How much of this log is below the surface?

Question

The figure shows the circular cross-section of a uniform log of radius $40\,\text{cm}$ floating in water. The points $A$ and $B$ are on the surface and the highest point $X$ is $8\,\text{cm}$ above the surface.

Show that $\widehat{AOB}$ is approximately $1.29$ radians.

Calculate

the length of the arc $AXB$ ,

the area of the cross-section below the surface,

the percentage of the volume of the log below the surface.

Solution

Let $M$ be the point on $AB$ vertically below $X$ . The length $OX$ is $40$ and $MX$ is $8$ , so $OM$ is $32\,\text{cm}$ .

We’ll compute the angle $\alpha$ , shown here, using its cosine.

We find $\alpha = \cos^{-1}\left(\frac{32}{40}\right) \approx 0.6435... \text{ radians.}$

Since the angle $\widehat{AOB}$ is $2 \alpha$ , we have $\widehat{AOB} \approx 1.2870... = 1.29$ radians (3sf).

The length of the arc is $s = r\theta$ . In our case, $s \approx 40 \times 1.287 = 51.5\,\text{cm}$ (3sf).

Let’s call the area of the cross-section below the surface $S$ . We can calculate it as the sum of the triangle $OAB$ and the major sector $OAB$ .

The area of the triangle is $\frac{1}{2}a b \sin C = \frac{1}{2} 40^2\sin1.287 \approx 768.0$ .

The area of the sector is $\frac{1}{2}r^2\theta = \frac{1}{2} 40^2(2\pi-1.287) \approx 3997.0$ .

So we have $S\approx768.0+3997.0 = 4765.0 = 4770\,\text{cm}^{2}$ (3sf). To calculate $S$ , we need to compute the area of the circle segment shown in red below. This is approximately the area of the sector $OAXB$ minus the area of triangle $OAB$ , which is $\frac{40^2}{2}(1.287 - \sin(1.287)) \approx 261.6.$ Given that the whole circle has an area of $40^2\pi \approx 5026.6\,\text{cm}^{2}$ , $S \approx 5026.6-261.6 = 4765.0=4770\,\text{cm}^{2}$ (3sf).

The percentage of the volume of the log below the surface is the same as the percentage of the area of the cross-section below the surface.

This is $\dfrac{4765.0}{\pi 40^2}\times100 \approx 94.80 = 94.8\%$ (3sf).