Answer :
We are given a grain silo made up of a cylinder and a hemisphere. The diameter of the silo is 4.4 meters, so the radius is
[tex]$$
r = \frac{4.4}{2} = 2.2\text{ m}.
$$[/tex]
The height of the cylindrical part is 6.2 meters. We use [tex]$\pi = 3.14$[/tex].
1. Volume of the Cylinder
The volume of a cylinder is given by
[tex]$$
V_{\text{cylinder}} = \pi r^2 h.
$$[/tex]
Substituting the values:
[tex]$$
V_{\text{cylinder}} = 3.14 \times (2.2)^2 \times 6.2.
$$[/tex]
Calculating the intermediate steps:
[tex]$$
(2.2)^2 = 4.84,
$$[/tex]
then
[tex]$$
V_{\text{cylinder}} \approx 3.14 \times 4.84 \times 6.2 \approx 94.225\text{ m}^3.
$$[/tex]
2. Volume of the Hemisphere
The volume of a full sphere is
[tex]$$
V_{\text{sphere}} = \frac{4}{3} \pi r^3.
$$[/tex]
Since we have a hemisphere, the volume is half that:
[tex]$$
V_{\text{hemisphere}} = \frac{1}{2} \times \frac{4}{3}\pi r^3 = \frac{2}{3} \pi r^3.
$$[/tex]
Substituting the value of [tex]$r$[/tex]:
[tex]$$
V_{\text{hemisphere}} = \frac{2}{3} \times 3.14 \times (2.2)^3.
$$[/tex]
First, compute:
[tex]$$
(2.2)^3 \approx 10.648.
$$[/tex]
Then the volume becomes:
[tex]$$
V_{\text{hemisphere}} \approx \frac{2}{3} \times 3.14 \times 10.648 \approx 22.2898\text{ m}^3.
$$[/tex]
3. Total Volume
The total volume of the silo is the sum of the volumes of the cylinder and the hemisphere:
[tex]$$
V_{\text{total}} = V_{\text{cylinder}} + V_{\text{hemisphere}} \approx 94.225 + 22.2898 \approx 116.5148\text{ m}^3.
$$[/tex]
Rounding to the nearest tenth, we get
[tex]$$
V_{\text{total}} \approx 116.5\text{ m}^3.
$$[/tex]
Thus, the approximate total volume of the silo is [tex]$\boxed{116.5\text{ m}^3}$[/tex].
[tex]$$
r = \frac{4.4}{2} = 2.2\text{ m}.
$$[/tex]
The height of the cylindrical part is 6.2 meters. We use [tex]$\pi = 3.14$[/tex].
1. Volume of the Cylinder
The volume of a cylinder is given by
[tex]$$
V_{\text{cylinder}} = \pi r^2 h.
$$[/tex]
Substituting the values:
[tex]$$
V_{\text{cylinder}} = 3.14 \times (2.2)^2 \times 6.2.
$$[/tex]
Calculating the intermediate steps:
[tex]$$
(2.2)^2 = 4.84,
$$[/tex]
then
[tex]$$
V_{\text{cylinder}} \approx 3.14 \times 4.84 \times 6.2 \approx 94.225\text{ m}^3.
$$[/tex]
2. Volume of the Hemisphere
The volume of a full sphere is
[tex]$$
V_{\text{sphere}} = \frac{4}{3} \pi r^3.
$$[/tex]
Since we have a hemisphere, the volume is half that:
[tex]$$
V_{\text{hemisphere}} = \frac{1}{2} \times \frac{4}{3}\pi r^3 = \frac{2}{3} \pi r^3.
$$[/tex]
Substituting the value of [tex]$r$[/tex]:
[tex]$$
V_{\text{hemisphere}} = \frac{2}{3} \times 3.14 \times (2.2)^3.
$$[/tex]
First, compute:
[tex]$$
(2.2)^3 \approx 10.648.
$$[/tex]
Then the volume becomes:
[tex]$$
V_{\text{hemisphere}} \approx \frac{2}{3} \times 3.14 \times 10.648 \approx 22.2898\text{ m}^3.
$$[/tex]
3. Total Volume
The total volume of the silo is the sum of the volumes of the cylinder and the hemisphere:
[tex]$$
V_{\text{total}} = V_{\text{cylinder}} + V_{\text{hemisphere}} \approx 94.225 + 22.2898 \approx 116.5148\text{ m}^3.
$$[/tex]
Rounding to the nearest tenth, we get
[tex]$$
V_{\text{total}} \approx 116.5\text{ m}^3.
$$[/tex]
Thus, the approximate total volume of the silo is [tex]$\boxed{116.5\text{ m}^3}$[/tex].
