Download Green`s Theorem. Curl and Divergence

Calculus 3
Lia Vas
Green’s Theorem. Curl and Divergence
Green’s Theorem.
Let C be a smooth curve ~r(t) = (x(t), y(t))
with endpoints ~r(a) = (x(a), y(a)) and ~r(b) =
(x(b), y(b)). A curve is called closed if ~r(a) =
~r(b). In this case, we say that C is positive
oriented if it is traversed single time in counterclockwise
orientation. If C is a closedR curve,
H
notation C is sometimes used instead of C .
Let C be a positive oriented, smooth, closed curve and P and Q be functions of x and y with
continuous derivatives.
H
In this case, the line integral C P dx+Qdy can
be reduced to a double integral over the interior
D of C.
I
P dx + Qdy =
Z Z
D
C
Z Z
D
∂Q ∂P
−
∂x
∂y
!
dxdy =
(Qx − Py )dxdy
This statement is known as Green’s Theorem.
In many cases it is easier to evaluate the line integral using Green’s Theorem than directly. The
integrals in practice problem 1. below are good examples of this situation.
Curl and Divergence. Curl and divergence are two operators that play an important role in
electricity and magnetism. Also, in chemistry and physics Green’s theorem is frequently encountered
in vector forms involving curl and divergence operators.
Curl and divergence are related to the gradient operator. Recall that the gradient operator ∇ is
defined as
∂ ∂
∂
∂~
∂~
∂
,
i = ~i +
j +
k.
∇=h ,
∂x ∂y ∂z
∂x
∂y
∂z
Let f~ = hP, Q, Ri be a vector field such that P, Q and R are differentiable for all their variables.
Divergence of f~ is defined as the scalar product of ∇ and f~ and curl of f~ is defined as the vector
product of ∇ and f~. Thus,
∂P
∂Q
∂R
div f~ = ∇ · f~ =
+
+
∂x
∂y
∂z
and
~i
!
!
!
~j ~k ∂R
∂P
∂Q
∂Q
∂R
∂P
∂
∂
∂
~k
~i +
~j +
curl f~ = ∇ × f~ = ∂x
=
−
−
−
∂y
∂z ∂y
∂z
∂z
∂x
∂x
∂y
P Q R 1
Note that div f~ = ∇ · f~ is a scalar function while curl f~ = ∇ × f~ is a vector function.
Vector Forms of Green’s Theorem. Let C be a positive oriented, smooth closed curve and
~
f = hP, Q, 0i a vector function such that P and Q have continuous derivatives. Using curl, the
Green’s Theorem can be written in the following vector form
I
P dx + Qdy =
C
I
f~ · d~r =
C
Z Z
curlf~ · ~k dxdy.
D
Sometimes the integral C P dy − Qdx is considered instead of the integral
case,
I
I
Z Z
~
P dy − Qdx =
f · ~n ds =
divf~ dxdy
H
C
C
H
C
P dx + Qdy. In this
D
where ~n is the unit normal vector to C at point (x, y). The product ~nds is hy 0 (t), −x0 (t), 0i.
Practice problems.
1. Evaluate the following integrals using Green’s theorem. Then compute them without using
Green’s Theorem.
a)
H
b)
H
C
x4 dx + xydy where C is the triangle with vertices (0, 0), (0, 1), and (1, 0).
C
xy 2 dx + x3 dy where C is the rectangle with vertices (0, 0), (2, 0), (2, 3), and (0, 3).
2. Evaluate the following integrals using Green’s theorem.
y 2 dx + 3xydy where C is the boundary of the region between the circles x2 + y 2 = 1
and x2 + y 2 = 4 above x-axis.
a)
H
b)
H
c)
xydx + 2x2 dy where C is the line segment from (-2,0) to (2,0) and the upper half of the
circle x2 + y 2 = 4.
C
C
ey dx + 2xey dy where C is the square with vertices (0, 0), (1, 0), (1, 1), and (0, 1).
H
C
3. Use Green’s Theorem to find the work done by the force f~(x, y) = x(x + y)~i + xy 2~j in moving
a particle along the triangle with vertices (0,0), (1,0) and (0,1) starting at the origin.
4. Find curl and divergence of the following vector fields.
a) f~ = hxz, xyz, −y 2 i
b) f~ = hex sin y, ex cos y, zi
i
c) f~ = h xz , yz , −1
z
Solutions.
1. Ha) Using Green’s Theorem:
Let P = x4 and Q = xy so that Py = 0 and Qx = y. Then
R
R
4
C x dx + xydy =
D (y − 0)dxdy where D is the interior of the triangle. The bounds are
R
R
R
3
2
0 ≤ x ≤ 1 and 0 ≤ y ≤ 1 − x. So, the integral is 01 dx 01−x ydy = 01 (1−x)
dx = −(1−x)
|10 = 16 .
