Question

The table below gives the depth of water across a river measured at one metre
intervals between banks.
Distance (m) 0 1 2 3 4
Water depth (m) 0 0.5 1.6 0.9 0
Use the Trapezium rule to estimate the cross-sectional area of the river.
A river hydrologist estimates that at the place where this cross sectional data was
measured the average speed of water flow is 0.6m/s. Estimate the volume of water
which passes this section of the river in one minute.

Answer 1

The table below gives the depth of water across a river measured at one metre intervals between banks. Distance (m) 0 1 2 3 4 Water depth (m) 0 0.5 1.6 0.9 0 Use the Trapezium rule to estimate the cross-sectional area of the river. A river hydrologist estimates that at the place where this cross sectional data was measured the average speed of water flow is 0.6 m/s. Estimate the volume of water which passes this section of the river in one minute.

Solution

1st trapezium points: (0,0), (1,0), (0,0), (1,0.5).

1st trapezium area: $S_{1} = (1 - 0) * \min\{0,0.5\} + (1 - 0) * \frac{0.5 - 0}{2} = 0.25 \, m^{2}$

2nd trapezium points: (1,0), (2,0), (1,0.5), (2,1.6).

2nd trapezium area: $S_{2} = (2 - 1) * \min\{0.5, 1.6\} + (2 - 1) * \frac{1.6 - 0.5}{2} = 1.05 \, m^{2}$

3rd trapezium points: (2,0), (3,0), (2,1.6), (3,0.9).

3rd trapezium area: $S_{3} = (3 - 2) * \min\{1.6, 0.9\} + (3 - 2) * \frac{1.6 - 0.9}{2} = 1.25 \, m^{2}$

4th trapezium points: (3,0), (4,0), (3,0.9), (4,0).

4th trapezium area: $S_{4} = (4 - 3) * \min\{0.9, 0\} + (4 - 3) * \frac{0.9 - 0}{2} = 0.45 \, m^{2}$

Total area: $S = S_{1} + S_{2} + S_{3} + S_{4} = 0.25 + 1.05 + 1.25 + 0.45 = 3 \, m^{2}$

So, the volume of water passes this section of the river in one minute is approximately

$$V = 3 \, m^{2} * \left(\frac{0.6 \, m}{s} * 60 \, s\right) = 108 \, m^{3}.$$

Answer: $3 \, m^{2}$; $108 \, m^{3}$.