l03 intscalar post

7/30/2019 L03 IntScalar Post

1/18

LECTURE 3 slide 1

Lecture 3

Total Charge:

Line, Surface and Volume Integrals

Sections: 1.8, 1.9, 2.3

Homework: D2.4; 1.23, 1.27, 2.14, 2.16


2/18

LECTURE 3 slide 2

Line Elements 1

metric increment due to a differential change in position along a line

a

bdl

xdxa

ydya

x yd ddx y= +l aa

a

d

ad a

d a

d dd = +l aa

dlxa

ya

a b


3/18

LECTURE 3 slide 3

Line Elements 2

line increment is a vector has direction: point of integration

moves from point a to point b

each of its components is a linear increment (in meters)

RCS: dx, dy, and dzare linear by default

CCS: d and dzare linear, d is not

SCS: dris linear, dand d are not

angular increment d corresponds to linear increment d

zd d d dz = + +l a a a

x y zd dx dy dz = + +l a a a

sinrd dr rd r d = + +l a a asin

d rdd r d

6

6

d d 6


4/18

LECTURE 3 slide 4

Line Elements 3

SCS:

ra

a

a

dr

drrd

dl

sinr

sinr d

d

x

y

zsinrd dr rd r d = + +l a a a


5/18

LECTURE 3 slide 5

the direction of integration does

not matter: charge is scalar

Line Integration: Charge on Lines 1

B

lQ dl=

straight line: choose RCS

axis along charged line

EASY SPECIAL CASES: CHARGE ON PRINCIPAL GRID LINES

( )Bx

l

x

Q x dx=

-line in CCS: circularcharges in thex-0-yplane 0( )

B

A

lQ d

=

-line and -line in SCS

0( )B

lQ r d

= 0 0( ) sinB

A

lQ r d

=


6/18

LECTURE 3 slide 6

Line Integration: Charge on Lines 2

GENERAL CASE curved line: needs line equation in parametric form( ) ( ) ( ) ( )x y zu x u y u z u= + +r a a a x y zd d dx dy dz = = + +l r a a a

x

y

z

B

( )ur dr

2 2 2 2dl dx dy dz = + +2 2 2 2

dl dx dy dz

du du du du

= + +

2dl d d

du du du

=

r r dl d d

du du du

=

r r d ddl du

du du =

r r

( )Bu

l

u

d dQ u du

du du=

r r

B

A

u

AB

u

d dL du

du du=

r r


7/18

LECTURE 3 slide 7

The surface element is defined by two line elements:

1 2d d d= s l l

surface elements in the RCS principal planes

plane:

plane:

plane:

z

y

x

z const d dxdy

y const d dxdz

x const d dydz

= = = =

= =

s a

s a

s a

Surface Elements 1

1dl2dlds

0

z

y

x

a

/ 2z a=

/ 2z a=

/ 2x a=

/ 2x a= / 2

ya

=/ 2

ya

=

dy

dz

dy

dx

dzdx


8/18

LECTURE 3 slide 8

surface elements in the CCS principal planes

plane:

plane:

plane: z

const d d dz

const d d dz

z const d d d

= =

= =

= =

s a

s a

s a

Surface Elements 2

z

a

dz

d

a

dd

za

const=

z const=

const

=

d

dz a


9/18

LECTURE 3 slide 9

surface elements in the SCS principal planes

Surface Elements 3

2

surface (sphere):

sin r

r const

d r d d

==s a

surface (cone):sin

constd r drd

==s a

surface (circular plane):constd rdrd ==s a


10/18

LECTURE 3 slide 10

Surface Integration: Charge on Surfaces 1

We will limit ourselves to SPECIAL CASES

charges on principal coordinate planes

charge on a plane2 2

1 1

( , )

y x

s

y x

Q x y dxdy=

charge on a circular disk or ringx

y

1x 2x

1y

2y

2 2

1 1

( , )sQ d d

=

s

S

Q ds=

1

21

2

x

y


11/18

LECTURE 3 slide 11


charge on a cylinder

charge on a sphere

22

1 1

0( , )

z

s

z

Q z d dz

= 1z

2z1 2

0

2 2

1 1

20( , ) sinsQ r d d

= 0

x

y

z

0r

12

1

2


12/18

LECTURE 3 slide 12


charge on a cone2 2

1 1

0( , ) sin

r

s

r

Q r r drd

= 1 2

1r

2r

0

NOTE: If you sets = 1 in the above formulas, you can compute

the area of the respective surfaces.


13/18

LECTURE 3 slide 13

1 2 3( )dv d d d = l l l

Volume Elements 1

The volume element is defined by three line elements

RCS: dv dxdydz= 2dl1

dl3dl


14/18

LECTURE 3 slide 14

CCS: dv d d dz =

Volume Elements 2


15/18

LECTURE 3 slide 15

2SCS: sindv r drd d =

Volume Elements 3


16/18

LECTURE 3 slide 16

v

V

Q dv=

Volume Integration: Volume Charges

parallelogram22 2

1 1 1

( , , )

yz x

v

z y x

Q x y z dxdydz =

cylindrical volume

1 2 2

1 1 1

( , , )

z

v

z

Q z d d dz

=

spherical volume

2 2 2

1 1 1

2( , , ) sin

r

v

r

Q r r drd d

=

NOTE: You can find the volume of the element by settingv = 1.


17/18

LECTURE 3 slide 17

Volume Integration: Example

Work in SCS. Assume thez-axis is along the axis of thesymmetrical conical light beam. The sine of half the subtended

angle of the beam is

2

2

0 0 0

sin

a

V r

V dv r drd d

= = =

= = 3 3

352 (1 cos ) 2 0.0202 0.21, m3 3

aV = = =

A light source shines onto a hemispherical dome of radius a = 5 m,and makes a round spot 2 m in diameter, d= 2 m. What is the

volume of the light cone from the light source to the dome?

/ 2sin 0.2

d

a = =2cos 1 sin 0.9798 = =

Homework: Find the area of the dome lit up by the beam. 23.173, mA =


18/18

LECTURE 3 slide 18

You have learned how to

find the total charge along a straight line or any other curved line

find the total charge on any portion of the surface of a plane, disk,

cylinder, sphere, cone

find the total charge on any portion of a parallelogram, cylinder,

sphere, cone

use integration to find length, area and volume

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