A building structure in the shape of a cylinder, topped by a hemispherical vaulted dome, contains $41\frac{19}{21}{ \mathrm{m}}^3$air. Find the height of the building if the internal diameter of the dome is equal to its total height above the floor.

Given solid is combination of cylindrical and hemisphere Because the volume of a hemisphere is half that of a sphere, it is written as Volume of hemisphere $\frac{2}{3}\pi R^3$, where r is the radius of the hemisphere.

Volume of the cylinder is $=\frac{1}{3}\pi r^2{h}_1+\pi r^2{h}_1+\frac{2}{3}\pi r^3$

Diameter of hemisphere $\left(\mathrm{D}\right)=$ Total height of building $\left(\mathrm{H}\right)$

Radius of hemisphere $\left(r\right)=\frac{\mathrm{H}}{2}=$ Height of the hemisphere

Height of cylinder $=\mathrm{H}-\frac{\mathrm{H}}{2}=\frac{\mathrm{H}}{2}$

Volume of building = Vol. of cylinder + Vol. of hemisphere

$41\frac{19}{21}$$=\pi r^2\text{H}+\frac{2}{3}\pi {\text{r}}^3$

$\frac{880}{21}=\pi {\left(\frac{\text{H}}{2}\right)}^3+\frac{2}{3}\pi {\left(\frac{\text{H}}{2}\right)}^3$

$\text{H}=4 \text{m}$