Answer :
To find the change in volume (\(dV\)) of the cylinder, we'll first calculate the initial volume (\(V_0\)) and the final volume (\(V_f\)), and then find the difference between them.
Given:
- Initial radius, \(r = 4 \, \text{cm}\)
- Initial height, \(h_0 = 3 \, \text{cm}\)
- Final height, \(h_f = 3.05 \, \text{cm}\)
1. **Initial Volume (\(V_0\))**:
\[V_0 = \pi r^2 h_0\]
\[V_0 = \pi (4 \, \text{cm})^2 \times 3 \, \text{cm}\]
\[V_0 = 16 \pi \, \text{cm}^3 \times 3 \, \text{cm}\]
\[V_0 = 48 \pi \, \text{cm}^3\]
2. **Final Volume (\(V_f\))**:
\[V_f = \pi r^2 h_f\]
\[V_f = \pi (4 \, \text{cm})^2 \times 3.05 \, \text{cm}\]
\[V_f = 16 \pi \, \text{cm}^3 \times 3.05 \, \text{cm}\]
\[V_f = 48.8 \pi \, \text{cm}^3\]
Now, let's find the change in volume (\(dV\)):
\[dV = V_f - V_0\]
\[dV = 48.8 \pi \, \text{cm}^3 - 48 \pi \, \text{cm}^3\]
\[dV = 0.8 \pi \, \text{cm}^3\]
Approximately,
\[dV \approx 2.512 \, \text{cm}^3\]
So, the change in volume of the cylinder is approximately \(2.512 \, \text{cm}^3\).
Given:
- Initial radius, \(r = 4 \, \text{cm}\)
- Initial height, \(h_0 = 3 \, \text{cm}\)
- Final height, \(h_f = 3.05 \, \text{cm}\)
1. **Initial Volume (\(V_0\))**:
\[V_0 = \pi r^2 h_0\]
\[V_0 = \pi (4 \, \text{cm})^2 \times 3 \, \text{cm}\]
\[V_0 = 16 \pi \, \text{cm}^3 \times 3 \, \text{cm}\]
\[V_0 = 48 \pi \, \text{cm}^3\]
2. **Final Volume (\(V_f\))**:
\[V_f = \pi r^2 h_f\]
\[V_f = \pi (4 \, \text{cm})^2 \times 3.05 \, \text{cm}\]
\[V_f = 16 \pi \, \text{cm}^3 \times 3.05 \, \text{cm}\]
\[V_f = 48.8 \pi \, \text{cm}^3\]
Now, let's find the change in volume (\(dV\)):
\[dV = V_f - V_0\]
\[dV = 48.8 \pi \, \text{cm}^3 - 48 \pi \, \text{cm}^3\]
\[dV = 0.8 \pi \, \text{cm}^3\]
Approximately,
\[dV \approx 2.512 \, \text{cm}^3\]
So, the change in volume of the cylinder is approximately \(2.512 \, \text{cm}^3\).
