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(D) Let the radius of the hemispherical dome be $r$ meters and the total height of the building be $h$ meters.Since the base diameter of the dome is e

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$6$

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(D) Let the radius of the hemispherical dome be $r$ meters and the total height of the building be $h$ meters.Since the base diameter of the dome is equal to $\frac{2}{3}$ of the total height,therefore $2r = \frac{2}{3}h$,which implies $r = \frac{h}{3}$.Let $H$ be the height of the cylindrical portion. Then $H = h - r = h - \frac{h}{3} = \frac{2}{3}h$ meters.The volume of air inside the building is the sum of the volume of the hemispherical dome and the volume of the cylindrical portion.Volume $= \frac{2}{3}\pi r^3 + \pi r^2 H = \frac{2}{3}\pi (\frac{h}{3})^3 + \pi (\frac{h}{3})^2 (\frac{2}{3}h) = \frac{2}{3}\pi (\frac{h^3}{27}) + \pi (\frac{h^2}{9}) (\frac{2}{3}h) = \frac{2\pi h^3}{81} + \frac{2\pi h^3}{27} = \frac{2\pi h^3 + 6\pi h^3}{81} = \frac{8\pi h^3}{81}$.Given the volume is $67 \frac{1}{21} = \frac{1408}{21} \, m^3$.So,$\frac{8}{81} 	imes \frac{22}{7} 	imes h^3 = \frac{1408}{21}$.$h^3 = \frac{1408}{21} 	imes \frac{81 	imes 7}{8 	imes 22} = \frac{1408}{21} 	imes \frac{567}{176} = 8 	imes 27 = 216$.$h = \sqrt[3]{216} = 6 \, m$.

$A$ building is in the form of a cylinder surmounted by a hemispherical dome (see figure). 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. (in $m$)

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