Arch Service Deck: Integrals and Design Constraints
A concrete arch is modelled by the region under y=9−x² and above the ground y=0, for −3≤x≤3, with x and y in metres. A horizontal service deck is to be placed at height y=k, where 0<k<9, across the arch. The architect wants the open area above the deck to be exactly 8 m², so that ventilation and light are adequate, while the deck itself remains as wide as possible. A contractor proposes estimating the deck width by using only the parabola's maximum height and ignoring the curved sides; the architect insists on an exact integral. The same cross-section will later be used to estimate concrete volume by multiplying the area under the arch by a constant width of 0.8 m, but only after the deck position is fixed. The design team must also report the average height of the arch above ground over its full width as a check against a simple midpoint estimate.
1. Find the deck height k so that the open area above the deck is exactly 8 m². Give an exact expression and an approximation.
Read the worked answer
Put h=9−k and a=√h. The open region above the deck exists for −a≤x≤a, because y=9−x² is at least k exactly when x²≤9−k=h. Its area is ∫ from −a to a of (9−x²−k) dx = ∫ from −a to a of (a²−x²) dx. By symmetry this is 2[a²x−x³/3] from 0 to a = 2(a³−a³/3)=4a³/3. Setting this equal to 8 gives 4a³/3=8, so a³=6 and a=∛6. Hence k=9−a²=9−6^(2/3). Numerically 6^(2/3)≈3.302, so k≈5.70 m. This lies between 0 and 9, as required.
2. Compute the average height of the arch above ground over its full width, and use the constant-width assumption to estimate the concrete volume.
Read the worked answer
The total area under the arch is ∫ from −3 to 3 of (9−x²) dx. An antiderivative is 9x−x³/3. At x=3 this is 27−9=18; at x=−3 it is −27+9=−18. The area is 18−(−18)=36 m². The full width is 6 m, so the average height is 36/6=6 m. This is below the maximum height 9 m, as expected for a concave-down arch. Under the constant-width assumption of 0.8 m, the estimated volume is 36×0.8=28.8 m³. This is an estimate for a prism-like arch; if the real structure tapers in width, 28.8 m³ is not automatically exact.
3. A safety rule is added requiring k≥6. Can the open-area requirement still be met? Justify your answer using the area formula.
Read the worked answer
If k=6, then h=9−6=3 and a=√3. The open area is 4a³/3=4(√3)³/3=4·3√3/3=4√3≈6.93 m². This is already below the required 8 m². Moreover, as k increases, h=9−k decreases, so a=√h decreases and the open area 4a³/3 decreases. Therefore for every k≥6 the open area is at most about 6.93 m², which is less than 8 m². The two requirements cannot both be satisfied; either the minimum k must be lowered or the required open area must be reduced.