Parabolic Stage Arch: Plywood Area, Paint Area and a Taller Banner
A stage designer is building a flat wooden arch for a school fest. The arch is modelled by the parabola y = 9 - x² for -3 ≤ x ≤ 3, with the x-axis as the floor and y measured in metres. A rectangular banner 4 m wide and 5 m high is to be fixed centrally on the arch, with its bottom on the floor. The designer needs the area of the whole arch face to order plywood, and the area of the arch face left visible above and around the banner to order paint. A helper suggests approximating the arch by a rectangle 6 m by 9 m, or by a triangle, but the designer wants an exact integral. Later, the banner is redesigned to be 4 m wide and 6 m high. The designer claims it will still fit because the arch is 9 m high at the centre; the helper is not sure. You are asked to compute the exact areas and to test the redesign.
1. Find the exact area of the whole arch face. Then find the area of the region between the parabola and the line y = 5 for -2 ≤ x ≤ 2.
Read the worked answer
The whole arch area is
∫ from -3 to 3 of (9 - x²) dx = [9x - x³/3] from -3 to 3 = (27 - 9) - (-27 + 9) = 18 - (-18) = 36 m².
The line y = 5 meets the parabola where 9 - x² = 5, so x² = 4 and x = ±2. The area between the parabola and y = 5 is
∫ from -2 to 2 of ((9 - x²) - 5) dx = ∫ from -2 to 2 of (4 - x²) dx = [4x - x³/3] from -2 to 2 = (8 - 8/3) - (-8 + 8/3) = 16/3 - (-16/3) = 32/3 ≈ 10.67 m².
This is the part of the arch face above the banner's top edge within the banner's width.
2. The banner is 4 m wide and 5 m high. Find the area of the arch face that remains visible, and verify it by splitting the visible region into the part above y = 5 and the two side pieces.
Read the worked answer
The banner covers 4×5 = 20 m², so the visible arch area is 36 - 20 = 16 m².
Check by splitting. Above the banner, between y = 5 and the parabola for -2 ≤ x ≤ 2, the area is 32/3 m², as computed. The two side pieces are the arch outside the banner:
2∫ from 2 to 3 of (9 - x²) dx = 2([9x - x³/3] from 2 to 3) = 2((27 - 9) - (18 - 8/3)) = 2(18 - 46/3) = 2(8/3) = 16/3 m².
Sum = 32/3 + 16/3 = 48/3 = 16 m², which matches 36 - 20. The exact integral confirms the helper's rectangular or triangular approximations would be inaccurate.
3. Test the 4 m × 6 m redesign. Does it fit as a rectangle with its base on the floor? Find the maximum height of a 4 m wide centred rectangle and the maximum width of a 6 m high centred rectangle.
Read the worked answer
For a centred rectangle 4 m wide, x runs from -2 to 2. The limiting height is set by the parabola at the vertical edges, not at the centre:
y(±2) = 9 - 4 = 5 m.
So the tallest 4 m wide rectangle that fits has height 5 m. A 6 m high rectangle would have top corners at (±2, 6), but the parabola is only at y = 5 there, so the rectangle protrudes above the arch and does not fit.
For a 6 m high rectangle, we need 9 - x² ≥ 6, so x² ≤ 3 and |x| ≤ √3 ≈ 1.73 m. The maximum width is 2√3 ≈ 3.46 m. The centre height of 9 m is misleading; the binding constraint is at the sides. The designer must lower the banner to 5 m, narrow it to about 3.46 m, or reshape its top corners.