Question

A building is in the form of cylinder surmounted 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

Height of building = height of cylinder + height of hemisphere

$H=h+r$

$r$ is radius of cylinder and dome

diameter of dome $=2/3\times$ height of building

$\therefore 2r=\frac{2}{3}\times (h+r)$

$\therefore 6r=2h+2r$ $\therefore 3r=h+r$

$\therefore h=2r$

Volume of air in building = Volume of cylinder+ Volume of dome

$\therefore 67\frac{1}{21}=\pi r^{2}h+\frac{2}{3}\pi r^3$

$\therefore \frac{1408}{21}=\pi r^2(h+\frac{2}{3}r)$

$=\pi r^2(2r+2/3r)$

$=\pi r^2\left(\frac{8}{3}r\right)$

$\therefore \frac{1408}{21}=\frac{8}{3}\pi r^3$

$\therefore \frac{1408}{7}=8\times \frac{22}{7}\times r^3$

$\therefore r^3=\frac{1408}{22\times 8}=8$

$\therefore r=2m\Rightarrow h=4m$

Height of building $=r+h=6m$