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YOU WILL LEARN ABOUT:

How to use Green’s Theorem to calculate a line integral, which may represent a physical quantity such as the area of a heart-shaped island.

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If D is a simply connected region that is bounded by a positively oriented, piecewise smooth closed curve C, and P and Q have continuous first-order partial derivatives on an open region containing D, then
∫CP dx+Q dy =∬D∂Q∂x-∂P∂ydA.

In the special case where P=-y2 and Q=-x2 , then Green’s Theorem gives

1 2 ∫ C x d y - y d x = ∬ D d A = Area of D

Green’s Theorem is important in many applications, such as the following.

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A computer engineer can use Green’s Theorem to find the center of mass of a platter being used to build a hard drive.
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An ecologist can use Green’s Theorem to find the area of a lake whose shoreline is described by a pair of parametric equations.
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An air conditioning repair person can use Green’s Theorem to determine what the heat flux through a duct should be, in order to determine if heat is flowing as it should.

GREEN'S THEOREM
Shape Parameter, a
1
A= 1 2∫ c xdy−ydx=
∫ 0 4 π 3 cos 2( 5 4t)dt=6π
Without Using Green's Theorem
Using Green's Theorem
∫ C − 2 y 2 d x + 4 x y d y =
0.00
0.00
42.67
0.00
∫ C − 2 y 2 d x + 4 x y d y =
∫ C 1 − 2 y 2 d x + 4 x y d y
+
∫ C 2 − 2 y 2 d x + 4 x y d y
+
∫ C 3 − 2 y 2 d x + 4 x y d y
+
∫ C 4 − 2 y 2 d x + 4 x y d y
=
∫ 0 0.0 0 d t
=
+
∫ 1 1.7 ( 32 π cos 2 ( π ( t − 1 ) ) sin ( π ( t − 1 ) ) + 16 π sin 3 ( π ( t − 1 ) ) ) d t
=
+
∫ 2 0.0 0 d t
=
+
∫ 3 3.7 ( − 4 π cos 2 ( π ( t − 3 ) ) sin ( π ( t − 3 ) ) − 2 π sin 3 ( π ( t − 3 ) ) ) d t
=
∫ C − 2 y 2 d x + 4 x y d y =
∫ ∫ D 8 y d A
=
∫ 0 π ∫ 1 2 ( 8 r sin θ ) r d r d θ
=
37.33
Without Using Green's Theorem
Using Green's Theorem
Work =
0
Work =
∫ C sin x d x+x cos ( π y )d y
=
∫ C 1 sin x d x + x cos ( π y ) d y
+
∫ C 2 sin x d x + x cos( π y ) d y
+
∫ C 3 sinx d x + x cos( π y ) d y
=
∫ 0 0 sin ( t ) d t
=
0
Work =
∫ C sin x d x + x cos ( π y ) d y
=
∫ ∫ D cos ( π y ) d A
=
∫ 0 1 ∫ 0 1 − x cos( π y ) d y d x
=
0.202642
Equation of the line tangent to the graph of (x)  : y = (x − )

INSTRUCTIONS

One application of Green’s Theorem is to find the area of a plane region. The shapes on the left are given by
C = { ( x , y ) : x ( t ) = ( − a − 2 )cos t
+ 2cos ( ( a 2 + 1 ) t ) , y ( t ) = ( − a − 2 )sint
− 2sin ( ( a 2 + 1 ) t ) } , where a = 1 , 2 , 4 , 6 , 8 , 10 is a “shape parameter.” Move the “Shape parameter” slider to find the area of various shapes and answer each of the following questions.

PRACTICE QUESTIONS

1 2 3
1. Let n represent the number of curved edges on each shape. If we did not have Green’s Theorem to determine the area of the shape on the left with n sides, we could approximate the area by taking advantage of the formula from geometry for the area of a regular polygon with n sides. Two possible ways we could do this would be to inscribe the shape with n curved edges into a regular polygon with the n sides or to inscribe a regular polygon with n sides into the shape with n curved edges. How would each of these approximations compare to the actual area?
2. Examine the areas of the shapes in the graph when the “Shape parameter” is 2,4,6,8, and 10. Based on this pattern, what would the next area of the shape with a “Shape parameter” of 12 be?
3. Let a represent the “Shape parameter” and
n represent the number of curved edges on each shape. For example, when a=6, n=5. What is general formula for the area of these shapes?

INSTRUCTIONS

One advantage of using Green’s Theorem to evaluate a line integral is it can greatly simplify the calculation. Select whether or not to use Green’s Theorem to evaluate the line integral. If you choose not to use Green’s Theorem, move the “Time” slider to answer each of the following questions.

In this example, C 1, C 2, C 3, and C 4 are parameterized as follows.
C 1 = { ( x , y ) :
x ( t ) = t + 1 ,
y ( t ) = 0 , 0 ≤ t ≤ 1 }
C 2 = { ( x , y ) :
x ( t ) = 2 cos ( π ( t − 1 ) ) ,
y ( t ) = 2 sin ( π ( t − 1 ) ) , 1 ≤ t ≤ 2 }
C 3 = { ( x , y ) :
x ( t ) = t − 4 ,
y ( t ) = 0 , 2 ≤ t ≤ 3 }
C 4 = { ( x , y ) :
x ( t ) = - cos ( π ( t − 3 ) ) ,
y ( t ) = sin ( π ( t − 3 ) ) , 3 ≤ t ≤ 4 }

PRACTICE QUESTIONS

1 2 3
1. What, if anything, can we say about the value of the line integral along just the top arc of the annulus?
2. What, if anything, can we say about the value of the line integral along just the bottom arc of the annulus?
3. Which of the following parameterizations of C 1, C 2, C 3, and C 4 is equivalent to our parameterization of these curves?

INSTRUCTIONS

In this example, we will look at the work done by the force
F → ( x , y ) =
sin x i ⇀
+
x cos ( π y ) j ⇀
on a particle moving around the triangle consisting of the parameterized lines
C 1 = { ( x , y ) :
x ( t ) = t ,
y ( t ) = 0 , 0 ≤ t ≤ 1 }
C 2 = { ( x , y ) :
x ( t ) = 2 - t ,
y ( t ) = t - 1 , 1 ≤ t ≤ 2 }
C 3 = { ( x , y ) :
x ( t ) = 0 ,
y ( t ) = 3 - t , 2 ≤ t ≤ 3 }
Select whether or not to use Green’s Theorem to calculate the work. If you choose not to use Green’s Theorem, move the “Time” slider to answer each of the following questions.

PRACTICE QUESTIONS

1 2 3
1. Without using the Fundamental Theorem for Line Integrals, what technique(s) of integration will be used to compute the work along C 2? Select all that apply.
2. How much work is done by F →along C 1?
3. How much work is done by F →along C 2?
Result

x can be any number from to