#### A building is in the form of a cylinder surmounted by a hemispherical vaulted dome and contains $41\frac{19}{21}m^3$ of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building?

Give: Volume of building

$41\frac{19}{21}=\frac{880}{21}$

Let total height above the floor = h

Hemisphere‘s diameter = h (given)

$Radius =\frac{h}{2}$

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

Height of cylinder = total height – the height of hemisphere

$h-\frac{h}{2}=\frac{h}{2}$

Volume $\pi r^2 h=\pi \times \left (\frac{h}{2} \right )^2 \times \frac{h}{2} =\pi \left (\frac{h}{2} \right )^3$

According to question

The volume of building = Volume of cylinder + volume of the hemisphere

$\frac{880}{21}=\left ( \frac{h}{2} \right )^3\left [ \pi + \frac{2}{3} \pi \right ]$

$\frac{880}{21}=\left ( \frac{h}{2} \right )^3\left [ \frac{3\pi +2\pi }{3} \right ]$

$\frac{880 \times 2^3}{21}=h^3 \left [ \frac{5\pi }{3} \right ]$

$\frac{880 \times 2\times 2\times 2\times 7\times 3}{21\times 5\times 22}=h^3$

$h^3 = {2 \times 2 \times 2\times 2\times 2\times 2$

$h = \sqrt[3]{2 \times 2 \times 2\times 2\times 2\times 2}$

$h = 2 \times 2$

h=4m