8.2 Find centroids of simple lines and areas by integration | University of Alberta ENGG 130: Engineering Mechanics: Statics

University of Alberta ENGG 130: Engineering Mechanics: Statics

Centroids and Centres of Gravity

ENGG 130 study topic 8.2

A centroid is the geometric balance point of a uniform line or plane area. It is found by averaging the positions of small pieces, weighted by their lengths or areas. Integration performs this continuous sum. In this lesson, the system is a geometric line or area; forces and material behaviour are not being analysed. Define the shape and axes, choose a small element, and express both its measure and location. Then integrate the total measure and first moments. The useful checks here are geometric first-moment checks, not force or structural equilibrium equations.

What you will learn

Explain a centroid as the geometric balance point of a uniform line or area.
Set up centroid coordinates by integrating small lengths or areas.
Choose coordinates and integration limits that match a shape.
Check a centroid using symmetry, units, bounds, and first moments.

Centroid coordinates and first moments

For an area made of small pieces

dA

, the horizontal centroid coordinate is the area-weighted average of the pieces’ horizontal positions. The vertical coordinate is the corresponding average of their vertical positions. For a line, use small lengths

dL

in place of areas.

The coordinates

x

and

y

locate a small piece, while

\bar{x}

and

\bar{y}

locate the centroid. The integrals

\int x\,dA

and

\int y\,dA

are first moments of area. Dividing by total area gives a length. For a line, divide the first moments of length by total length.

Centroid coordinates depend on the chosen axes, but the geometric centroid does not. Choose axes that make the element description and integration limits simple. If a shape is symmetric about an axis, its centroid lies on that axis; symmetry can determine one coordinate without integration.

\bar{x}=\frac{\int x\,dA}{\int dA},\quad \bar{y}=\frac{\int y\,dA}{\int dA};\quad \bar{x}=\frac{\int x\,dL}{\int dL},\quad \bar{y}=\frac{\int y\,dL}{\int dL}

Use $dA$ for area and $dL$ for line length.
A centroid coordinate is a first moment divided by total area or length.
Centroid coordinates have units of length.

Choose an element that matches the shape

For an area, a thin strip is often convenient. A vertical strip of width

dx

between upper and lower boundaries has area equal to its width times its height. If its lower boundary is the

x

-axis, the strip’s centroid is halfway up its height. Use that midpoint when calculating a vertical first moment.

For a curved line, the length element must follow the curve, rather than merely measuring its horizontal or vertical projection. On a circular arc of radius

R

, a small angle change

d\theta

, measured in radians, gives

dL=R\,d\theta

. For another curve, obtain the length element from that curve’s geometry.

Write the size and centroid location of the element before integrating. Use matching limits for total measure and first moments. Integrate the total area or length, then divide each first moment by that total.

dA=(y_{\mathrm{upper}}-y_{\mathrm{lower}})\,dx,\quad dL=R\,d\theta

Choose an element whose measure describes the actual shape.
Use the element’s centroid location, not an arbitrary point on it.
Use consistent axes and limits in the numerator and denominator.

Integrate, interpret, and check

A reliable procedure is to define the line or area, mark axes and limits, choose a small element, and state its measure and centroid location. Integrate to find the total measure and first moments, then divide. Keep dimensions visible: an area first moment has units of length cubed, while a line first moment has units of length squared.

Check whether the result lies within the shape’s bounds and agrees with any symmetry. A direct check is to multiply the centroid coordinate by the total measure and compare it with the corresponding integrated first moment. This checks the weighted-average calculation.

A uniform quarter-circle arc illustrates why the correct length element matters: its points are described by angle, and its small length is

R\,d\theta

. Treating its length as

dx

would omit the changing relationship between horizontal position and distance along the arc.

\bar{x}\int dA=\int x\,dA,\quad \bar{y}\int dA=\int y\,dA

For an area, total area times a centroid coordinate equals its first moment.
Use symmetry as an independent check where it applies.
A first moment divided by total area or length must have units of length.