2
6
Without Green’s Theorem, you have to evaluate three line integrals because each side of the
triangle has a different parametrization. Let
C1 be the side Rconnecting (0,0) and (1,0). Then
R
x = x, and y = 0, dy = 0 and 0 ≤ x ≤ 1. C1 x4 dx + xydy = 01 x4 dx + 0 = 15 .
Let C2 be the
side connectingR (1,0) and (0,1). Then x = x, and y = 1 − x, dy = −dx and
R
1 ≤ x ≤ 0. C2 x4 dx + xydy = 10 x4 dx − x(1 − x)dx = −1
+ 12 − 13 = −1
.
5
30
2
Let C3 be the side Rconnecting (0,1) and (0,0). Then x = 0, and y = y, dx = 0 and 1 ≤ y ≤ 0.
0
1
1
1
4
C3 x dx + xydy = 1 0 + 0 = 0. So, the sum of the three integrals is 5 − 30 = 6 .
R
b)
Using Green’s Theorem:
Let P = xy 2 and Q = x3 so that Py = 2xy and Qx = 3x2 . Then
H
R
R
2
3
2
interior of the rectangle.
The bounds
C xy dx + x dy =
D (3x − 2xy)dxdy where D isR the
R2
2R3
2
2
are 0 ≤ x ≤ 2 and 0 ≤ y ≤ 3. So, the integral is 0 0 (3x − 2xy)dxdy = 0 (9x − 9x)dx =
24 − 18 = 6.
Without Green’s Theorem, you have to evaluate four line integrals because each side of the
rectangle has a different parametrization.
Let
C1 be the side connecting
(0,0) and (2,0). Then x = x, and y = 0, dy = 0 and 0 ≤ x ≤ 2.
R
R2
2
3
C1 xy dx + x dy = 0 0 + 0 = 0.
Let C2 be the side connecting
(2,0) and (2,3). Then x = 2, and y = y, dx = 0 and 0 ≤ y ≤ 3.
R3
2
3
xy
dx
+
x
dy
=
8dy
=
24.
C2
0
R
Let
C3 be the side connecting
(2,3) and (0,3). Then x = x, and y = 3, dy = 0 and 2 ≤ x ≤ 0.
R
R0
2
3
C3 xy dx + x dy = 2 9xdx + 0 = −18.
Let C4 be the side connecting
(0,3) and (0,0). Then x = 0, and y = y, dx = 0 and 3 ≤ y ≤ 0.
R0
2
3
xy
dx
+
x
dy
=
0
+
0
=
0.
So, the sum of the four integrals is 0 + 24 − 18 + 0 = 6.
C4
3
R
2. a) Let P = y 2 and Q = 3xy so that Py = 2y and Qx = 3y. Then C y 2 dx + 3xydy = D (3y −
2y)dxdy where D is the interior of the
region bounded
by C. Using polar
coordinates,
0≤θ≤π
R
R
R R
RR
and 1 ≤ r ≤ 2. So, the integral is D ydxdy = 0π 12 r sin θrdrdθ = 0π sin θdθ 12 r2 dr = 2 73 =
14
.
3
H
RR
y
b)
Let P = ey and Q
= 2xe
so that Py = ey and Qx = 2ey . Then
RR
R1
R1 y
y
y
1
D (2e − e )dxdy = 0 dx 0 e dy = e − 1 = 1.718.
H
C
ey dx + 2xey dy =
c) Let P =R xy
and Q = 2x2 so that Py = x and Qx = 4x. Then C xydx + 2x2 dy =
(4x −
R
x)dxdy
=
3xdxdy.R Using polar
coordinates, 0 ≤ θ ≤ π and 0 ≤ r ≤ 2. The integral is
Rπ R2
R2 2
π
3r
cos
θrdrdθ
=
cos
θdθ
3r
dr = 0(8) = 0.
0 0
0
0
H
3. The work is W = C f~d~r = C x(x + y)dx + xy 2 dy =
R 1 (1−x)3
1
− x(1 − x))dx = 12
− 12 + 13 = −1
.
0(
3
12
R
R
RR
D (y
2
RR
− x)dxdy =
R 1 R 1−x
0
0
(y 2 − x)dxdy =
4. a) Let P = xz, Q = xyz, and R = −y 2 . Then divf~ = Px + Qy + Rz = z + xz and curlf~ =
~i
~k ~j
∂
∂
∂
∂x
= h−y(x + 2), x, yzi.
∂y
∂z xz xyz −y 2 b) Let P = ex sin y, Q = ex cos y, and R = z. Then divf~ = Px +Qy +Rz = ex sin y−ex sin y+1 =
~k ~i
~j
∂
∂ ∂
1 and curlf~ = ∂x
= h0, 0, 0i.
∂y
∂z ex sin y ex cos y z . Then divf~ = Px + Qy + Rz =
c) Let P = xz , Q = yz , and R = −1
z
~i
~j ~k y
∂
∂
∂ curlf~ = ∂x
= h z 2 , −x
, 0i.
z2
∂y
∂z
x
z
y
z
−1
z
3
1
z
+
1
z
+
1
z2
=
2
z
+
1
,
z2
and