---
title: "Radius of gyration; centroid by integration y = kx²"
subject: "Applied Mechanics"
chapter: 4
key: "semester_1/applied_mechanics:radius-gyration-centroid-integration-y-kx2"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/radius-gyration-centroid-integration-y-kx2
site: BCT Engineering
---

# Radius of gyration; centroid by integration y = kx²

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

- Asked in: 2071 Chaitra, 3+8 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/radius-gyration-centroid-integration-y-kx2
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-4

## Question

Explain radius of gyration. Determine the centroid of the shaded area shown in figure below, using direct integration method. [Figure: area bounded by the y axis, region between the y axis and the curve y = kx², from O up to height b, with base length a along the x axis.]

## Short answer

Radius of gyration k=√(I/A). For the shaded area above the parabola y=kx² between y-axis and y=b, using vertical strips, centroid is at x̄=3a/8, ȳ=3b/5 from O.

## Answer

#### Radius of gyration
The radius of gyration of an area about an axis is the distance k from the axis at which the whole area, if concentrated at a point, would give the same moment of inertia: I = A k², k = √(I/A). It measures how far the area is spread from the axis, and it is used in column design (slenderness ratio).

#### Centroid by direct integration
**Reading of the figure:** the area lies between the y axis, the top line y = b and the curve y = kx² (which goes from O to the point (a, b)). It is the area above the curve. So k = b/a². Use a vertical strip at x with height from y = kx² to y = b, i.e. h = b − kx².

**Area**
A = ∫₀ᵃ (b − kx²) dx = ab − k a³/3 = ab − ab/3 = **2ab/3**

**x̄**
∫ x dA = ∫₀ᵃ x (b − kx²) dx = b a²/2 − k a⁴/4 = a²b/2 − a²b/4 = a²b/4
x̄ = (a²b/4) / (2ab/3) = **3a/8**

**ȳ** (strip centroid at the middle of the strip, (b + kx²)/2)
∫ ȳ dA = ∫₀ᵃ ½ (b + kx²)(b − kx²) dx = ½ ∫₀ᵃ (b² − k² x⁴) dx = ½ (ab² − ab²/5) = 2ab²/5
ȳ = (2ab²/5) / (2ab/3) = **3b/5**

**Answer: centroid at (3a/8, 3b/5) from O.**
(Check: the area under the curve alone has centroid (3a/4, 3b/10); taking moments of the two complementary areas about the y axis: (ab/3)(3a/4) + (2ab/3)(3a/8) = a²b/4 + a²b/4 = a²b/2 = (ab)(a/2) ✓.)

## Check yourself

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

1. What does radius of gyration measure?
   A. The maximum dimension of the area
   B. The centroid distance from the origin
   C. The area divided by its perimeter
   D. How far the area is spread from the axis; k = sqrt(I/A)
   Answer: D. k is the distance at which concentrating the area gives the same I. (Key point: Radius of gyration)

2. For the area above y = kx^2 up to y = b, the strip height is:
   A. b - kx^2
   B. b + kx^2
   C. kx^2
   D. b
   Answer: A. The strip runs from the curve to the top line. (Key point: Strip height)

3. The centroid height of that strip is:
   A. (b - kx^2)/2
   B. kx^2/2
   C. (b + kx^2)/2
   D. b/2
   Answer: C. It lies midway between kx^2 and b. (Key point: Strip ȳ)

4. What is x̄ of the area above the curve?
   A. 3a/4
   B. a/4
   C. 3a/8
   D. a/2
   Answer: C. Integral of x dA = a^2 b/4 and A = 2ab/3, so x̄ = 3a/8. (Key point: Centroid x̄)

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