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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 1/25
Example:
For a mechanical part with a shape of one half turn of a circular helix given by the
position vector from (1,0,0) to (–1,0,π), whose mass density
per unit length changes linearly with its height as ρ = z, determine
a) its mass, M; and
b) the vertical location of its center of mass.
The mass of this part can be found as
ktjsinticost(t)rrrrr
++=
kjcostisint(t)rrrrr
++−=′
2(1)(cost)sint)((t)r 222 =++−=′r
6.982
π
2t
2dt2tdt(t)rz(t)zdsρdsM2πt
0t
2πt
0t
πt
0tCC
≅===′====
=
=
=
=
= ∫∫∫∫r
The vertical location of its center of mass can be found as
2.093
2π3t
π
2dt2t
π
2dt(t)r(t)z
M1
ρzdsM1
z
πt
0t
3
2
πt
0t
22
πt
0t
2
C
CM ≅===′===
=
=
=
=
= ∫∫∫r
ππ
11
xx
yy
zz
––11
00
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 2/25
Evaluate the line integral
from A(1,1) to B(2,4) along the path
a) Parabola y = x2
b) Straight line passing through A and B
c) Straight line AP where P(1,4), then PB
d) Straight line AQ where Q(2,1), then QB
Example:
∫ +C
dxyx
1
a) Using the curve a expression y = x2 in f(x,y,z) gives
2xy
xy xx
1yx
1z)y,f(x,
2
2
+=
+=
==
0.28834
lnx)]ln(1[ln(x)dxx1
1x1
dxxx
1dx
yx1 2
1
2
1
2
12
C
≅
=+−=
+−=
+=
+ ∫∫∫
a
bb
AA
44
11
yy
00 11 22 xx
BB
d
Q
P
c
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 3/25
b) Using the curve b (straight line passing through A and B)
expression y = 3x–2 in f(x,y,z) gives
( ) 0.2753ln41
2)]-[ln(4x41
dx24x
1dx
yx1 2
1
2
1C
≅==−
=+ ∫∫
24x1
23xx1
yx1
z)y,f(x,23xy
23xy −=
−+=
+=
−=−=
c) Along AP (straight line parallel to y-axis): dx = 0
Along PB (straight line parallel to x-axis): y = 4
0.18256
ln4)ln(xdx4x
10dx
yx1
dxyx
1dx
yx1 2
1
2
1
P
A
B
PC
≅
=+=+
+=+
++
=+ ∫∫ ∫∫
d) Along AQ (straight line parallel to x-axis): y = 1
Along QB (straight line parallel to y-axis): dx = 0
0.40523
ln1)ln(x0dx1x
1dx
yx1
dxyx
1dx
yx1 2
1
2
1
P
Q
Q
AC
≅
=+=++
=+
++
=+ ∫∫∫∫
a
bb
AA
44
11
yy
00 11 22 xx
BB
d
Q
P
c
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 4/25
Evaluate
from A(1, , 0) to along the arc of
the circle x2 + y2 = 4, with z = 0.
y + z ds
xC∫
B( 2, 2, 0)
One may solve the problem by defining the position vector of the curve C as
Alternative solution (evaluation over x):
( ) 2 cos(t) 2 sin(t) r t i j= +r rr
2 2 2ds = dx + dy2 2
2 222
- x dx x dxy = 4 - x dy = dy =
4 - x4 - x2
2 2 22 2
x 2ds = dx + dx => ds = dx
4 - x 4 - x
Example:
( ) 0.6932ln2ln(x)dxx-4
2x
x-4ds
xzy 2
1
2
12
2
C
≅==⋅=+∫∫
3
A
B
2
2 x
y
32
210
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 5/25
Properties of Line Integrals
k is a constantValid also for dx, dy, dz next to ds∫∫ =
B
A
B
A
z)dsy,f(x,kz)dsy,kf(x,
∫∫∫ ±=±B
A
2
B
A
1
B
A
21 z)dsy,(x,fz)dsy,(x,fz)]dsy,(x,fz)y,(x,[f Valid also for dx, dy, dznext to ds
∫∫∫ =+B
A
B
P
P
A
z)dsy,f(x,z)dsy,f(x,z)dsy,f(x,Valid if and only if P is between A and B on the path of integration.Valid also for dx, dy, dz next to ds
∫∫ =A
B
B
A
z)dsy,f(x,z)dsy,f(x,
∫∫ −=A
B
B
A
z)dxy,f(x,z)dxy,f(x,
Valid only when ds is used, because s is positively defined variable regardless of the direction of travel along a curve
Valid only when dx, dy or dz is used, because x, y, z possess positive and negative directions
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 6/25
Geometrical interpretation of line integrals along planar curves as areas
It is possible, as in ordinary integration, to interpret line integrals along planar
curves as areas.
