Problem A closed cylindrical tank measures 12 ft. long and 5 ft. in diameter. It has to contain water to a depth of 3 ft when lying in the horizontal position. Find the depth of water when it is in vertical position.

Solution

$\cos \frac{1}{2}\theta = \dfrac{0.5}{2.5}$

$\frac{1}{2}\theta = 78.463^\circ$

$\theta = 156.926^\circ$

$\alpha = 360^\circ - \theta = 360^\circ - 156.926^\circ$

$\alpha = 203.074^\circ$

Area of cross section: $A_b = \frac{1}{2} r^2(\alpha_{rad} + \sin \theta_{deg})$

$A_b = \frac{1}{2} (2.5^2)\left( 203.074^\circ \cdot \dfrac{\pi}{180^\circ} + \sin 156.926^\circ \right)$

$A_b = 12.3 ~ \text{ft}^2$

Volume of water: $V_w = A_b L = 12.3(12)$

$V_w = 147.61 ~ \text{ft}^3$

Depth of water in vertical position:

$V_w = \pi r^2 h$

$147.61 = \pi (2.5^2)h$

$h = 7.518 ~ \text{ft}$ answer

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