Question

How do you see cubic graphy or palobora

Answer 1

Answer on Question#42693 – Math – Geometry

Suppose a cylindrical tank is 12 feet long and 2 feet in diameter. The tank is laid on its side and filled with water to a height of 7 inches. Given that the volume of water in the tank is equal to the length of the tank times the cross-sectional area of the water, find the tank's volume.

Solution:

The volume of water in the tank is:

$$V = A l$$

Where $A$ is the area of a segment and $l$ is the length of tank

Given $r = 1$ foot, since the radius is half the diameter, $h = 7/12$ feet, $l = 12$ feet

The area of a circular segment is:

$$A = 2 \left(A _ {M O R} - A _ {M O P}\right)$$ $$A _ {M O R} = \pi r ^ {2} \frac {\angle M O R}{2 \pi} = r ^ {2} \frac {\angle M O R}{2}$$ $$\angle M O R = \cos^ {- 1} \frac {O P}{r} = \cos^ {- 1} \frac {r - h}{r} = \cos^ {- 1} \frac {1 - \frac {7}{1 2}}{1} = \cos^ {- 1} \frac {5}{1 2}$$ $$A _ {M O R} = 1 ^ {2} \frac {\cos^ {- 1} \frac {5}{1 2}}{2} = \frac {\cos^ {- 1} \frac {5}{1 2}}{2}$$ $$A _ {M O P} = \frac {1}{2} O P \cdot P M$$ $$O P = r - h = 1 - \frac {7}{1 2} = \frac {5}{1 2}$$ $$P M = \sqrt {r ^ {2} - (r - h) ^ {2}} = \sqrt {1 - \left(\frac {5}{1 2}\right) ^ {2}} = \frac {\sqrt {1 1 9}}{1 2}$$ $$A _ {M O P} = \frac {1}{2} \cdot \frac {5}{1 2} \cdot \frac {\sqrt {1 1 9}}{1 2}$$ $$A = 2 \left(\frac {\cos^ {- 1} \frac {5}{1 2}}{2} - \frac {1}{2} \cdot \frac {5}{1 2} \cdot \frac {\sqrt {1 1 9}}{1 2}\right) = \cos^ {- 1} \frac {5}{1 2} - \frac {5 \sqrt {1 1 9}}{1 4 4} = 0. 7 6 2 2$$

So the volume of water in the tank is:

$$V = 0.7622 \cdot 12 = 9.147\ \text{feet}^3$$

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