---
title: "Centroid by integration, y = K(x−a)²"
subject: "Applied Mechanics"
chapter: 4
key: "semester_1/applied_mechanics:centroid-direct-integration-y-k-x-a-squared"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/centroid-direct-integration-y-k-x-a-squared
site: BCT Engineering
---

# Centroid by integration, y = K(x−a)²

Applied Mechanics · Chapter 4: Center of Gravity, Centroid and Moment of Inertia

- Asked in: 2071 Shrawan, 5 marks
- Video: [How to find Centroid of a Parabolic Spandrel by Integration](https://www.youtube.com/watch?v=UQ5r0-dBi3s) — Manas Patnaik
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/centroid-direct-integration-y-k-x-a-squared
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-4

## Question

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.]

## 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.

## Answer

**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).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the constant K found for y = K(x - a)^2?
   A. Use y = a at x = b, so K = a/b^2
   B. Use y = b at x = 0, so K = b/a^2
   C. Use y = 0 at x = 0, so K = 0
   D. K is taken as 1
   Answer: B. At x = 0 the height is b, so K a^2 = b. (Key point: Finding K)

2. For a vertical strip under the curve, what are the strip's centroid coordinates?
   A. x̃ = x/2 and ỹ = y/2
   B. x̃ = x/2 and ỹ = y
   C. x̃ = x and ỹ = y
   D. x̃ = x and ỹ = y/2
   Answer: D. A vertical strip has its centroid at its own x and halfway up its height. (Key point: Strip centroid)

3. What is x̄ for the area under y = K(x - a)^2?
   A. a/4
   B. a/3
   C. 3a/4
   D. a/2
   Answer: A. Integral of x dA = a^2 b/12 and A = ab/3, so x̄ = a/4. (Key point: Result)

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