---
title: "Parallel axis theorem; centroid of area under y = Kx²"
subject: "Applied Mechanics"
chapter: 4
key: "semester_1/applied_mechanics:parallel-axis-centroid-y-kx2-ashad"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/parallel-axis-centroid-y-kx2-ashad
site: BCT Engineering
---

# Parallel axis theorem; centroid of area under y = Kx²

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

- Asked in: 2070 Asar, 3+8 marks
- 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-y-kx2-ashad
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-4

## Question

State and prove parallel axis theorem. Also determine the centroidal X and Y coordinate of the hatched area. [Figure: area under the curve y = Kx² from x = 0 to x = a, with height h at x = a.]

## Short answer

Parallel axis theorem: I=I_G+Ad². For the hatched area under y=Kx² from x=0 to a, area=ah/3. Centroid: x̄=3a/4 from y-axis, ȳ=3h/10 from x-axis.

## 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 under y = Kx²
**Reading of the figure:** the shaded area is under the curve y = Kx² between x = 0 and x = a, where the height is h, so K = h/a². Use a vertical strip of height y = Kx² and width dx.

**Area**
A = ∫₀ᵃ Kx² dx = K a³/3 = **ah/3**

**x̄**
∫ x dA = ∫₀ᵃ x · Kx² dx = K a⁴/4 = a²h/4
x̄ = (a²h/4)/(ah/3) = **3a/4**

**ȳ** (strip centroid at y/2)
∫ ȳ dA = ∫₀ᵃ ½ (Kx²)(Kx²) dx = K² a⁵/10 = a h²/10
ȳ = (a h²/10)/(ah/3) = **3h/10**

**Answer: x̄ = 3a/4, ȳ = 3h/10 (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, where d is the distance between the axes
   B. I = I_G + A d^2, where I_G is about the centroidal axis and d is the distance between the axes
   C. I = I_G - A d^2, because the centroidal value is the largest
   D. I = A d^2 only, since I_G is zero for any area
   Answer: B. 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. Among all parallel axes, where is the moment of inertia of an area smallest?
   A. Largest about the centroidal axis
   B. Smallest about the axis farthest from the centroid
   C. Smallest about the centroidal axis, since A d^2 is never negative
   D. The same for all parallel axes
   Answer: C. I = I_G + A d^2 and A d^2 >= 0, so the centroidal axis (d = 0) gives the smallest value. (Key point: Minimum at centroid)

3. How is K determined for y = Kx^2 with height h at x = a?
   A. K = h/a^2
   B. K = h/a
   C. K = a/h^2
   D. K = ha^2
   Answer: A. h = K a^2. (Key point: Finding K)

4. What is the centroid of the area under y = Kx^2?
   A. (3a/4, h/2)
   B. (a/4, 3h/10)
   C. (a/2, h/3)
   D. (3a/4, 3h/10)
   Answer: D. x̄ = (a^2 h/4)/(ah/3) = 3a/4 and ȳ = (a h^2/10)/(ah/3) = 3h/10. (Key point: Centroid)

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