Question

A factory's pressure tank rests on the upper base of a vertical pipe whose inside diameter is 1 1/2 ft. and whose length is 40 ft. The tank is a vertical cylinder surmounted by a cone, and it has a hemispherical base. If the altitudes of the cylinder and the cone are respectively 6 ft. and 3 ft. and if all three parts of the tank have an inside diameter of 6 ft. find the volume of the water in the tank and pipe when full. 

Answer 1

The volume of the water:

$$V=V_P+V_T$$

where the volume of the pipe (the volume of a cylinder 40 ft high and 1.5 ft in diameter):

$$V_P =\pi H_p \frac {D^2_p}{4}=\pi \cdot 40ft \cdot \frac {2.25} 4 ft^2 =22.5\pi \space ft^3$$

Tank volume:

$$V_T =V_{hs}+V_{cyl}+V_{con}$$

volume of a hemisphere with a diameter of 6 ft:

$$V_{hs}=\frac 1 2 \frac 4 3 \pi \frac {D^3_{hs} }{8}=\frac 1 {12} \pi \cdot216 ft^3=18 \pi \space ft^3$$

volume of a cylinder 6 ft high and 6 ft in diameter:

$$V_{cyl} =\pi H_{cyl} \frac {D^2_{cyl}}{4}=\pi \cdot 6ft \cdot \frac {36} 4 ft^2 =54\pi \space ft^3$$

volume of a cone 3 ft high and 6 ft in diameter:

$$V_{con}=\frac{H_{con}}{3} \pi \frac {D^2_{con}}{4}=\pi \frac{3ft \cdot 36 ft^2}{12}=9 \pi \space ft^3$$

So

$$V=22.5\pi+18\pi+54\pi+9\pi=103.5\pi \approx 325.155 ft^3$$

Answer: $103.5\pi \approx 325.155 ft^3$