Centroid, Area, and Moments of Inertia for Laminas
1 Definition
Definitions of an area, centroid, and moments of inertia for a lamina may be found in
most text books of engineering mechanics. For completeness, we give a brief review of
these definitions in this section.
Let D be a region consisting of the boundary of the lamina and its interior.
Then,
 Area: A=òò_{D} dxdy.
 Mass: M=òò_{D} r(x,y)dxdy.
 Moment about the xaxis: M_{x}=òò_{D} yr(x,y)dxdy.
 Moment about yaxis: M_{y}=òò_{D} xr(x,y)dxdy.
 Moment of inertia about the xaxis: I_{xx}=òò_{D} y^{2}r(x,y)dxdy.
 Moment of inertia about the yaxis: I_{yy}=òò_{D} x^{2}r(x,y)dxdy.
 Polar moment of inertia: I_{O}=I_{x2+y2} = òò_{D} (x^{2}+y^{2})r(x,y)dxdy=I_{xx}+I_{yy}.
 Product of inertia: I_{xy}=òò_{D} xyr(x,y)dxdy.
The centre of mass (or centroid) is given by
x_{c}= 
M_{y} M

, y_{c}= 
M_{x} M

. 

In most engineering applications, the density of mass of lamina is constant so
that r can be taken out of the integration sign. We may further assume
that r = 1 by disregarding the constant factor. Therefore, we have the
following simpler formulas:
 Mass: M=òò_{D} dxdy.
 Moment about the xaxis: M_{x}=òò_{D} y dxdy.
 Moment about yaxis: M_{y}=òò_{D} x dxdy.
 Moment of inertia about the xaxis: I_{xx}=òò_{D} y^{2} dxdy.
 Moment of inertia about the yaxis: I_{yy}=òò_{D} x^{2} dxdy.
 Polar moment of inertia: I_{O}=I_{xx}+I_{yy}.
 Product of inertia: I_{xy}=òò_{D} xy dxdy.
In engineering mechanics, we often want to know the moments of inertia about
the axes passing through the centroid. Let a user defined coordinate system be
(O,x,y) and the centroid of a lamina be C=(x_{c},y_{c}). Assume the new
coordinate system is (C,x,h) with x and haxis parallel to the
x and yaxis. Then, we have
I_{xx}=òòy^{2} dxdy=òò(y_{c}+h)^{2}dxdy = y_{c}^{2}òòdxdy+2y_{c}òòhdxdy+òòh^{2}dxdy
=y_{c}A+2y_{c}òòhdxdh+òòh^{2}dxdh.
It is noted that the third term represents the moment of inertia about the centroidal axis x. The second term is zero since the x and haxis pass through the centroid, i.e., òòhdxdh = 0. Thus, we have
I_{xx}=y_{c}^{2}A+I_{xx} or I_{xx}=I_{xx} y_{c}^{2}A. 

Similarly, we can drive:
 I_{xh}=I_{xy}x_{c}y_{c}A
 I_{hh}=I_{yy}x_{c}^{2}A
 I_{x2+h2}=I_{xx}+I_{hh}.
It should be pointed out that I_{xh}=0 if a lamina is symmetric to either
x or haxis.
2 Principal moments of inertia
Let I_{xx}, I_{hh}, I_{x2+h2}, and I_{xh} be
moments of inertia defined in the coordinate system (C,x,h), where C
is the centroid of lamina. We now want to compute moments of inertia about a
new system (C,u,v) which is obtained by rotating (C,x,h) by q.
It is noted that
Therefore, we can derive that
 I_{uu}=òv^{2} dA=I_{xx} cos^{2}q+I_{hh} sin^{2}q 2I_{xh} sinqcos q.
 I_{vv}=òu^{2} dA=I_{xx} sin^{2}q+I_{hh} cos^{2}q +2I_{xh} sinqcosq.
 I_{uv}=òuv dA=I_{xx} sinqcosqI_{hh} sinqcosq+2I_{xh}(cos^{2}qsin^{2}q).
 I_{u2+v2}=I_{x2+h2}.
The above equations may be simplified by using the trigonometric identities
sin(2q)=2sinqcosq and cos(2q)=cos^{2}qsin^{2}q, 