Worked example

Centroid of a quarter-circular line

Find the centroid of a uniform quarter-circle arc of radius

R

in the first quadrant, running from

(R,0)

to

(0,R)

. Place the origin at the circle centre.

Set coordinates and limits
Describe a point on the arc using angle $\theta$ measured from the positive horizontal axis. Its coordinates are $x=R\cos\theta$ and $y=R\sin\theta$ , with limits from zero to $\pi/2$ .
$x=R\cos\theta,\quad y=R\sin\theta,\quad 0\leq\theta\leq\frac{\pi}{2}$
Find the length element and total length
For a circular arc, a small angular change in radians corresponds to a length equal to the radius times that change. Integrating gives the quarter-arc length.
$dL=R\,d\theta,\quad L=\int_0^{\pi/2}R\,d\theta=\frac{\pi R}{2}$
Find the first moments
Multiply each coordinate by the length element and integrate over the arc. The sine and cosine integrals are equal over these limits.
$\int x\,dL=R^2,\quad \int y\,dL=R^2$
Divide by total length
Divide both first moments by the arc length. Equal positive coordinates agree with the arc’s symmetry about the line $y=x$ .
$\bar{x}=\bar{y}=\frac{2R}{\pi}$

Answer: The centroid is at

(2R/\pi,\,2R/\pi)

relative to the circle centre.

Check: Each coordinate is between zero and

R

. The first-moment check gives

L\bar{x}=L\bar{y}=(\pi R/2)(2R/\pi)=R^2

, matching the integrated first moments.

Worked example

Centroid of a triangular area

Find the centroid of the area below the straight boundary

y=h(1-x/b)

and above the

x

-axis for

0\leq x\leq b

. Use

b=120\,\mathrm{mm}

and

h=60\,\mathrm{mm}

.

Define a vertical strip
At horizontal position $x$ , the strip height is $y(x)$ . Its area is $y(x)\,dx$ . Since it extends from the $x$ -axis to height $y(x)$ , its centroid is halfway up, at height $y(x)/2$ .
$dA=h\left(1-\frac{x}{b}\right)dx,\quad y_{\mathrm{strip}}=\frac{y(x)}{2}$
Find the total area
Integrate strip areas across the full base. This verifies that the chosen element covers the whole triangle.
$A=\int_0^b h\left(1-\frac{x}{b}\right)dx=\frac{bh}{2}$
Find the horizontal coordinate
For the horizontal first moment, weight each strip by its horizontal coordinate and divide by total area. The result lies toward the wider, left side of the triangle.
$\bar{x}=\frac{\int_0^b xh(1-x/b)\,dx}{bh/2}=\frac{b}{3}$
Find the vertical coordinate and substitute
For the vertical first moment, use the strip centroid height. Dividing by total area gives $h/3$ . Substituting the dimensions produces the coordinates in millimetres.
$\bar{y}=\frac{\int_0^b [y(x)/2]y(x)\,dx}{bh/2}=\frac{h}{3},\quad (\bar{x},\bar{y})=(40,20)\,\mathrm{mm}$

Answer: The centroid is

40\,\mathrm{mm}

to the right of the vertical axis and

20\,\mathrm{mm}

above the base.

Check: The area is

3600\,\mathrm{mm^2}

. Thus

A\bar{x}=120000\,\mathrm{mm^3}

and

A\bar{y}=72000\,\mathrm{mm^3}

. These equal the strip-integral first moments

b^2h/6

and

bh^2/6

, respectively.

Worked example

Centroid of an upper semicircular area

Find the centroid of a uniform upper semicircular area of radius

R=45\,\mathrm{mm}

. Its diameter lies on the

x

-axis and its centre is at the origin.

