Question

A building is in the form of a cylinder surrounded by a hemispherical dome. The base diameter of the dome is equal to $\frac{2}{3}$ of the total height of the building . Find the height of the building , if it contains $67\frac{1}{21}{m}^3$ of air.

Answer 1

Let the radius of the dome be r.
Diameter be d.
Let the height of the building be H.
Given $d=\frac{2}{3}H$
$$\begin{gathered} \Rightarrow 2r=\frac{2}{3}H \\ \Rightarrow r=\frac{H}{3} \\ \Rightarrow 3r=H \end{gathered}$$
Also, h + r = H
$$\begin{gathered} \Rightarrow 3r=h+r \\ \Rightarrow 2r=h \\ \Rightarrow r=\frac{h}{2} \end{gathered}$$
Volume of air = Volume of air in the cylinder + Volume of air int he hemispherical dome
$$\begin{gathered} \Rightarrow {\mathrm{\pi r}}^2\mathrm{h}+\frac{2}{3}{\mathrm{\pi r}}^3=67\frac{1}{21} \\ \Rightarrow \mathrm{\pi }{\left(\frac{\mathrm{h}}{2}\right)}^2\mathrm{h}+\frac{2}{3}\mathrm{\pi }{\left(\frac{h}{2}\right)}^3=\frac{1408}{21} \\ \Rightarrow h^3=\frac{11264}{176}=64 \\ \Rightarrow h=4m \end{gathered}$$
Hence, the radius will be $r=\frac{h}{2}=\frac{4}{2}=2m$
Height of the building, H = 3r = $3\times 2=6m$