---
title: "Centroid of y = kx² area; parallel axes; define centroid"
subject: "Applied Mechanics"
chapter: 4
key: "semester_1/applied_mechanics:centroid-y-kx2-parallel-axis-define-centroid-cg"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/centroid-y-kx2-parallel-axis-define-centroid-cg
site: BCT Engineering
---

# Centroid of y = kx² area; parallel axes; define centroid

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

- Asked in: 2070 Chaitra, 7+3+2 marks
- Video: [Centroid of parabolic spandrel || centroid 14 || Engineering Mechanics || Sumit Dwivedi ||](https://www.youtube.com/watch?v=_gIuTjDqQ04) — Learning with Sumit
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/centroid-y-kx2-parallel-axis-define-centroid-cg
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-4

## Question

Determine centroid of the given plane figure. State and prove parallel axes theorem for moment of inertia. Define centroid and center of gravity. [Figure: area under the curve y = kx² from x = 0 to x = a, with height h at x = a.]

## Short answer

By integration, centroid is (3a/4, 3h/10). The parallel axis theorem states I = I_G + A d²; proof: ∫(y' + d)² dA = I_G + 0 + A d². Centroid is geometric center; center of gravity is weight's resultant point.

## Answer

#### Centroid of the plane figure
**Reading of the figure:** the area lies under the curve y = kx² from x = 0 to x = a, where the height is h, so k = h/a². Strip: width dx, height y = kx².

**Area:** A = ∫₀ᵃ kx² dx = k a³/3 = **ah/3**

**x̄:** ∫ x dA = ∫₀ᵃ k x³ dx = k a⁴/4 = a²h/4
x̄ = (a²h/4)/(ah/3) = **3a/4**

**ȳ:** ∫ ȳ dA = ∫₀ᵃ ½ k² x⁴ dx = k² a⁵/10 = a h²/10
ȳ = (a h²/10)/(ah/3) = **3h/10**

**Centroid = (3a/4, 3h/10) from O.**

#### Parallel axis theorem
**Statement:** the moment of inertia of an area about any axis equals its moment of inertia about a parallel axis through its centroid plus the product of the area and the square of the distance between the two axes.

I = I_G + A d²

**Proof:** let G be the centroid of the area A, and let X′ be the centroidal axis. Take a parallel axis X at distance d from X′. For a small element dA at distance y′ from X′, its distance from X is y = y′ + d.
```text
I_X = ∫ y² dA = ∫ (y′ + d)² dA
    = ∫ y′² dA + 2d ∫ y′ dA + d² ∫ dA
```
- ∫ y′² dA = I_G, the moment of inertia about the centroidal axis.
- ∫ y′ dA = A·ȳ′ = 0, because the distance of the centroid from the centroidal axis is zero.
- ∫ dA = A.

So **I_X = I_G + A d²**. Similarly, I_Y = I_Gy + A d_x². Since A d² is never negative, the moment of inertia about the centroidal axis is the least of the moments of inertia about all parallel axes.

#### Centroid and centre of gravity
- **Centroid:** the geometric centre of an area (or line or volume). It is the point at which the whole area can be assumed concentrated; x̄ = ∫x dA/A, ȳ = ∫y dA/A. It depends only on shape.
- **Centre of gravity:** the point through which the resultant of the weights of all the particles of a body acts, whatever the orientation of the body. It depends on the weight distribution. For a body of uniform density the centre of gravity and the centroid coincide.

## Check yourself

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

1. What is the area under y = kx^2 from 0 to a?
   A. ah/3
   B. ah/2
   C. 2ah/3
   D. ah/4
   Answer: A. A = k a^3/3 = ah/3 using k = h/a^2. (Key point: Area under curve)

2. In finding ȳ, the strip centroid is taken at:
   A. y/2
   B. x/2
   C. y/3
   D. y
   Answer: A. A vertical strip's centroid is halfway up. (Key point: Strip ȳ)

3. In the proof, why does the term 2d * integral(y' dA) drop out?
   A. Because the integral of y' dA is A times the centroid's distance from the centroidal axis, which is zero
   B. Because the integral of dA equals zero
   C. Because d is zero for every parallel axis
   D. Because y'^2 is always negative
   Answer: A. The integral of y' dA equals A times the centroid's distance from the centroidal axis, and that distance is zero, so the cross term vanishes. (Key point: Cross term vanishes)

4. Which statement is correct?
   A. They differ even for uniform bodies
   B. CG depends only on shape
   C. CG depends on weight distribution; centroid depends only on shape
   D. Centroid depends on weight distribution
   Answer: C. They coincide only for uniform density. (Key point: Centroid vs CG)

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