16.4

Green’s Theorem

:

Suppose that is a simple piecewise smooth closed curve, traversed

counter clockwise, and C is the boundary of a region .

If , , , and are continuous in , then

y x

C

R

P Q P Q R

Green's Theorem

x yR C

Q P dA Pdx Qdy

We can use Green's theorem to compute areas:

1( )

2R R R

area R xdy ydx xdy ydx

Review:

: (a) If is defined in a connected and simply connected region,

then if and only if ( ) 0f

Theorem F

F curl F

b) If , then ( ) ( ) C

f dr f B f A F F

2

2

2, 2arctan , 0

1

x

x yP e Q x Q Px

2

2

22arctan

1

GThmx

C R

e dx x dy dAx

0 1

2

1

2

1x

dydxx

y x

10

2

1

2

1 xy dx

x

0

2 2

1

2 2

1 1

xdx

x x

02

12arctan ln 1x x

0 2arctan 1 ln 2 42 ln 2 2 ln 2

2

Compute the closed line intgeral 2arctan

where C is the triangle with vertices 0,0 , 0,1 ,and 1,1 .

x

C

e dx x dy

Example :

x yR C

Q P dA Pdx Qdy

Regions with Holes

x yR

Q P dA 1 2

x y x y

R R

Q P dA Q P dA

1 2C C

Pdx Qdy Pdx Qdy C

Pdx Qdy

1 2C C C

We can use Green's Theorem when there is a hole (or holes) in the interior.still

line integrals along the

connecting lines cancel!

But we need to keep the interior region on the left!

1 2divide into two regions and R R R

1 2

now use Green's theorem

on and :R R

So the interior curve is traversed clockwise, while the exterior curve counterclockwise.

1 2If in the region R between and we have:

, and , , , are continuous, then

we can use Green's Theorem:

y x

C C

P Q P Qx yQ = P

C

Pdx Qdy 1 2C C

Pdx Qdy Pdx Qdy

0

2 2

but C C

Pdx Qdy Pdx Qdy

1 2

1 2

1 2

If in the region and

we have , then

with and both traversed counter clockwise

x y

C C

C C

Q P Pdx Qdy Pdx Qdy

C C

Theorem : between

1 2

Under these circumstances, the line integral on

can be traded in for the line integral on .C C

1

2 1

2

Do this when the line integral is too involved

and you can find a totally contained in the

with the line integral being much easier to calculate.

C

C C

C

1 2Evaluate where C

Pdx Qdy C C C Example :

2 2 2 2 2 2 2 2

'

So C C

y x y xdx dy dx dy

x y x y x y x y

2 2

22 2

As we saw in previous section: x yy x

Q Px y

2 2: 1C x y

: cos , sin , 0 2C x t y t t

2 2 2 2

'C

y xdx dy

x y x y

2

0

1 2dt

1 2 3 42 2 2 2Evaluate where

C

y xdx dy C C C C C

x y x y

Example :

2

0

= sin( )( sin(t)) cos( )cos(t)t t dt

2 2 2 2

We can use the same argument to say that

2 for closed

curved C that contains the origin in its interior.

C

y xdx dy any

x y x y

2 2 2 2and 0 if the origin is not in the interior!

C

y xdx dy

x y x y

We can introduce a new curve ' that avoids the origin!C

but , are not continuous at (0,0)!

(so we cannot use Green's theorem on the square)

P Q

2 2

Evaluate 3 where is the path from

( 2,0) to (2,0) along the upper portion of the ellipse 2 4.

xy xy

C

ye y dx xe x dy C

x y

Example :

Notice that 1 xy xy xyye y e xyey

and 3 3 xy xy xyxe x e xye

x

x yC R

Pdx Qdy Q P dA

hence is an exact differential (not ).xy xyye y dx xe x dy not df

But we can also use Green's theorem by "closing up" the half of the ellipse with

a convenient path. Choose a straight line from ( 2,0) to (2,0).C'

along ' : , 0, 1, 0 hence 0! xy xyC x t y dx dy ye y dx xe x dy

' 2

By Green's theorem 2 2 ( ) 2 2 2xy xy

C C

ye y dx xe x dy dA area R

Notice that is a closed curve going counter clockwise

enclosin g a semicircle R.

C' C

' '

But also .... .... ....xy xy

C C C C C

ye y dx xe x dy

Hence 4 2xy xyC

ye y dx xe x dy

16.5,16.6

Surface Integrals

Goal: define the surface integral , , where is a surface in 3-space.S

G x y z d S If 1, we get ( )

S

G d area S To compute a line integral '( )

C C

d t dt F r F r

we choose a parametrization: ( ) ( ), ( ), ( )t x t y t z t r

What is a parametrization for a surface? , ( , ), ( , ), ( , )u v x u v y u v z u vr

where ( , ) lie in some region in the planeu v uv

Examples: 2 2 21) A cylinder: x y a Set cos( ), sin( ) x a u y a u

2 2 22) A cone: z x y

use the paramter to describe the heightv

cos( ),

sin( )

x v u

y v u

2 2 2x y v

Set z v 2 2 2 2 2

Here:

a z h x y

sin( )cos( )

sin( )sin( )

z cos( )

x

y

Example: 2 2 2 2P arametrize the sphere: x y z a

0

0 2

0

Recall spherical coordinates: , ,

The sphere is described by a

Use the angles as , parameter: u v sin( )cos( ), sin( )sin( ), z cos( )x a u v y a u v a u

, ( , ), ( , ), ( , )u v x u v y u v z u vr

area element in the

plane: uv u v

, , , , ,u vx y z x y z

u u u v v v

r r

span the tangent plane

area of the rectangle with sides and u vu v r r

= u v u vu v u v r r r r

area element is u vd dudv r r

What is the area element?

a surface S is called smooth if and are

linearly indepenedent, i.e. 0

u v

u v

r r

r r

Surface area: ( ) u vS

area S d dudv r r

where , ( , ), ( , ), ( , )u v x u v y u v z u vr

2 2 2 2Compute the surface area of a sphere: .x y z a Example:

, sin( )cos( ), sin( )sin( ), cos( )u v a u v a u v a ur

u v r r cos( )cos( ) cos( )sin( ) sin( )

sin( )sin( ) sin( )cos( ) 0

a u v a u v a u

a u v a u v

i j k

cos( )cos( ), cos( )sin( ), sin( )u a u v a u v a u r

sin( )sin( ), sin( )cos( ), 0v a u v a u v r

2 2 2 2 2 2 2 2sin ( )cos( ) sin ( )sin( ) sin( )cos( )cos ( ) sin( ) cos( )sin ( )a u v a u v a u u v a u u v i j k

2 2 2 2 2sin ( )cos( ) sin ( )sin( ) sin( )cos( )a u v a u v a u u i j k2

u v r r4 4 2 4 4 2 4 2 2sin ( )cos ( ) sin ( )sin ( ) sin ( )cos ( )a u v a u v a u u

4 4 4 2 2sin ( ) sin ( )cos ( )a u a u u 4 2sin ( )a u2 sin( )u v a u r r

, u v

2 2 2 2Surface area of a sphere: .x y z a

( ) u vS

area S d dudv r r

2 sin( )u v a u r r

2

2

0

sin( )o

a u dudv

2

2 2

0

0 0

sin( ) cos( ) 2a u du dv a u

2 21 1 2 4a a

2Area of a sphere of radius is 4a a

34Recall: volume of a sphere of radius is 3

a a

, u v