Area, Volume and a Polygonal Approximation for a Stamped Plate
A metal badge is stamped from sheet 0.8 mm thick. On the design drawing the face of the badge is the plane region enclosed between the parabola y = x² and the line y = 2x + 3, with x and y in centimetres, and the region is later rotated about the x-axis to model the polymer coating applied to the domed face. The coating bath is sized from the volume of that solid of revolution. The die that will actually cut the badges is cheap: it produces straight edges between the points of the region at x = −1, 0, 1, 2 and 3, so the delivered badge is a polygon, not the drawn shape. Quality control allows the stamped area to differ from the drawn area by no more than 5 per cent. Before the die is ordered, the toolmaker must know the exact area and volume from the drawing, the area the five-point polygonal die would actually produce, and whether a finer set of cut points is needed.
1. Find the points of intersection of the two curves and the exact area of the badge face.
Read the worked answer
Set x² = 2x + 3: x² − 2x − 3 = 0, so (x − 3)(x + 1) = 0, giving x = −1 and x = 3. The intersection points are (−1, 1) and (3, 9).
Between these values the line is above the parabola (at x = 0, the line gives 3 and the parabola 0), so the height of a strip is (2x + 3) − x².
Area = ∫ from −1 to 3 of (2x + 3 − x²) dx = [x² + 3x − x³/3] from −1 to 3.
At x = 3: 9 + 9 − 9 = 9. At x = −1: 1 − 3 + 1/3 = −5/3.
Area = 9 − (−5/3) = 32/3 ≈ 10.67 cm².
2. Find the volume of the solid formed when the region is rotated completely about the x-axis, giving an exact value.
Read the worked answer
Use washers: at each x between −1 and 3, the outer radius is the line value 2x + 3 and the inner radius is the parabola value x². Both are non-negative on [−1, 3] (the line runs from 1 to 9, the parabola from 1 down to 0 and back to 9), and x² ≤ 2x + 3 throughout, so the annulus has area π[(2x + 3)² − (x²)²].
V = π ∫ from −1 to 3 of [(2x + 3)² − x⁴] dx = π ∫ (−1 to 3) (4x² + 12x + 9 − x⁴) dx = π[(4/3)x³ + 6x² + 9x − x⁵/5] from −1 to 3.
At x = 3: 36 + 54 + 27 − 243/5 = 117 − 48.6 = 68.4 = 342/5. At x = −1: −4/3 + 6 − 9 + 1/5 = −62/15.
Difference = 342/5 + 62/15 = 1026/15 + 62/15 = 1088/15.
V = 1088π/15 ≈ 227.9 cm³.
Note the inner radius is x², not zero: the badge is a ring-shaped solid near the ends of the interval, and treating it as a plain solid of revolution about the x-axis with radius 2x + 3 would overstate the volume.
3. The five-point die uses straight chords at x = −1, 0, 1, 2, 3. Estimate the stamped area by the trapezium rule, compare it with the exact area, and decide whether a finer set of cut points is required.
Read the worked answer
Let f(x) = 2x + 3 − x², the height of the region. At the five cut points: f(−1) = 0, f(0) = 3, f(1) = 4, f(2) = 3, f(3) = 0.
Trapezium rule with h = 1: T = (h/2)[f(−1) + f(3) + 2(f(0) + f(1) + f(2))] = (1/2)[0 + 0 + 2(3 + 4 + 3)] = (1/2)(20) = 10.0 cm².
Exact area = 32/3 ≈ 10.667 cm², so the error is T − I ≈ −0.667 cm², a relative error of 0.667/10.667 ≈ 6.25 per cent. This exceeds the 5 per cent tolerance (0.533 cm²), so the five-point die as specified must be rejected.
Why the error is negative: f″(x) = −2 < 0 everywhere, so f is concave down and every chord lies below the curve; the polygon always underestimates.
Size of the error: with constant second derivative, E ≈ (b − a)h²f″/12 = (4)(h²)(−2)/12 = −2h²/3. For h = 0.5 this gives E ≈ −0.167 cm², a relative error of about 1.6 per cent, comfortably inside tolerance. The requirement |E| ≤ 0.05 × 32/3 = 0.533 gives h ≤ 0.894, so h = 1 just fails while h = 0.5 passes with margin.
Recommendation: cut at 0.5 cm intervals (nine points from −1 to 3) rather than five, or retain the price of the five-point die only if tolerances are relaxed to about 6.5 per cent.