes of a softball and a baseball, we will use the formula for the volume of a sphere:

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

where \( r \) is the radius of the sphere.

First, we need to find the radius of each ball by dividing their diameters by 2.

1. **Softball:**
- Diameter = 3.5 inches
- Radius \( r_{\text{softball}} = \frac{3.5}{2} = 1.75 \) inches

2. **Baseball:**
- Diameter = 2.9 inches
- Radius \( r_{\text{baseball}} = \frac{2.9}{2} = 1.45 \) inches

Now we can calculate the volumes:

1. **Volume of the softball:**
\[
V_{\text{softball}} = \frac{4}{3} \cdot 3.14 \cdot (1.75)^3
\]

First, calculate \( (1.75)^3 \):
\[
(1.75)^3 = 1.75 \times 1.75 \times 1.75 = 5.359375
\]

Now, substitute back into the volume formula:
\[
V_{\text{softball}} = \frac{4}{3} \cdot 3.14 \cdot 5.359375 \approx \frac{4 \cdot 3.14 \cdot 5.359375}{3} \approx \frac{67.41425}{3} \approx 22.47142 \text{ in}^3
\]

2. **Volume of the baseball:**
\[
V_{\text{baseball}} = \frac{4}{3} \cdot 3.14 \cdot (1.45)^3
\]

First, calculate \( (1.45)^3 \):
\[
(1.45)^3 = 1.45 \times 1.45 \times 1.45 = 3.052625
\]

Now, substitute back into the volume formula:
\[
V_{\text{baseball}} = \frac{4}{3} \cdot 3.14 \cdot 3.052625 \approx \frac{4 \cdot 3.14 \cdot 3.052625}{3} \approx \frac{38.20326}{3} \approx 12.73442 \text{ in}^3
\]

Finally, we find the difference in volume between the softball and the baseball:

\[
\text{Difference} = V_{\text{softball}} - V_{\text{baseball}} = 22.47142 - 12.73442 \approx 9.737
\]

Rounding this to the nearest tenth gives us:

\[
\text{Approximate difference in volumes} \approx 9.7 \text{ in}^3
\]

Thus, the approximate difference in their volumes is **9.7 in³**.