Answer :
The silo is made up of a cylindrical portion and a hemispherical cap. We can compute the volumes of each part separately and then add them together.
1. First, note that the diameter of the silo is given as [tex]$4.4$[/tex] m, so the radius is
[tex]$$
r = \frac{4.4}{2} = 2.2 \text{ m}.
$$[/tex]
2. The volume of the cylinder is found using the formula
[tex]$$
V_{\text{cylinder}} = \pi r^2 h,
$$[/tex]
where [tex]$h = 6.2$[/tex] m is the height of the cylindrical portion. Using [tex]$\pi = 3.14$[/tex], we have
[tex]$$
V_{\text{cylinder}} = 3.14 \times (2.2)^2 \times 6.2 \approx 94.22512 \text{ m}^3.
$$[/tex]
3. The volume of a complete sphere would be
[tex]$$
V_{\text{sphere}} = \frac{4}{3} \pi r^3.
$$[/tex]
Since we have a hemisphere (which is half of a sphere), its volume is
[tex]$$
V_{\text{hemisphere}} = \frac{1}{2} \times \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3.
$$[/tex]
Plugging in the radius and [tex]$\pi = 3.14$[/tex], we get
[tex]$$
V_{\text{hemisphere}} = \frac{2}{3} \times 3.14 \times (2.2)^3 \approx 22.2898 \text{ m}^3.
$$[/tex]
4. Adding the two volumes gives the total volume of the silo:
[tex]$$
V_{\text{total}} = V_{\text{cylinder}} + V_{\text{hemisphere}} \approx 94.22512 + 22.2898 \approx 116.5 \text{ m}^3.
$$[/tex]
Thus, the approximate total volume of the silo is [tex]$\boxed{116.5\text{ m}^3}$[/tex].
1. First, note that the diameter of the silo is given as [tex]$4.4$[/tex] m, so the radius is
[tex]$$
r = \frac{4.4}{2} = 2.2 \text{ m}.
$$[/tex]
2. The volume of the cylinder is found using the formula
[tex]$$
V_{\text{cylinder}} = \pi r^2 h,
$$[/tex]
where [tex]$h = 6.2$[/tex] m is the height of the cylindrical portion. Using [tex]$\pi = 3.14$[/tex], we have
[tex]$$
V_{\text{cylinder}} = 3.14 \times (2.2)^2 \times 6.2 \approx 94.22512 \text{ m}^3.
$$[/tex]
3. The volume of a complete sphere would be
[tex]$$
V_{\text{sphere}} = \frac{4}{3} \pi r^3.
$$[/tex]
Since we have a hemisphere (which is half of a sphere), its volume is
[tex]$$
V_{\text{hemisphere}} = \frac{1}{2} \times \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3.
$$[/tex]
Plugging in the radius and [tex]$\pi = 3.14$[/tex], we get
[tex]$$
V_{\text{hemisphere}} = \frac{2}{3} \times 3.14 \times (2.2)^3 \approx 22.2898 \text{ m}^3.
$$[/tex]
4. Adding the two volumes gives the total volume of the silo:
[tex]$$
V_{\text{total}} = V_{\text{cylinder}} + V_{\text{hemisphere}} \approx 94.22512 + 22.2898 \approx 116.5 \text{ m}^3.
$$[/tex]
Thus, the approximate total volume of the silo is [tex]$\boxed{116.5\text{ m}^3}$[/tex].
