16.4 Green’s Theorem - University of Pennsylvaniawziller/math114s14/Ch16-4-5.pdfBut we can also...
Transcript of 16.4 Green’s Theorem - University of Pennsylvaniawziller/math114s14/Ch16-4-5.pdfBut we can also...
-
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