To demonstrate this fact, consider an integrand function f(x,y) as defining the
surface S by z = f(x,y), extending above some region in xy-plane as illustrated in the figure.
z
x
y
Surface S
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 7/25
The vertical cylindrical surface standing on a curve C as base will cut the surface S
in some curve PQ.
The area A1 of this cylindrical surface ABQP will then be written as
which is a line integral of f(x,y) along C.
z
x
y
Surface S
Curve C
dsA
P
Q
B
Area ABQP = A1
∫∫ ==CC
1 y)dsf(x,zdsA
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 8/25
Similarly, if the projection of the cylindrical surface ABQP on the xz-plane is
considered,
which is another line integral of f(x,y) along C, obviously different than the first one.
∫∫ ==CC
2 y)dxf(x,zdxA
z
x
y
Surface S
Curve C
dsA
P
Q
BB’
A’dx
P’
Q’
Area A’B’Q’P’ = A2
the area A2 of this planar surface A’B’Q’P’ will then be written as
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 9/25
Similarly, if the projection of the cylindrical surface ABQP on the yz-plane is
considered,
which is another line integral of f(x,y) along C, obviously different than the first two.
∫∫ ==CC
3 y)dyf(x,zdyA
z
x
y
Surface S
Curve C
dsA
P
Q
B
Q”
dy
A”
P”
B”
Area A”B”Q”P” = A3
the area A3 of this planar surface A”B”Q”P” will then be written as
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 10/25
Evaluate from A(0, 0) to B(1, 1) along a straight line path.
Example:
Note that the surface S corresponding to the
integrand function is a plane defined by
z = 2 - x - y or
which crosses all x, y, and z axes at 2 as shown in
the figure.
12z
2y
2x =++
Along C from (0,0) to (1,1), y = x. Therefore,
112x2x2x)dx(2I1x
0x
21x
0x=−=−=−=
=
=
=
=∫Note that this line integral could also be evaluated by using the fact that its values is equal to the area of the AB’P triangle (the projection of ABP triangle) as
1(1)(2)21
P)Area(AB'y)dxx(2I
C
===−−= ∫
A
z
x Curve C
y
Plane S
B’
P
B1
1
2
2
2
∫ −−C
y)dxx(2
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 11/25
Line Integrals of Vector Functions
In many physical problems, line integrals involving vectors occur.
along a curve C given by
r(t) x(t) i y(t) j z(t) k= + +r r rr
from a point A to another point B
The most common line integral of a vector function
is given by
Such line integrals arise naturally in mechanics, giving the work done by a force in
a displacement along a curve C (that is, its point of application moves
along a curve C), thus leading to call them as the work integrals.
z)y,(x,Fr
kz)y,(x,Fjz)y,(x,Fiz)y,(x,Fz)y,(x,F 321
rrrr++=
)dtzFyFx(FdzFdyFdxFdtrFrdFB
A321
C
321
B
AC
∫∫∫∫ ′+′+′=++=′⋅=⋅rrrr
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 12/25
Example:
A particle is attracted towards the origin by a force whose magnitude is proportional to the distance of
the particle from the origin; that is,
where is the position vector of the particle and k
is a known proportionality constant.
