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Answer :
We are given that the silo has two parts: a cylinder and a hemisphere. The diameter of each part is given as 4.4 meters, so the radius is
[tex]$$
r = \frac{4.4}{2} = 2.2 \text{ meters}.
$$[/tex]
The height of the cylinder is 6.2 meters.
1. To find the volume of the cylindrical part, we use the formula
[tex]$$
V_{\text{cylinder}} = \pi r^2 h.
$$[/tex]
Substitute the values:
[tex]$$
V_{\text{cylinder}} = 3.14 \times (2.2)^2 \times 6.2.
$$[/tex]
Calculating further:
[tex]$$
(2.2)^2 = 4.84,
$$[/tex]
so
[tex]$$
V_{\text{cylinder}} = 3.14 \times 4.84 \times 6.2 \approx 94.22512 \text{ cubic meters}.
$$[/tex]
2. Next, we calculate the volume of the hemispherical part. The volume of a full sphere is
[tex]$$
V_{\text{sphere}} = \frac{4}{3} \pi r^3.
$$[/tex]
Since the silo has a hemisphere, its volume is half of that:
[tex]$$
V_{\text{hemisphere}} = \frac{1}{2} \times \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3.
$$[/tex]
Substitute the values:
[tex]$$
V_{\text{hemisphere}} = \frac{2}{3} \times 3.14 \times (2.2)^3.
$$[/tex]
Calculating further:
[tex]$$
(2.2)^3 \approx 10.648,
$$[/tex]
so
[tex]$$
V_{\text{hemisphere}} \approx \frac{2}{3} \times 3.14 \times 10.648 \approx 22.28981 \text{ cubic meters}.
$$[/tex]
3. Finally, we find the total volume of the silo by adding the volumes of the cylinder and the hemisphere:
[tex]$$
V_{\text{total}} = V_{\text{cylinder}} + V_{\text{hemisphere}} \approx 94.22512 + 22.28981 \approx 116.51493.
$$[/tex]
Rounded to the nearest tenth, the total volume is
[tex]$$
116.5 \text{ cubic meters}.
$$[/tex]
Thus, the approximate total volume of the silo is [tex]$\boxed{116.5\ m^3}$[/tex].
[tex]$$
r = \frac{4.4}{2} = 2.2 \text{ meters}.
$$[/tex]
The height of the cylinder is 6.2 meters.
1. To find the volume of the cylindrical part, we use the formula
[tex]$$
V_{\text{cylinder}} = \pi r^2 h.
$$[/tex]
Substitute the values:
[tex]$$
V_{\text{cylinder}} = 3.14 \times (2.2)^2 \times 6.2.
$$[/tex]
Calculating further:
[tex]$$
(2.2)^2 = 4.84,
$$[/tex]
so
[tex]$$
V_{\text{cylinder}} = 3.14 \times 4.84 \times 6.2 \approx 94.22512 \text{ cubic meters}.
$$[/tex]
2. Next, we calculate the volume of the hemispherical part. The volume of a full sphere is
[tex]$$
V_{\text{sphere}} = \frac{4}{3} \pi r^3.
$$[/tex]
Since the silo has a hemisphere, its volume is half of that:
[tex]$$
V_{\text{hemisphere}} = \frac{1}{2} \times \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3.
$$[/tex]
Substitute the values:
[tex]$$
V_{\text{hemisphere}} = \frac{2}{3} \times 3.14 \times (2.2)^3.
$$[/tex]
Calculating further:
[tex]$$
(2.2)^3 \approx 10.648,
$$[/tex]
so
[tex]$$
V_{\text{hemisphere}} \approx \frac{2}{3} \times 3.14 \times 10.648 \approx 22.28981 \text{ cubic meters}.
$$[/tex]
3. Finally, we find the total volume of the silo by adding the volumes of the cylinder and the hemisphere:
[tex]$$
V_{\text{total}} = V_{\text{cylinder}} + V_{\text{hemisphere}} \approx 94.22512 + 22.28981 \approx 116.51493.
$$[/tex]
Rounded to the nearest tenth, the total volume is
[tex]$$
116.5 \text{ cubic meters}.
$$[/tex]
Thus, the approximate total volume of the silo is [tex]$\boxed{116.5\ m^3}$[/tex].
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