r r r r ′=− + r M - METUcourses.me.metu.edu.tr/courses/me210/Transparencies/ME210-14S … · ME...

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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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