Geometric application: arc length Problem: Find the length s of the arc defined by the curve y=f(x)...

Geometric application: arc length Problem: Find the length s of the arc defined by the curve

y=f(x) from a to b. Solution: Use differential element method, take any

element [x,x+dx], the length of the sub-arc corresponding to

[x,x+dx] is:

Therefore the total arc length is21 [ ( )] .

b b

a as ds f x dx

2 2 2 2 2( ) ( ) ( ) ( ) 1 [ ( )] ,s x y dx dy f x dx ds

Arc length formula If a curve has the equation x=g(y), then the length of the

arc corresponding is

Ex. Find the length of the arc of the parabola from

(0,0) to (1,1). Sol.

2 2 2( ) ( ) 1 [ ( )] .d d d

c c cs ds dx dy g y dy

c y d

2y x

1 12 2

0 01 [ ( )] 1 4s g y dy y dy

arctan 2 2 arctan 200

1 1sec sec (sec tan ln | sec tan |)

2 4t tdt t t t t

1 5 ln( 5 2){sec(arctan 2) 2 ln[sec(arctan 2) 2]} .

4 2 4

Differential of arc length function A smooth curve is defined by y=f(x) and let s(x) be the

length of the arc corresponding to the part from a to x. Then

s(x) is called the arc length function, and we have formula

The differential of arc length is

2( ) 1 [ ( )] .x x

a as x ds f t dt

2 2( ) ( ) .ds dx dy

Geometric application: surface area A surface of revolution is generated by rotating a planar

curve about a line.

Problem: find the area of the surface obtained by rotating

the curve about the x-axis. ( ) ( )y f x a x b

Surface area formula Use differential element method, take an element [x,x+dx],

the small surface is approximately a portion of a circular cone,

thus its surface area is

and therefore the total surface area is

( )2 2

2

y y dyS ds yds

22 2 ( ) 1 [ ( )] .b b b

a a aS dS yds f x f x dx

22 ( ) 1 [ ( )] ,f x f x dx dS

Surface area formula If the curve is given by and rotate

the curve about y-axis, then the surface area formula is

Ex. The curve about x-axis, find the surface area.

Sol.

22 2 ( ) 1 [ ( )] .d d d

c c cS dS xds g y g y dy

( )( ),x g y c y d

24 , ( 1 1)y x x

1 2

12 ( ) 1 [ ( )]S f x f x dx

21 1221 1

2 4 1 4 84

xx dx dx

x

Surface area formula If the curve is given by and rotate

it

about x-axis, the surface area formula is

If the curve is and rotate about y-axis,

the formula is

22 2 1 [ ( )] .d d

c cS yds y g y dy

22 2 1 [ ( )] .b b b

a a aS dS xds x f x dx

( ) ( ),x g y c y d

( )( )y f x a x b

Example Ex. Find the area of the surface generated by rotating the

curve segment from (1,1) to (2,4), about the y-axis. Sol I.

Sol II.

2y x2 2 2

1 12 2 1 (2 ) (17 17 5 5).

6S xds x x dx

4 4 2

1 1

12 2 1 ( )

2S yds y dy

y

44 3/ 2

11

1 4 12 ( ) (17 17 5 5)

4 3 4 6y dy y

Physical application: hydrostatic pressure

Background: when an object is submerged in a fluid, the

fluid will exert a force upon the object. The hydrostatic

pressure is defined to be the force per unit area. Suppose the object is a thin horizontal plate. Its area is A

and is submerged into depth d of a fluid with density Then

we have

By the principle of fluid pressure, at any point in a liquid

the pressure is the same in all directions, we have

.V Ad m V Ad

.F

F mg gAd P gdA

.P gd

Hydrostatic force Ex. A cylindrical drum with radius 3m is submerged in

water 10m deep. Find the hydrostatic force on one end of it. Sol. Use differential element method. First build up the

coordinate frame with origin placed at the center of the

drum. Take any infinitesimal element [x,x+dx], the pressure

on this part is and the force is

Therefore the total force is

9800(7 ),P gd x

3 2

39800(7 )2 9 617400 .F x x dx N

29800(7 ) 2 9 ,F P A x x dx

Moments and centers of mass Center of mass: a point on which a thin plate balances

horizontally A system of two masses and which lie at and

respectively, the center of mass of the system is

and are called the moments of the masses

and (with respect to the origin) respectively.

