Lawn Area Between Two Paths and a Raised Basin Volume
A landscape architect is designing a curved lawn between two intersecting footpaths on a plan where units are metres. The upper boundary is the path y = 6x − x² and the lower boundary is y = x² − 2x; the lawn is the region enclosed between them. The lower path dips below the x-axis for part of its length, and a colleague has already warned that 'you cannot just integrate the difference when the lower curve goes negative - you must split at x = 2 and take absolute values, otherwise the area comes out wrong'. Separately, a raised water basin is to be formed by rotating the curve y = √x, between x = 0 and x = 4, about the horizontal line y = −1, which is the top of a stone plinth, giving a solid of revolution. The architect needs the lawn area, a ruling on the colleague's warning, and the basin volume, all to 3 significant figures.
1. Find where the two paths meet and hence the area of the lawn.
Read the worked answer
Solve 6x − x² = x² − 2x: 8x − 2x² = 0 → 2x(4 − x) = 0 → x = 0 or x = 4.
Between the intersections, test x = 1: upper = 5, lower = −1, so the upper curve is 6x − x² throughout (0, 4).
Difference = (6x − x²) − (x² − 2x) = 8x − 2x² = 2x(4 − x), which is non-negative on [0, 4] and zero only at the endpoints. That non-negativity is what licenses a single integral with no splitting.
Area = ∫₀⁴ (8x − 2x²) dx = [4x² − (2/3)x³]₀⁴ = 64 − 128/3 = 192/3 − 128/3 = 64/3 = 21.3 m² (3 s.f.).
2. Rule on the colleague's warning: is splitting at x = 2 and using absolute values correct, and what does it produce?
Read the worked answer
The colleague is confusing two different events: a curve crossing the x-axis, and the two curves crossing each other. The formula Area = ∫(upper − lower) dx is valid whenever 'upper' really is the upper curve on the whole interval, because then the integrand is non-negative and the integral is already a genuine area. Nothing about the lower curve's sign enters.
Check the claim numerically. Signed integral of the lower curve: ∫₀⁴ (x² − 2x) dx = [x³/3 − x²]₀⁴ = 64/3 − 16 = 16/3, which is positive. Its pieces are ∫₀² = 8/3 − 4 = −4/3 and ∫₂⁴ = 20/3. Under the colleague's method those become |−4/3| + 20/3 = 24/3 = 8.
Also ∫₀⁴ (6x − x²) dx = [3x² − x³/3]₀⁴ = 48 − 64/3 = 80/3.
Correct subtraction of signed integrals: 80/3 − 16/3 = 64/3 ✓. The colleague's method: 80/3 − 8 = 56/3 = 18.7 m² ✗.
The error is an undercount of 8/3 m² ≈ 2.67 m², exactly twice the 'negative pocket' - the colleague double-counts the region between x = 0 and x = 2. Splitting is required only if the curves cross inside the interval, which they do not here.
3. Find the volume of the basin formed by rotating y = √x, for 0 ≤ x ≤ 4, about the line y = −1.
Read the worked answer
The region bounded by y = √x, the x-axis and x = 0, x = 4 is rotated about the horizontal line y = −1, so horizontal slices become washers.
Distances from the line of rotation: outer radius = √x − (−1) = √x + 1; inner radius = 0 − (−1) = 1.
V = π∫₀⁴ [(√x + 1)² − 1²] dx = π∫₀⁴ (x + 2√x + 1 − 1) dx = π∫₀⁴ (x + 2x^(1/2)) dx.
Antiderivative: x²/2 + (4/3)x^(3/2).
Evaluate at 4: 8 + (4/3)(8) = 8 + 32/3 = 56/3. At 0: 0.
V = (56/3)π = 58.6 m³ (3 s.f.).
The judgement the marks depend on is the inner radius. Forgetting it gives π∫₀⁴ (√x + 1)² dx = π[x²/2 + (4/3)x^(3/2) + x]₀⁴ = π(8 + 32/3 + 4) = (68/3)π ≈ 71.2 m³, an overestimate of 12.6 m³, because the plinth itself is not part of the water.