in which case
 I_{uu}=[(I_{xx}+I_{yy})/2]+[(I_{xx}I_{yy})/2]cos(2q) I_{xy}sin(2q) (1)
 I_{vv}=[(I_{xx}+I_{yy})/2][(I_{xx}I_{yy})/2]cos(2q) +I_{xy}sin(2q)
 I_{uv}=[(I_{xx}I_{yy})/2]sin(2q)+I_{xy}cos(2q)
Different q results in different moments of inertia. We now want to
determine the q_{p} such that I_{uu} and I_{vv} are maximum and
minimum. This particular set of axes is called the principal axes, and the
corresponding moments of inertia with respect to these axes are called the
principal moments of inertia. The q_{p} may be found by differentiating,
for example, I_{uu} with respect to q and setting the result equal to
zero. Thus,

dI_{uu} dq

=2 
I_{xx}Ihh 2

sin2q 2I_{xh} cos2q = 0. 

Consequently, at q = q_{p}, we have
tan2q_{p}= 
2I_{xh} I_{xx}Ihh


 (2) 
and
I_{max}, I_{min}= 
I_{xx}+I_{hh} 2

± 
æ ú
Ö


æ ç
è

I_{xx}I_{hh} 2

ö ÷
ø

2

+I_{xh}^{2} 

. 

It is noted that equation (2) gives two solutions, q_{1} and q_{2}, which are 90^{o} apart. To find which one is correct, we may substitute q into equation (1) and check if I_{uu} matches I_{max}. If not, we should set q = q+p/2.
We now consider two principal axes. Let the x and yaxis be defined respectively by two unit vectors
x=(x_{1},x_{2}) and y=(y_{1},y_{2}). Then, the unit vectors that define the
principal axes are
u=x cosq_{p}+y sinq_{p} and v=y cosq_{p}+x sinq_{p}. 

Replacing q in (1), we can obtain I_{uu} and I_{vv} which
correspond to I_{min} and I_{max}.
3 Green Theorem
We start our discuss by considering the Green's theorem.
Theorem 1 [Green Theorem]
Let C be a piecewise smooth simple closed curve and D be the region
consisting of C and its interior. If P(x,y) and Q(x,y) are functions that
are continuous and have continuous first partial derivatives throughout an open
region R containing D, then

ó õ


ó õ

D


æ ç
è

dQ dx

 
dP dy

ö ÷
ø

dxdy =  ó (ç) õ

C

Pdx+Qdy 

The sign of the right side integration is associated with the orientation of
the curve. If we walk along the curve and see the region D on our left side,
then the integration is positive. This requirement coincides with the
definition of the ACIS about an orientation of a curve. Green's theorem can
also be applied to a region D which contains holes, provided we integrated
over entire boundary and always keep the region D to the left. This
requirement again matches the definition of the ACIS. Hence, Green's theorem
will be used to compute properties of lamina which have holes inside.
In CAD systems, curves are often given parametrically as r(t)=(x(t),y(t))
with t Î [t_{0},t_{1}]. Let P(x,y)=y and Q(x,y)=0. From Green theorem, a
formula of computing an area for a lamina is given by
A= 
ó õ


ó õ

D

dxdy= 
ó õ


ó õ

D


æ ç
è

dQ dx

 
dP dy

ö ÷
ø

dxdy=  ó (ç) õ

C

ydx =  
ó õ

t_{1}
t_{0}

y(t)x¢(t)dt. 

As an example, we compute an area of a circle x(q)=Rcos(q)
and y(q)=Rsin(q):
A= 
ó õ

2p
0

y(q)x^{¢}(q)dq = R^{2} 
ó õ

2p
0

(sinq)^{2}dq = R^{2} 
æ ç
è

q 2

 
1 4

sin2q 
ö ÷
ø

2p
0

=R^{2}p. 

