---
title: "MOI of common area between y = K√x and y = mx"
subject: "Applied Mechanics"
chapter: 4
key: "semester_1/applied_mechanics:moi-common-area-y-k-sqrt-x-y-mx"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/moi-common-area-y-k-sqrt-x-y-mx
site: BCT Engineering
---

# MOI of common area between y = K√x and y = mx

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

- Asked in: 2068 Shrawan, 12 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/moi-common-area-y-k-sqrt-x-y-mx
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-4

## Question

Determine the moment of inertia of the common area as shown in figure below about x and y axis. [Figure: area between the curve y2 = K√x and the straight line y1 = mx from the origin O to x = a, y = b.]

## Short answer

Common area = ab/6. Using strip method, moment of inertia about x-axis is ab³/20, about y-axis is a³b/28.

## Answer

**Setting up:** the common area lies between the curve y₂ = K√x (upper) and the straight line y₁ = mx (lower), as labelled in the figure. Both pass through O and through the point (a, b), so
K = b/√a, m = b/a.

#### Area
A = ∫₀ᵃ (K√x − mx) dx = (2/3) K a^{3/2} − m a²/2 = (2/3)ab − ab/2 = **ab/6**

#### Moment of inertia about the x axis
Strip of width dx between the two curves: dI_x = (y₂³ − y₁³) dx / 3
```text
I_x = (1/3) ∫0^a (K³ x^(3/2) − m³ x³) dx
    = (1/3) [ K³ (2/5) a^(5/2) − m³ a⁴/4 ]
    = (1/3) [ (2/5) a b³ − (1/4) a b³ ]
    = (1/3)(3/20) a b³ = a b³ / 20
```
#### Moment of inertia about the y axis
dI_y = x² (y₂ − y₁) dx
```text
I_y = ∫0^a (K x^(5/2) − m x³) dx
    = K (2/7) a^(7/2) − m a⁴/4
    = (2/7) a³ b − (1/4) a³ b = a³ b / 28
```
(Radii of gyration: k_x = √(I_x/A) = √(3/10) b = 0.548 b, k_y = √(I_y/A) = √(3/14) a = 0.463 a.)

**Answer: I_x = ab³/20 and I_y = a³b/28.**

## Check yourself

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

1. K and m are found from:
   A. Both curves passing through (a, b): K = b/sqrt(a), m = b/a
   B. Both curves passing through the origin only
   C. K = m = 1
   D. K = a/b, m = b/a
   Answer: A. Substituting (a, b) gives K and m. (Key point: Constants)

2. For I_x, the strip between curves contributes:
   A. (y1^3 - y2^3) dx/3
   B. x^2 (y1 - y2) dx
   C. (y1 - y2) dx
   D. (y1^2 - y2^2) dx/2
   Answer: A. Each strip about the x axis has I = y^3 dx/3; difference of upper and lower. (Key point: Strip I_x)

3. For I_y, the strip contributes:
   A. y^2 dx
   B. x (y1 - y2) dx
   C. (y1^3 - y2^3) dx/3
   D. x^2 (y1 - y2) dx
   Answer: D. I_y = integral of x^2 dA with dA = (y1 - y2) dx. (Key point: Strip I_y)

4. I_x and I_y are:
   A. ab^3/28 and a^3 b/20
   B. ab^3/12 and a^3 b/12
   C. ab^3/6 and a^3 b/6
   D. ab^3/20 and a^3 b/28
   Answer: D. The integrals give I_x = ab^3/20 and I_y = a^3 b/28. (Key point: Results)

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