---
title: "Parallel axis theorem; centroid of area between parabolas"
subject: "Applied Mechanics"
chapter: 4
key: "semester_1/applied_mechanics:parallel-axis-centroid-curves-x-ky2"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/parallel-axis-centroid-curves-x-ky2
site: BCT Engineering
---

# Parallel axis theorem; centroid of area between parabolas

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

- Asked in: 2072 Chaitra, 4+8 marks
- Video: [Statics: Lesson 42 - Centroids by Calculus: Flip the Strip Method](https://www.youtube.com/watch?v=hUBhwZ6Xj24) — Jeff Hanson
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/parallel-axis-centroid-curves-x-ky2
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-4

## Question

State and prove parallel axes theorem for moment of inertia. Determine centroid of the given plane in figure below. [Figure: area enclosed between the curves x = ky² and y = kx² in a square 500 mm × 500 mm, axes x, y from origin O.]

## Short answer

Parallel axis theorem: I=I_G+Ad². The area between x=ky² and y=kx² within a 500 mm square has area 83,333 mm². By symmetry and integration, centroid is at (225 mm, 225 mm) from O.

## Answer

#### 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 of the area between x = ky² and y = kx²
**Reading of the figure:** the two parabolas meet at O and at (500, 500) mm, so for both curves k = 1/500 mm⁻¹ (500 = k × 500²). The area lies between them, above y = kx² and to the right of x = ky² (i.e. below y = √(x/k)). Let a = 500 mm.

Upper curve y_u = √(x/k) = √(a x), lower curve y_l = kx² = x²/a.

**Area**
A = ∫₀ᵃ (√(a x) − x²/a) dx = √a (2/3) a^{3/2} − a³/(3a) = (2/3)a² − (1/3)a² = a²/3 = 500²/3 = **83 333 mm²**

**x̄** = (1/A) ∫ x (y_u − y_l) dx
∫₀ᵃ x √(a x) dx = √a (2/5) a^{5/2} = (2/5) a³
∫₀ᵃ x · x²/a dx = a³/4
x̄ = (3/a²) · a³ (2/5 − 1/4) = (3/a²) · (3/20) a³ = 9a/20 = 9 × 500/20 = **225 mm**

**ȳ**: the figure is symmetrical about the line y = x (the two curves are mirror images), so ȳ = x̄ = **225 mm**.

(Check by integration: ȳ = (1/2A) ∫ (y_u² − y_l²) dx = (3/(2a²)) (a³/2 − a³/5) = (3/(2a²)) (3a³/10) = 9a/20 ✓.)

**Answer: centroid at (225 mm, 225 mm) from O.**

## Check yourself

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

1. Which statement of the parallel axis theorem is correct?
   A. I = I_G + A d^2, where I_G is about the centroidal axis and d is the distance between the parallel axes
   B. I = I_G - A d^2, because the centroidal value is the largest
   C. I = I_G + A d, where d is the distance between the axes
   D. I = A d^2 only, since I_G is zero for any area
   Answer: A. The moment of inertia about any axis equals the centroidal value plus A d^2 (area times distance squared). (Key point: Parallel axis theorem)

2. 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 d is zero for every parallel axis
   C. Because y'^2 is always negative
   D. Because the integral of dA equals zero
   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)

3. For the region between the curves, what is the strip height used for the area integral?
   A. only the lower curve y = x^2/a
   B. y_upper + y_lower
   C. y_upper * y_lower
   D. y_upper - y_lower, i.e. sqrt(a x) - x^2/a
   Answer: D. A vertical strip between the curves has height equal to the difference of the curves. (Key point: Strip area)

4. Why does ȳ equal x̄ = 225 mm?
   A. Because k = 1
   B. The curves are mirror images about the line y = x
   C. Because both integrals are identical by chance
   D. Because the area is a square
   Answer: B. x = ky^2 and y = kx^2 are reflections in y = x, so the centroid lies on that line. (Key point: Using symmetry)

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