---
title: "MOI and radius of gyration between two parabolas"
subject: "Applied Mechanics"
chapter: 4
key: "semester_1/applied_mechanics:moi-radius-gyration-between-parabolas"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/moi-radius-gyration-between-parabolas
site: BCT Engineering
---

# MOI and radius of gyration between two parabolas

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

- Asked in: 2069 Asar, 12 marks; 2067 Asar, 11 marks (asked 2 times)
- Video: [Statics: Lesson 70 - Area Moment of Inertia, Calculus Method](https://www.youtube.com/watch?v=f7slKV01ax0) — 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/moi-radius-gyration-between-parabolas
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-4

## Question

Determine the moment of inertia and the radius of gyration of the shaded area given below. [Figure: lens-shaped area between the curves y1 = K1x² and y2 = K2√x in the first quadrant, extending to x = a and y = b (another paper gives a = 80 cm, b = 30 cm).]

## Short answer

Integrate dI_x = (y2³−y1³)dx/3 and dI_y = x²(y2−y1)dx using a=80 cm, b=30 cm. I_x = 1.85×10⁵ cm⁴, I_y = 1.32×10⁶ cm⁴; radii of gyration k_x = 15.2 cm, k_y = 40.6 cm.

## Answer

**Setting up:** the area lies between y₁ = K₁x² (lower) and y₂ = K₂√x (upper). Both curves pass through the origin and through the point (a, b), so
K₁ = b/a², K₂ = b/√a.
The figure of the second paper gives a = 80 cm and b = 30 cm. I derive the general formulae and then put in these numbers.

#### Area
A = ∫₀ᵃ (y₂ − y₁) dx = ∫₀ᵃ (K₂ x^½ − K₁ x²) dx = (2/3) K₂ a^{3/2} − K₁ a³/3 = (2/3)ab − (1/3)ab = **ab/3**

#### Moment of inertia about the x axis
Strip of width dx, between y₁ and y₂: dI_x = (y₂³ − y₁³) dx / 3
```text
I_x = (1/3) ∫ (K2³ x^(3/2) − K1³ x^6) dx
    = (1/3) [ K2³ (2/5) a^(5/2) − K1³ a^7 / 7 ]
    = (1/3) [ (2/5) a b³ − (1/7) a b³ ]
    = (1/3)(9/35) a b³ = 3 a b³ / 35
```

#### Moment of inertia about the y axis
dI_y = x² (y₂ − y₁) dx
```text
I_y = ∫ (K2 x^(5/2) − K1 x^4) dx
    = K2 (2/7) a^(7/2) − K1 a^5 / 5
    = (2/7) a³ b − (1/5) a³ b = 3 a³ b / 35
```

#### Radii of gyration
k_x = √(I_x/A) = √( (3ab³/35) / (ab/3) ) = √(9/35) · b = 0.507 b
k_y = √(I_y/A) = √(9/35) · a = 0.507 a

#### Numbers (a = 80 cm, b = 30 cm)
```text
A   = 80 × 30 / 3                 = 800 cm²
I_x = 3 × 80 × 30³ / 35           = 185 143 cm⁴  (1.851 × 10⁵ cm⁴)
I_y = 3 × 80³ × 30 / 35           = 1 316 571 cm⁴ (1.317 × 10⁶ cm⁴)
k_x = 0.5071 × 30                 = 15.21 cm
k_y = 0.5071 × 80                 = 40.57 cm
```
**Answer: I_x = 3ab³/35 = 1.85 × 10⁵ cm⁴, I_y = 3a³b/35 = 1.32 × 10⁶ cm⁴, k_x = 15.21 cm, k_y = 40.57 cm.**

## Check yourself

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

1. Why K1 = b/a^2 and K2 = b/sqrt(a)?
   A. Both curves pass through (a, b)
   B. Both curves start at (a, b)
   C. To make the area equal to 1
   D. Because x and y are interchangeable
   Answer: A. Substituting x = a, y = b into each curve equation. (Key point: Constants)

2. What is the area between the curves?
   A. A = ab/3 = (2/3)ab - (1/3)ab
   B. A = ab
   C. A = 2ab/3
   D. A = ab/6
   Answer: A. Integrate (K2 x^(1/2) - K1 x^2) from 0 to a. (Key point: Area)

3. Which is the element for I_x of a vertical strip between y1 and y2?
   A. dI_x = (y2^3 - y1^3) dx / 3
   B. x^2 (y2 - y1) dx
   C. (y2 - y1)^3 dx / 3
   D. y^2 dx
   Answer: A. Each strip's moment of inertia about the x axis is the difference of the b h^3/3 values. (Key point: Strip for I_x)

4. Which results are correct?
   A. I_y = 3a^3 b/35 and k_y = sqrt(9/35) a = 0.507 a
   B. I_y = 3ab^3/35 and k_y = 0.507 b
   C. I_y = ab^3/3 and k_y = a/3
   D. I_y = 3a^3 b/35 and k_y = 3a/35
   Answer: A. I_y = ∫ x^2 (y2 - y1) dx and k = sqrt(I/A). (Key point: I_y and k)

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