Answer
612 π m³
1922 m³
Explanation
The body of the rocket is a cylinder; to find its volume, we are going to use the formula for the volume of a cylinder:
[tex]V=\pi r^2h[/tex]
where
[tex]r[/tex] is the radius of the cylinder
[tex]h[/tex] is the height
We know from our problem that the radius of the cylinder is 6 m and its height is 16 m, so let's replace the values:
[tex]V=\pi (6m)^2(16m)[/tex]
[tex]V=\pi (36m^2)(16m)[/tex]
[tex]V=576\pi m^3[/tex]
[tex]V=1809m^3[/tex]
The volume of the cylinder is [tex]1810m^3[/tex]
To find the volume of the cone, we are going to use the formula:
[tex]V=\pi r^2\frac{h}{3}[/tex]
where
[tex]r[/tex] is the radius
[tex]h[/tex] is the height
We know form our problem that the radius of the cone is 6 m and its height is 3 m, so let't replace the values:
[tex]V=\pi (6m)^2(\frac{3m}{3} )[/tex]
[tex]V=\pi (36m^2)(1m)[/tex]
[tex]V=36\pi m^3[/tex]
[tex]V=113m^3[/tex]
Now, we just need to add the volumes in terms of [tex]\pi[/tex] and the volumes rounded to the nearest whole:
Volume of the item in terms of π = [tex]576\pi m^3+36\pi m^3=612\pi m^3[/tex]
Volume of the item rounded = [tex]1809m^3+113m^3=1922m^3[/tex]
