how to calculate volume in(liters) for a specific height in horizontal cylinder I have horizontal cylinder with diameter = 193.04 inches and length = 548.64 inches
I want to find the volume of an oil at specific height with the help of some formula.
I came across below formula to calculate volume at specific height
$$A = \pi a^2/2 - a^2\arcsin(1-h/a) - (a-h)\sqrt{h(2a-h)}.$$
I have a dip chart available for this tank
Dip(cms)   Volume (in liters)
2          27.31
4          77.05  
6          147.44
8          223.36
10         309.81
12         412.40
14         515.66 
16         625.08

Above formula is giving me some other values. Can you please help me with this?
 A: Here is how the volume is calculated. Let $L$ be the length of the cylinder, let $a$ be the radius, and let $A$ be the area of the shaded cross-section in the picture below:

The volume $V$ will be:
$$V = L \cdot A$$
To calculate $A$ we just need to calculate the difference between the "pie section" and the triangle that is cut off by the chord:
$$A = \theta a^2 - \frac{1}{2}a \cdot 2 sin \theta \cdot a \cdot cos \theta = a^2 (\theta - sin \theta \cdot cos \theta) $$
$\theta$ can be calculated from $a \cdot cos \theta = a - h$, but in order to derive the formula in the original question, let's use the angle $\varphi = \frac{\pi}{2}-\theta$ instead. Substituting $\theta = \frac{\pi}{2}-\varphi$ gives:
$$A = \frac{\pi a^2}{2} - \varphi a^2 - a^2 cos \varphi \cdot sin \varphi $$
Eliminate $cos\varphi$:
$$A = \frac{\pi a^2}{2} - \varphi a^2 - a^2 sin \varphi \cdot \sqrt{1-sin^2 \varphi}$$
Since $a \cdot sin \varphi = a-h$, we have $sin\varphi = 1-\frac{h}{a}$ and $\varphi = arcsin(1-\frac{h}{a})$. So the above formula becomes:
$$A = \frac{\pi a^2}{2} - a^2 arcsin(1-\frac{h}{a}) - a^2(1-\frac{h}{a}) \sqrt{1-(1-\frac{h}{a})^2} $$
After simplification of the last term this is identical to the original formula:
$$A = \frac{\pi a^2}{2} - a^2 arcsin(1-\frac{h}{a}) - (a-h)\sqrt{h(2a-h)}$$