1m 1x2 ,m

2x :x

1 1m x 2 2m x 1m

2m

1 1 2 21 1 2 2

1 2

( ) ( ) 0 .m x m x

m x x m x x xm m

Moment and center of mass A system of n particles located at the points

on the x-axis, then

is the center of mass of the system, is the total mass

of the system, and the sum of individual moments

is called the moments of the system about the origin.

1 2, , , nm m m

1 2, , , nx x x

1 11 1

1

( ) ( ) 0 .

n n

i i i ii i

n n n

ii

m x m xm x x m x x x

mm

m

i iM m x

x

Moment and center of mass A system of n particles located at the points

in the xy-plane, then we define

the moment of the system about the y-axis to be

and the moment of the system about the x-axis as

the coordinates of the center of mass of the system

where is the total mass.

1 2, , , nm m m1 1 2 2( , ), ( , ), , ( , )n nx y x y x y

im m

( , )x y

y i iM m x

x i iM m y

/ , /y xx M m y M m

Moment and center of mass Consider a flat plate (called a lamina) with uniform density

that occupies a region R of the plane. Center of mass is called the centroid The symmetry principle says that if R is symmetric about

a line l, then the centroid of R lies on l. The moments are defined so that if the entire mass of a regi

on is concentrated at the center of mass, then its moments remain unchanged.

The moment of the union of two nonoverlapping regions is the sum of the moments of the individual regions.

Moment of a planar region Suppose the region R lies between the lines x=a and x=b, a

bove the x-axis and below the curve y=f (x). Use differential element method: the moment of the subreg

ion corresponding to [x,x+dx], about y-axis, is

So the moment of R about y-axis is

similarly, the moment of R about x-axis is

( ) ( )y yM f x dx x xf x dx dM

( )b

y aM xf x dx

212 ( )

b

x aM f x dx

Center of mass of a planar region The center of mass of R is defined by

the total mass is

so the coordinates of the centroid are

( )b

am A f x dx

( ) 1( )

( )

b

by ab a

a

xf x dxMx xf x dx

m Af x dx

21

2 212

( ) 1( )

( )

b

bx a

b a

a

f x dxMy f x dx

m Af x dx

/ , /y xx M m y M m

Example Ex. Find the centroid of the region bounded by the curves

y=cosx, y=0, x=0, and Sol.

/ 2.x

/ 2

0cos 1A xdx

/ 2

0

1( ) cos 1

2

b

ax xf x dx x xdx

A

/ 22 212 0

1 1( ) cos

2 8

b

ay f x dx xdx

A

Center of mass of a planar region Suppose now the region R lies between the lines x=a and

x=b, above the curve y=g(x) and below the curve y=f (x)

where The coordinates of the centroid are

1[ ( ) ( )]

b

ax x f x g x dx

A

2 212

1[ ( ) ( )]

b

ay f x g x dx

A

( ) ( ).f x g x

Example Ex. Find the centroid of the region bounded by the line y=x

and Sol.

2.y x1 2

0

1( )

6A x x dx

1 1 2

0 0

1 1[ ( ) ( )] 6 ( )

2x x f x g x dx x x dx

A

1 12 2 2 4120 0

1 2[ ( ) ( )] 3 ( )

5y f x g x dx x x dx

A

Homework 20 Section 7.8: 25, 37, 40, 54, 55, 56, 58, 77

Section 8.1: 20, 37