Similarly, we can derive
M_{x}=òòydxdy=^{1}/_{2}(ò)y^{2}dx=^{1}/_{2}ò_{t0}^{t1} y^{2}(t)x^{¢}(t)dt (P=^{1}/_{2}y^{2}, Q=0)
M_{y}=òòxdxdy=^{1}/_{2}(ò)x^{2}dy=^{1}/_{2}ò_{t0}^{t1}x^{2}(t)y^{¢}(t)dt (P=0, Q=^{1}/_{2} x^{2})
I_{xx}=òòy^{2}dxdy=^{1}/_{3}(ò)y^{3}dx=^{1}/_{3}ò_{t0}^{t1}y^{3}(t)x^{¢}(t)dt (P=^{1}/_{3}y^{3}, Q=0)
I_{xy}=òòxydxdy=^{1}/_{2}(ò)xy^{2}dx=^{1}/_{2}ò_{t0}^{t1}y^{2}(t)x(t)x^{¢}(t)dt (P=^{1}/_{2}xy^{2}, Q=0)
I_{yy}=òòx^{2}dxdy=^{1}/_{3}(ò)x^{3}dy=^{1}/_{3}ò_{t0}^{t1}x^{3}(t)y^{¢}(t)dt (P=0, Q=^{1}/_{3}x^{3})
4 GaussLegendre integration
It is seen that computation of an area and moments is actually a requirement of
computing the integration of the type:
To compute the above integration numerically, the wellknown method is the
GaussLegendre integration rule which has the following form:

ó õ

b
a

f(t)dt @ 
n å
i=0

A_{i}f(x_{i}) 

The above integration rule is exact if f(x) is a polynomial of degree not
more than 2n+1. In our case, f(t) is either polynomial or rational
function. If it is former case, f(t) is a polynomial and thus the integration
rule is exact. If it is the latter case, f(t) can be approximated by a
polynomial function since f(t) is an analytic function. Therefore, we can
usually obtain accurate results very efficiently using the GaussLegendre
integration rule.
5 Applying Green's theorem to a composite curve
It is noted that Green's theorem do not require C to be a single closed curve.
Instead, C could be a piecewise closed curve. If C is made of piecewise
curves C_{i} (C_{i} may be lines, arcs, and Bspline curves), then the Green's
formula becomes
 ó (ç) õ

C

P(x,y)dx+Q(x,y)dy= 
å
 
ó õ

C_{i}

P(x,y)dx+Q(x,y)dy. 

As an extreme case where C_{i} are all lines defined by points (x_{i},y_{i}) and
(x_{i+1},y_{i+1}), then the Green formula is given by
 ó (ç) õ

C

P(x,y)dx+Q(x,y)dy= 
å
 
ó õ

x_{i+1}
x_{i}

[P(x,y)+Q(x,y)y^{¢}(x)]dx. 

We now consider area, moments, and moments of inertia. For the purpose of
simplifying notation, let any two adjacent points be denoted by (x_{1},y_{1}) and
(x_{2},y_{2}). Then, a line joining the two points is given by
y=y_{1}+ 
y_{2}y_{1} x_{2}x_{1}

(xx_{1})=y_{1}+l(xx_{1}). 

Referring to previous
sections, the area of lamina is given by
A=  ó (ç) õ
 ydx= 
å
 
ó õ

x_{2}
x_{1}

y dx= 
1 2


å
 (y_{1}+y_{2}) (x_{2}x_{1}) 

We note that (y_{1}+y_{2})(x_{2}x_{1})/2 is the area of trapezoid. The moments of
lamina are given by
M_{x}= 
1 2

 ó (ç) õ
 y^{2}dx= 
1 2


å
 
ó õ

x_{2}
x_{1}

[y_{1}+l(xx_{1})]^{2}dx= 
1 6


å
 (x_{2}x_{1})(y_{1}^{2}+y_{1}y_{2}+y_{2}^{2}) 

M_{y}= 
1 2

 ó (ç) õ
 x^{2}dy= 
1 2


å
 
ó õ

y_{2}
y_{1}

[x_{1}+ 
1 l

(yy_{1})]^{2}dy= 
1 6


å
 (y_{2}y_{1})(x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}) 

The moments of inertia are given by
I_{xx}= 
1 3

 ó (ç) õ
 y^{3}dx= 
1 3


å
 
ó õ

x_{2}
x_{1}

[y_{1} +l(xx_{1})]^{3}dx= 
1 12


å
 (x_{2}x_{1})(y_{1}^{3}+y_{1}^{2}y_{2}+y_{1}y_{2}^{2} +y_{2}^{3}) 