Use symmetry
The shape is symmetric about the vertical axis. Matching area to the left and right of that axis cancels the horizontal first moment, so the horizontal centroid coordinate is zero.
$\bar{x}=0$
Set up vertical strips
At position $x$ , the upper boundary has height $y=\sqrt{R^2-x^2}$ . The vertical strip has area y dx and centroid height $y/2$ above the diameter.
$dA=y\,dx,\quad y_{\mathrm{strip}}=\frac{y}{2}$
Integrate area and vertical first moment
Integrate from the left edge to the right edge. The area is half a circle. For the vertical first moment, multiply strip area by its centroid height, then use $y^2=R^2-x^2$ .
$A=\frac{\pi R^2}{2},\quad \int y\,dA=\frac{1}{2}\int_{-R}^{R}(R^2-x^2)dx=\frac{2R^3}{3}$
Calculate the vertical coordinate
Divide the vertical first moment by the area. Substituting $R=45\,\mathrm{mm}$ gives a coordinate measured upward from the diameter.
$\bar{y}=\frac{4R}{3\pi}=19.1\,\mathrm{mm}$

Answer: The centroid is at

(0,19.1\,\mathrm{mm})

relative to the circle centre.

Check: The coordinate is positive and less than the radius. Also,

A\bar{y}=(\pi R^2/2)(4R/3\pi)=2R^3/3

, equal to the integrated vertical first moment.

Common mistakes and how to avoid them

Using

dx

as the length element for a curved line.

Correction: Use the actual length along the curve. For a circular arc, this is

dL=R\,d\theta

.

Using the top of an area strip as its vertical location.

Correction: Use the strip’s centroid. For a rectangular strip starting at the

x

-axis, that is halfway up its height.

Dividing an area first moment by total length, or a line first moment by total area.

Correction: Use the total measure that matches the element: total area for

dA

, total length for

dL

.

Assuming the centroid is always at the centre of the outline.

Correction: Use the weighted integrals or a valid symmetry argument; a boundary’s visual centre is not generally the area’s centroid.

Lesson summary

Centroid coordinates are first moments divided by total length or area.
For an area, choose strips with a simple area and a known strip-centroid location.
For a line, integrate along the actual curve length.
Check the result using units, symmetry, bounds, and first moments.

Check your understanding

Question 1

A uniform straight line extends from

x=0

to

x=L

. What is its centroid coordinate?

$L/4$
$L/2$
$2L/3$
$L$

Show answer and explanation

L/2

The line has constant length per unit coordinate, so its centroid is the midpoint:

\int_0^L x\,dx/\int_0^L dx=L/2

.

Question 2

For a vertical strip of height

y(x)

starting at the

x

-axis, what is its contribution to the vertical first moment of area?

$y(x)\,dx$
$xy(x)\,dx$
$\frac{1}{2}y(x)^2\,dx$
$\frac{1}{2}x^2\,dx$

Show answer and explanation

\frac{1}{2}y(x)^2\,dx

The strip area is y dx and its centroid height is

y/2

. Their product, the contribution to the vertical first moment, is

\frac{1}{2}y^2\,dx

.

Question 3

A uniform area is symmetric about the vertical axis

x=0

. What is its horizontal centroid coordinate?

$\bar{x}=0$
$\bar{x}$ equals the maximum width
$\bar{x}$ must be positive
It cannot be found without integrating both coordinates

Show answer and explanation

\bar{x}=0

For every area element at positive

x

, a matching element at negative

x

has an equal and opposite contribution to the horizontal first moment.

Key terms

Centroid: The geometric balance point of a uniform line or area.
Differential element: A very small piece of a line or area, written as $dL$ or $dA$ .
First moment: A small length or area multiplied by its distance from an axis, summed over the shape.
Parameter: A variable, such as an angle, used to describe points along a curve.

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8.2 · Find centroids of simple lines and areas by integration

University of Alberta ENGG 130: Engineering Mechanics: Statics

ENGG 130 study topic 8.2

What you will learn

Centroid coordinates and first moments

Choose an element that matches the shape

Integrate, interpret, and check

Centroid of a quarter-circular line

Centroid of a triangular area

Centroid of an upper semicircular area

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through ENGG 130

About this lesson and its review