Answer :
We begin by noting that the silo consists of two parts: a cylindrical portion and a hemispherical top.
1. First, determine the radius. The diameter is given as [tex]$4.4$[/tex] meters, so the radius is
[tex]$$
r = \frac{4.4}{2} = 2.2 \text{ meters}.
$$[/tex]
2. Next, calculate the volume of the cylindrical part. The formula for the volume of a cylinder is
[tex]$$
V_{\text{cyl}} = \pi r^2 h,
$$[/tex]
where the height of the cylinder is [tex]$6.2$[/tex] meters. Substituting the values (with [tex]$\pi = 3.14$[/tex]), we get
[tex]$$
V_{\text{cyl}} = 3.14 \times (2.2)^2 \times 6.2.
$$[/tex]
3. Now, calculate the volume of the hemispherical top. The formula for the volume of a hemisphere is
[tex]$$
V_{\text{hemi}} = \frac{2}{3}\pi r^3.
$$[/tex]
Again, substituting [tex]$r = 2.2$[/tex] meters and [tex]$\pi = 3.14$[/tex], we have
[tex]$$
V_{\text{hemi}} = \frac{2}{3} \times 3.14 \times (2.2)^3.
$$[/tex]
4. The total volume of the silo is the sum of the volumes of the cylinder and hemisphere:
[tex]$$
V_{\text{total}} = V_{\text{cyl}} + V_{\text{hemi}}.
$$[/tex]
5. After performing the calculations, we get:
- Volume of the cylinder: approximately [tex]$94.225$[/tex] cubic meters.
- Volume of the hemisphere: approximately [tex]$22.290$[/tex] cubic meters.
Therefore, the total volume is
[tex]$$
V_{\text{total}} \approx 94.225 + 22.290 = 116.5 \text{ cubic meters}.
$$[/tex]
Rounding to the nearest tenth, the approximate total volume of the silo is [tex]$\boxed{116.5 \, m^3}$[/tex].
1. First, determine the radius. The diameter is given as [tex]$4.4$[/tex] meters, so the radius is
[tex]$$
r = \frac{4.4}{2} = 2.2 \text{ meters}.
$$[/tex]
2. Next, calculate the volume of the cylindrical part. The formula for the volume of a cylinder is
[tex]$$
V_{\text{cyl}} = \pi r^2 h,
$$[/tex]
where the height of the cylinder is [tex]$6.2$[/tex] meters. Substituting the values (with [tex]$\pi = 3.14$[/tex]), we get
[tex]$$
V_{\text{cyl}} = 3.14 \times (2.2)^2 \times 6.2.
$$[/tex]
3. Now, calculate the volume of the hemispherical top. The formula for the volume of a hemisphere is
[tex]$$
V_{\text{hemi}} = \frac{2}{3}\pi r^3.
$$[/tex]
Again, substituting [tex]$r = 2.2$[/tex] meters and [tex]$\pi = 3.14$[/tex], we have
[tex]$$
V_{\text{hemi}} = \frac{2}{3} \times 3.14 \times (2.2)^3.
$$[/tex]
4. The total volume of the silo is the sum of the volumes of the cylinder and hemisphere:
[tex]$$
V_{\text{total}} = V_{\text{cyl}} + V_{\text{hemi}}.
$$[/tex]
5. After performing the calculations, we get:
- Volume of the cylinder: approximately [tex]$94.225$[/tex] cubic meters.
- Volume of the hemisphere: approximately [tex]$22.290$[/tex] cubic meters.
Therefore, the total volume is
[tex]$$
V_{\text{total}} \approx 94.225 + 22.290 = 116.5 \text{ cubic meters}.
$$[/tex]
Rounding to the nearest tenth, the approximate total volume of the silo is [tex]$\boxed{116.5 \, m^3}$[/tex].
