Answer :
The grain silo consists of two parts: a right circular cylinder and a hemisphere. We are given:
- Diameter of the silo: [tex]$4.4$[/tex] m (which gives the radius [tex]$r$[/tex] as half of the diameter)
- Height of the cylindrical portion: [tex]$6.2$[/tex] m
- Use [tex]$3.14$[/tex] for [tex]$\pi$[/tex]
Step 1. Calculate the radius.
Since the diameter is [tex]$4.4$[/tex] m, the radius is
[tex]$$
r = \frac{4.4}{2} = 2.2 \text{ m}.
$$[/tex]
Step 2. Volume of the cylindrical portion.
The formula for the volume of a cylinder is
[tex]$$
V_{\text{cyl}} = \pi r^2 h.
$$[/tex]
Substitute [tex]$r = 2.2$[/tex] m and [tex]$h = 6.2$[/tex] m into the formula:
[tex]$$
V_{\text{cyl}} = 3.14 \times (2.2)^2 \times 6.2.
$$[/tex]
First, compute the square of the radius:
[tex]$$
(2.2)^2 = 4.84.
$$[/tex]
Thus,
[tex]$$
V_{\text{cyl}} = 3.14 \times 4.84 \times 6.2 \approx 94.22512 \text{ m}^3.
$$[/tex]
Step 3. Volume of the hemispherical portion.
The volume of a sphere is
[tex]$$
V_{\text{sphere}} = \frac{4}{3} \pi r^3.
$$[/tex]
Since we have a hemisphere, its volume is half of that:
[tex]$$
V_{\text{hemi}} = \frac{1}{2} \times \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3.
$$[/tex]
Substitute [tex]$r = 2.2$[/tex] m:
[tex]$$
V_{\text{hemi}} = \frac{2}{3} \times 3.14 \times (2.2)^3.
$$[/tex]
First, compute the cube of the radius:
[tex]$$
(2.2)^3 = 10.648.
$$[/tex]
Then,
[tex]$$
V_{\text{hemi}} \approx \frac{2}{3} \times 3.14 \times 10.648 \approx 22.28981 \text{ m}^3.
$$[/tex]
Step 4. Total Volume of the Silo.
Add the volume of the cylinder and the hemisphere:
[tex]$$
V_{\text{total}} = V_{\text{cyl}} + V_{\text{hemi}} \approx 94.22512 + 22.28981 \approx 116.51493 \text{ m}^3.
$$[/tex]
When rounded to the nearest tenth, the total volume is:
[tex]$$
116.5 \text{ m}^3.
$$[/tex]
Thus, the approximate total volume of the silo is [tex]$\boxed{116.5 \text{ m}^3}$[/tex].
- Diameter of the silo: [tex]$4.4$[/tex] m (which gives the radius [tex]$r$[/tex] as half of the diameter)
- Height of the cylindrical portion: [tex]$6.2$[/tex] m
- Use [tex]$3.14$[/tex] for [tex]$\pi$[/tex]
Step 1. Calculate the radius.
Since the diameter is [tex]$4.4$[/tex] m, the radius is
[tex]$$
r = \frac{4.4}{2} = 2.2 \text{ m}.
$$[/tex]
Step 2. Volume of the cylindrical portion.
The formula for the volume of a cylinder is
[tex]$$
V_{\text{cyl}} = \pi r^2 h.
$$[/tex]
Substitute [tex]$r = 2.2$[/tex] m and [tex]$h = 6.2$[/tex] m into the formula:
[tex]$$
V_{\text{cyl}} = 3.14 \times (2.2)^2 \times 6.2.
$$[/tex]
First, compute the square of the radius:
[tex]$$
(2.2)^2 = 4.84.
$$[/tex]
Thus,
[tex]$$
V_{\text{cyl}} = 3.14 \times 4.84 \times 6.2 \approx 94.22512 \text{ m}^3.
$$[/tex]
Step 3. Volume of the hemispherical portion.
The volume of a sphere is
[tex]$$
V_{\text{sphere}} = \frac{4}{3} \pi r^3.
$$[/tex]
Since we have a hemisphere, its volume is half of that:
[tex]$$
V_{\text{hemi}} = \frac{1}{2} \times \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3.
$$[/tex]
Substitute [tex]$r = 2.2$[/tex] m:
[tex]$$
V_{\text{hemi}} = \frac{2}{3} \times 3.14 \times (2.2)^3.
$$[/tex]
First, compute the cube of the radius:
[tex]$$
(2.2)^3 = 10.648.
$$[/tex]
Then,
[tex]$$
V_{\text{hemi}} \approx \frac{2}{3} \times 3.14 \times 10.648 \approx 22.28981 \text{ m}^3.
$$[/tex]
Step 4. Total Volume of the Silo.
Add the volume of the cylinder and the hemisphere:
[tex]$$
V_{\text{total}} = V_{\text{cyl}} + V_{\text{hemi}} \approx 94.22512 + 22.28981 \approx 116.51493 \text{ m}^3.
$$[/tex]
When rounded to the nearest tenth, the total volume is:
[tex]$$
116.5 \text{ m}^3.
$$[/tex]
Thus, the approximate total volume of the silo is [tex]$\boxed{116.5 \text{ m}^3}$[/tex].