rkFrr
−=rr
Determine how much work is done by when the particle is moved from the
point A(1,2) to the point B(0,1) along a straight line as shown in the figure. Fr
The position vector of the path can be written parametrically as
jt)(2it)(1(t)rrrr
−+−= ji(t)rrrr
−−=′&jt)k(2it)k(1rkFrrrr
−−−−=−=
[ ] ( )
[ ] [ ] 2kt3tkdt2t)(3kdtt)k(2t)k(1
dtjijt)k(2it)k(1dtrFdsFrdFW
1t
0t
21t
0t
1t
0t
1t
0t
B
AC
t
C
=−=−=−+−=
−−⋅−−−−=′⋅=⋅=⋅=
=
=
=
=
=
=
=
=
∫∫
∫∫∫∫rrrrrrrrr
y
Curve C
x
A
0
B1
1
2
tFr
Fr
nFr
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 13/25
For the same problem, determine how much
work is done against friction if the coefficient of
friction between the particle and the path is µ.
dt2dtrrdds =′==r
and the magnitude of the normal component of the applied force isnFr
Fr
[ ]2
kt)2t1(
2
k
2
jijt)k(2it)k(1nFFn =−++−=
−⋅−−−−=⋅=rr
rrrrr
( ) µkdt22
kµW
1t
0tf =
= ∫=
=
The work done against friction can be obtained as
yielding
∫=C
nf dsFµWr
where
y
Curve C
x
A
0
B1
1
2
tFr
Fr
nFr
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 14/25
is independent of path C if and only if is the gradient of some scalar
function (hence it is conservative vector field); i.e.,i.e.,
oror
Line Integrals Independent of Path
A line integral
This is because the integrand becomes an exact differential
Under these conditions, leading to the evaluation of integral with the sole
information of φ(A) and φ(B) regardless of the path joining them as
∫∫ ++=⋅B
A321
C
z)dzy,(x,Fz)dyy,(x,Fz)dxy,(x,FrdFrr
φ∇=z)y,(x,Fr
zz)y,(x,
z)y,(x,F;y
z)y,(x,z)y,(x,F;
xz)y,(x,
z)y,(x,F 321 ∂φ∂=
∂φ∂=
∂φ∂=
φ=∂φ∂+
∂φ∂+
∂φ∂=++ ddz
zdy
ydx
xdzFdyFdxF 321
(A)(B)z)dzy,(x,Fz)dyy,(x,Fz)dxy,(x,FrdFB
A321
C
φ−φ=++=⋅ ∫∫rr
z)y,(x,Fr
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 15/25
Therefore, if a force field in a mechanical system can be expressed as a
gradient of a scalar field f(x,y,z), then the net force done by this force field over a
closed path will be zero, indicating that through such a motion the total energy of the system will be conserved. This is the main reason why such vector fields are
referred to as the conservative vector fields.
Another example for such a vector field is the gravitational force, where r is the
distance between two bodies.
One example for such a vector field is the attraction force used in the previous
example.
∇=−=rc
rr
cF
3
rr
−∇=−= 2kr21
rkFrr
z)y,(x,Fr
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 16/25
One end of a spring is attached to a fixed point (origin) while an external force is applied to the other end as shown in the figure. The force needed to hold this spring at equilibrium is given as
Example:
rkFrr
=Determine how much work is done by when the
free end of the spring is is moved from the point
A(1,0,0) to the point B(0,2,1).
Fr
Note that the force field generated by extending the spring along any direction by any amount is a conservative one, and it can be expressed as
Fr
A
z
x
y
B
12
1
AFr
BFr
∇== 2kr21
rkFrr
to give a potential function: 2kr
21=φ
Then, the work done by the force to bring the free end of the spring from A to B is independent the path followed and it can be obtained as
2k1)k(521
kr21
kr21
(A)(B)rdF 2A
2B
B
A=−=−=φ−φ=⋅∫
rr
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 17/25
Surface and Volume Integrals
The concept of line integral can be generalized to
� surface integral
� volume integral
by simply increasing geometrical dimension involved in defining the respective
integrals as
C is a curve in space
∫C
z)dsy,f(x, ∫∫S
z)dSy,f(x, ∫∫∫V
z)dVy,f(x,→ →
S is a surface in space V is a volume in space
ds
C
SdS
dVV
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 18/25
In a similar manner, the concept of single integral of a function f(x) of a single
variable x over an interval X is generalized to
� double integral of a function f(x,y) over a planar region R in xy-plane
� triple integral of a function f(x,y,z) over a volumetric region V in space
as
It should be noted that the volume integral and triple integral obtained in the
last steps of these generalizations turn out to be the same concepts.
