- 2071 Shrawan · 5 marks
Centroid by integration, y = K(x−a)²
Short answer
By vertical strip integration, area A=ab/3. Centroid coordinates: x̄=a/4 from the y-axis, ȳ=3b/10 from the x-axis.
Determine by direct integration method, the centroid of the area shown in figure below. [Figure: area under the curve y = K(x − a)² in the first quadrant with height b on the y axis and base length a on the x axis.]
Setting up: the curve y = K(x − a)² meets the y axis at height b (at x = 0), so b = K a², K = b/a². It falls to zero at x = a. The area lies under the curve, between x = 0 and x = a. Use a vertical strip of width dx and height y.
Area
A = ∫₀ᵃ K (x − a)² dx = K [ (x − a)³/3 ]₀ᵃ = K a³/3 = ab/3
x̄
x̃ = x (strip centroid x), so ∫ x dA = ∫₀ᵃ x K (a − x)² dx = K ∫₀ᵃ (a²x − 2a x² + x³) dx = K (a⁴/2 − 2a⁴/3 + a⁴/4) = K a⁴/12 = a²b/12 x̄ = (a²b/12) / (ab/3) = a/4
ȳ
The centroid of the strip is at y/2, so ∫ (y/2) dA = ½ ∫₀ᵃ K² (x − a)⁴ dx = K² a⁵/10 = b² a/10 ȳ = (b² a/10) / (ab/3) = 3b/10
Answer: centroid at x̄ = a/4 and ȳ = 3b/10, measured from O (the y axis and x axis).
✓ Answer checked by Claude Opus 5.5 (high effort)
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