I_{yy}= 
1 3

 ó (ç) õ
 x^{3}dy= 
1 3


å
 
ó õ

y_{2}
y_{1}

[x_{1}+ 
1 l

(yy_{1})]^{3}dy= 
1 12


å
 (y_{2}y_{1}) (x_{1}^{3}+x_{1}^{2}x_{2}+x_{1}x_{2}^{2}+x_{2}^{3}) 

I_{xy}=^{1}/_{2}(ò)xy^{2}dx=^{1}/_{2}åò_{x1}^{x2} x[y_{1}+l(xx_{1})]^{2}dx
= [1/(6l)]å{x[y_{1}+l(xx_{1})]^{3}ò_{x1}^{x2} [y_{1}+l(xx_{1})]^{3}dx}
= [1/(6l)][x_{2}y_{2}^{3}x_{1}y_{1}^{3}^{1}/_{4}(x_{2}x_{1}) (y_{2}^{3}+y_{2}^{2}y_{1}+y_{2}y_{1}^{2}+y_{1}^{3})]
It is noted that if y_{1}=y_{2}, then [1/(l)]=¥. In
this case, we represent the line parametrically as
y=y_{1} and x=x_{1}+(x_{2}x_{1})t, t Î [0,1]. 

Thus,
I_{xy}= 
1 2

 ó (ç) õ
 xy^{2}dx= 
1 4

y_{1}^{2}(x_{2}^{2}x_{1}^{2}). 

6 Ellipse: a special case
Let u and v denote the unit vectors of the major and minor axes of an
ellipse, and a, b the length of major and minor axes respectively. Then, we
may write an ellipse parametrically as
u(q)=acosq and v(q)=bsinq. 

Let f be the angle measured anticlockwisely from the xaxis to the
major axis u. Then, the x and y coordinates are given by

æ ç
è

 
ö ÷
ø

= 
æ ç
è

 
ö ÷
ø


æ ç
è

 
ö ÷
ø

= 
æ ç
è

 
ö ÷
ø

. 

In calculus, we have learned that
 (ò)cos^{2}qdq = (ò)sin^{2}qdq = p,
 (ò)cos^{3}qdq = (ò)sin^{3}qdq = 0,
 (ò)sin^{2}qcos^{2}qdq = p/4,
 (ò)cos^{4}qdq = (ò)sin^{4}qdq = 3p/4,
 (ò)cos^{3}qsinqdq = (ò)sin^{3}qcosqdq = 0.
Let a_{s}=asinf, a_{c}=acosf, b_{s}=bsinf, and b_{c}=bcosf.
Then, x(q)=a_{c}cosqb_{s}sinq and y(q)=a_{s}cosq+b_{c}sinq. Accordingly, we can derive
 I_{xx}=^{1}/_{3}(ò)y^{3}(q)dx(q) = [(p)/4](a_{c}b_{c}+a_{s}b_{s})(a_{s}^{2}+b_{c}^{2}),
 I_{xy}=^{1}/_{2}(ò)y^{2}(q)x(q)dx(q) = [(p)/4](a_{c}b_{c}+a_{s}b_{s})(a_{c}a_{s}b_{c}b_{s}),
 I_{yy}=^{1}/_{3}(ò)x^{3}(q)dy(q) = [(p)/4](a_{c}b_{c}+a_{s}b_{s})(a_{c}^{2}+b_{s}^{2}).
7 General organisation
In applications, a lamina may be defined by a curve made of lines, arcs, and
Bspline curves. For efficient purpose, we may compute line/curve integrals of
area and moments along lines and curves. Due to the additive property of areas
and moments, we can obtain the total area and moments of lamina by simply add
individuals together. To do so, it is however important to ensure that the
orientation of the curve is same.
It is easy to implement line integrals of area and moments. In the subsequent
sections, we discuss only the implementation of curve integrals of which a
curve is represented in Bspline form. The procedures are summarised below:
 Convert Bspline curve into Bézier curves.
 With respect to each Bézier curve, compute signed area A_{i}, moments
(M_{x} and M_{y}) and moments of inertia (I_{xx}, I_{xy}, I_{yy}).
 Compute åA_{i}, åM_{x}, åM_{y}, etc..
RETURN