∫X
f(x)dx ∫∫R
y)dAf(x, ∫∫∫V
z)dVy,f(x,→ →
dVV
R
dAxA
X
dxxB
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 19/25
Surface integrals are special integrals of scalar functions over surfaces.
Given a scalar field f(x,y,z) and a surface S by its parametric representation
the surface integral of f on S is defined as
where dS is an infinitesimal area element of the surface S.
Surface integrals
kv)z(u,jv)y(u,iv)x(u,v)(u,rrrrr
++=
∫∫S
z)dSy,f(x,
dSS
x
y
z
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 20/25
Evaluation theorem for surface integrals
For a continuous scalar field f(x,y,z) and a smooth surface* of integration S
described by
This theorem can be proven by expressing the infinitesimal area element dS of the
surface S in the following manner:
kv)z(u,jv)y(u,iv)x(u,v)(u,rrrrr
++=
dudvvr
ur
v)]z(u,v),y(u,v),f[x(u,dSz)y,f(x,
SS∫∫∫∫ ∂
∂×∂∂=
rr
* For a smooth surface S, should be continuous or piecewise continuous
and S should have a unique tangent plane everywhere within its domain of
definition, that is, and should exist.
v)(u,rr
uv)/(u,r ∂∂r
vv)/(u,r ∂∂r
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 21/25
spanning dS as
dS
v),(ur or
v)du,(ur o +
r )v(u,r o
r
dv)v(u,r o +r
duur
∂∂r
dvvr
∂∂r
dv)v(u,rr
)v(u,rr
v)du,(urr
v),(urr
o
o
o
o
+==
+==
rr
rr
rr
rr
dvvr
&duur
∂∂
∂∂
rr
dudvvr
ur
dS∂∂×
∂∂=
rr
Referring to the figure, the
area dS of the infinitesimal
parallelogram on the
surface S formed by four curves (mesh lines based on u and v on the surface
S) can be expressed as the magnitude of the cross product of two vectors
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 22/25
Given a surface S by determine the surface integral
for a region defined by 0 ≤ u ≤ 1 and 0 ≤ v ≤ 1.
Let us first discover the type of the surface S.
The second and third components of the position vector suggest that
y = 1 - v & z = v → y = 1 - z or y + z = 1
and the first component of the position vector indicates that x is free to choose.
Therefore, the corresponding surface S is an oblique plane, formed by extending
y + z = 1 line along x direction.
Example:
kvjv)(1iuv)(u,r 2rrrr
+−+=
∫∫ ++=S
z)dSy(xI
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 23/25
Note that
� parametric curves with respect
to u for v = constant are family of lines parallel to x-axis and
� parametric curves with respect
to v for u = constant are family
of oblique lines parallel to yz-plane.
The partial derivatives arekvjv)(1iuv)(u,r 2rrrr
+−+=kj
vr
&i2uur rr
rr
r
+−=∂∂=
∂∂
giving
u22)jk2u()kj(i2uvr
ur =+−=+−×=
∂∂×
∂∂ rrrrr
rr
Surface S
x
z1
1
1
y0
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 24/25
Hence, the surface integral becomes
∫∫ ++=S
z)dSy(xI
∫ ∫=
=
=
=+−+=
1u
0u
1v
0v
2 ududv2v]2v)(1[uI
2
3=
∫ ∫=
=
=
=+=
1u
0u
1v
0v
2 1)ududv(u22I
∫=
=+=
1u
0u
3 u)du(u22I1u
0u
24
2u
4u
22
=
=
+=
+=21
41
22 2.12≅
becomes
u = constant lines
x
z1
1
1
y0
kvjv)(1iuv)(u,r 2rrrr
+−+=
v = constant lines
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 25/25
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