This is the cross section of a truck tank, which is three metres long. All angles are in radians.
In a previous part to this question, we were asked to determine an expression for the volume of the left hand side of the tank.
This is what we obtained:
$$V=\begin{cases} \frac{3}{4}(2\beta - \sin2\beta), &0\le \beta \le \frac{\pi}{2}\\ \frac{3}{4}(2\beta + \sin2\beta), & \frac{\pi}{2} < \beta \le \frac{5\pi}{6} \end{cases}$$
Now, the question is, if the left part of the tank is being filled at a constant rate at $0.001m^3/s$, how much longer will it take to fill the left hand side of the tank.
Now, the given solution plugs in the value of beta when the tank is full (i.e. $5\pi/6$ radians) as follows:
However, I adopted a different approach. I calculated the area of the entire circle ($\pi$) and from this subtracted the minor segment EC (not actually shown in the diagram above). All of this gave me the following as the volume of the left hand side of the tank:
$$3\cdot\frac{\pi - \frac{\pi}{6} + \frac{\sqrt{3}}4}2$$
which is approximately equal to $4.58m^3$. This is not the same answer as the method above. Can anyone see where I am going wrong? I am sure it must be something very silly indeed!
The formula obtained in the the question's previous part is wrong: $V$ actually turns out to have the same expression for both the given cases of $\beta.$ This wrong formula also fails a sanity check: according to it, $V$ undergoes a sharp change as $\beta$ increases past $\frac{\pi}2.$
Let G be the intersection of AO produced and the water level. For $\beta\in\left(\frac{\pi}2,\frac{5\pi}6\right],$ the water’s cross-sectional area is $$\text{area of sector } OAF+\text{area of } OFG\\=\frac12\left(1^2\right)\beta+\frac12(1)(OG)\sin(\pi-\beta)\\=\frac12\beta+\frac12\cos(\pi-\beta)\sin(\pi-\beta)\\=\frac14\left(2\beta+2(-\cos\beta)\sin\beta\right)\\=\frac14\left(2\beta-\sin2\beta\right).$$
Thus, the full left-side volume (given by $\beta=\frac{5\pi}6$ and cylinder length $3\mathrm{m}$) is $4.58\mathrm{m}^3,$ which is consistent with the value obtained using your alternative approach.