Answer:
(1). [tex]V=\pi r^2l+\frac{1}{3}\pi r^2 h[/tex]
(2).a. [tex]25\pi[/tex] square feet
b. [tex]158.33\pi[/tex] cubic feet
Step-by-step explanation:
(1).
The rocket is made up of a cylinder and a cone, this means its total volume [tex]V[/tex] is
[tex]V =V_{cylinder}+V_{cone}[/tex]
Now
[tex]V_{cylinder} = \pi r^2l[/tex]
[tex]V_{cone} = \frac{1}{3}\pi r^2h[/tex]
Thus
[tex]\boxed{V=\pi r^2l+\frac{1}{3}\pi r^2 h}[/tex]
(2).
a. The floor is a disk with diameter 10/2 = 5 feet; therefore, its area [tex]A[/tex] is
[tex]A=\pi r^2=\pi (5ft)^2[/tex]
[tex]\boxed{A=25\pi }[/tex]
b.
The volume of the space inside the observatory [tex]V_{obs}[/tex] is equal to the volume of the cylinder and the hemisphere:
[tex]V_{obs} =V_{hemisphere}+V_{cylinder}[/tex]
[tex]V_{hemisphere}=\frac{1}{2}* \frac{4}{3}\pi r^3=83.33\pi ft^3[/tex]
[tex]V_{cylinder}=\pi (5)^2(3)=75\pi[/tex]
Thus
[tex]V_{obs}=83.33\pi ft^3+75\pi ft^3=\boxed{158.33ft^3}[/tex]
