Evaluate by Green’s theorem ∮ (cos sin − ) + sin cos y where is the circle x^{2} +y^{2}=1
Answers
Answer:
Let
R
be a region in the
x
y
-plane that is bounded by a closed, piecewise smooth curve
C
,
and let
F
=
P
(
x
,
y
)
i
+
Q
(
x
,
y
)
j
be a continuous vector function with continuous first partial derivatives
∂
P
∂
y
,
∂
Q
∂
x
in a some domain containing
R
.
Then Green’s theorem states that
∬
R
(
∂
Q
∂
x
−
∂
P
∂
y
)
d
x
d
y
=
∮
C
P
d
x
+
Q
d
y
,
where the symbol
∮
C
indicates that the curve (contour)
C
is closed and integration is performed counterclockwise around this curve.
If
Q
=
x
,
P
=
−
y
,
Green’s formula yields:
S
=
∬
R
d
x
d
y
=
1
2
∮
C
x
d
y
−
y
d
x
,
where
S
is the area of the region
R
bounded by the contour
C
.
We can also write Green’s Theorem in vector form. For this we introduce the so-called curl of a vector field. Let
F
=
P
(
x
,
y
,
z
)
i
+
Q
(
x
,
y
,
z
)
j
+
R
(
x
,
y
,
z
)
k
be a vector field. Then the curl of the vector field
F
is called the vector denoted by
rot
F
or
∇
×
F
, which is equal to
rot
F
=
∇
×
F
=
∣
∣
∣
∣
∣
i
j
k
∂
∂
x
∂
∂
y
∂
∂
z
P
Q
R
∣
∣
∣
∣
∣
=
(
∂
R
∂
y
−
∂
Q
∂
z
)
i
+
(
∂
P
∂
z
−
∂
R
∂
x
)
j
+
(
∂
Q
∂
x
−
∂
P
∂
y
)
k
.
In terms of curl, Green’s Theorem can be written as
∬
R
(
rot
F
)
⋅
k
d
x
d
y
=
∮
C
F
⋅
d
r
.
Note that Green’s Theorem is simply Stoke’s Theorem applied to a
2
-dimensional plane.
Solved Problems
Click or tap a problem to see the solution.
Example 1
Using Green’s theorem, evaluate the line integral
∮
C
x
y
d
x
+
(
x
+
y
)
d
y
,
where
C
is the curve bounding the unit disk
R
.
Example 2
Using Green’s formula, evaluate the line integral
∮
C
(
x
−
y
)
d
x
+
(
x
+
y
)
d
y
,
where
C
is the circle
x
2
+
y
2
=
a
2
.
Example 3
Using Green’s theorem, calculate the integral
∮
C
x
2
y
d
x
−
x
y
2
d
y
.
The curve
C
is the circle
x
2
+
y
2
=
a
2
(Figure
1
), traversed in the counterclockwise direction.
Example 4
Using Green’s formula, evaluate the integral
∮
C
(
x
+
y
)
d
x
−
(
x
−
y
)
d
y
,
where the curve
C
is the ellipse
x
2
a
2
+
y
2
b
2
=
1
(Figure
2
).
Example 5
Using Green’s formula, calculate the line integral
∮
C
y
2
d
x
+
(
x
+
y
)
2
d
y
,
where the contour
C
is the triangle
A
B
D
with vertices
A
(
a
,
0
)
,
B
(
a
,
a
)
,
D
(
0
,
a
)
(Figure
3
).
Example 6
Using Green’s theorem, evaluate the line integral
∮
C
(
y
−
x
2
)
d
x
−
(
x
+
y
2
)
d
y
,
where the contour
C
encloses the sector of the circle with radius
a
lying in the first quadrant (Figure
4
).
Example 7
Calculate the integral
∮
C
d
x
−
d
y
x
+
y
using Green’s theorem. The contour
C
is the boundary of the square with the vertices
A
(
1
,
0
)
,
B
(
0
,
1
)
,
D
(
−
1
,
0
)
,
E
(
0
,
−
1
)
(Figure
5
).
Example 8
Calculate the line integral
∮
C
√
x
2
+
y
2
d
x
+
y
[
x
y
+
ln
(
x
+
√
x
2
+
y
2
)
]
d
y
using Green’s theorem. The contour of integration
C
is the circle
x
2
+
y
2
=
a
2
(Figure
7
).
Example 9
Calculate the area of the region
R
bounded by the astroid
x
=
a
cos
3
t
,
y
=
a
sin
3
t
,
0
≤
t
≤
2
π
.
Example 1.
Using Green’s theorem, evaluate the line integral
∮
C
x
y
d
x
+
(
x
+
y
)
d
y
,
where
C
is the curve bounding the unit disk
R
.
Solution.
The components of the vector field are
P
(
x
,
y
)
=
x
y
,
Q
(
x
,
y
)
=
x
+
y
.
Using the Green’s formula
∬
R
(
∂
Q
∂
x
−
∂
P
∂
y
)
d
x
d
y
=
∮
C
P
d
x
+
Q
d
y
we transform the line integral into the double integral:
I
=
∮
C
x
y
d
x
+
(
x
+
y
)
d
y
=
∬
R
(
∂
(
x
+
y
)
∂
x
−
∂
(
x
y
)
∂
y
)
d
x
d
y
=
∬
R
(
1
−
x
)
d
x
d
y
.
Converting the double integral into polar coordinates, we have
I
=
∫
R
(
1
−
x
)
d
x
d
y
=
2
π
∫
0
1
∫
0
(
1
−
r
cos
θ
)
r
d
r
d
θ
=
2
π
∫
0
⎡
⎢
⎣
1
∫
0
(
r
−
r
2
cos
θ
)
d
r
⎤
⎥
⎦
d
θ
=
2
π
∫
0
[
(
r
2
2
−
r
3
3
cos
θ
)
∣
∣
∣
1
r
=
0
]
d
θ
=
2
π
∫
0
(
1
2
−
cos
θ
3
)
d
θ
=
(
θ
2
−
sin
θ
3
)
∣
∣
∣
2
π
0
=
π
.
Example 2.
Using Green’s formula, evaluate the line integral
∮
C
(
x
−
y
)
d
x
+
(
x
+
y
)
d
y
,
where
C
is the circle
x
2
+
y
2
=
a
2
.
Solution.
First we identify the components of the vector field:
P
=
x
−
y
,
Q
=
x
+
y
and find the partial derivatives:
∂
Q
∂
x
=
∂
(
x
+
y
)
∂
x
=
1
,
∂
P
∂
y
=
∂
(
x
−
y
)
∂
x
=
−
1.
Hence, the line integral can be written in the form
I
=
∮
C
(
x
−
y
)
d
x
+
(
x
+
y
)
d
y
=
∬
R
(
1
−
(
−
1
)
)
d
x
d
y
=
2
∬
R
d
x
d
y
.
In the last expression the double integral
∬
R
d
x
d
y
is equal numerically to the area of the disk
x
2
+
y
2
=
a
2
,
which is
π
a
2
.
Then the integral is
I
=
2
∬
R
d
x
d
y
=
2
π
a
2
.
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Problems 1-2