49. Green s Theorem. The following table will help you plan your calculation accordingly. C is a simple closed loop 0 Use Green s Theorem

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1 49. Green s Theorem Let F(x, y) = M(x, y), N(x, y) be a vector field in, and suppose is a path that starts and ends at the same point such that it does not cross itself. Such a path is called a simple closed loop, and it will enclose a region. Assume M and N and its first partial derivatives are defined within including its boundary. Furthermore, the path is to be traversed (circulated) in a counterclockwise direction, called the positive orientation. If these conditions are met, then the line integral around the simple loop path may be evaluated by a double integral. This is called Green s Theorem, and is written = (N x M y ) da. If F is a conservative vector field, then M y = N x, so that the integrand N x M y =. Thus, in a conservative vector field, all line integrals along a simple closed loop path evaluate to. In a physical sense, there is no net circulation around the loop, and a conservative vector field is often called a rotation-free (or irrotational) vector field. When calculating a line integral, you should check two things: Is the vector field conservative? Is the path a simple closed loop? The following table will help you plan your calculation accordingly. F is conservative F is not conservative is a simple closed loop Use Green s Theorem is not a loop of any kind (it has different start and end points). Find the potential function φ(x, y) and calculate the line integral by the Fundamental Theorem of Line Integrals (The FTLI) Parameterize the path(s) in variable t, and calculate the line integral directly. 8

2 Example 49.: Evaluate, where F(x, y) = y, 4x and is a triangle, traversed from (,) to (,) to (,4) back to (,). Solution: We sketch and verify that it is a simple closed loop that is traversed counter-clockwise: To evaluate as a sequence of line integrals, we would need to divide the path into three smaller paths: being the line from (,) to (,), being the line from (,) to (,4), and being the line from (,4) to (,). For, we have r (t) = t, with t, so that r (t) =, and F(t) =,8t. Thus, in this case, F dr =,8t, =, so that the vector field elements are orthogonal to the segment. =. In the above image, note that For, we have r (t) =,4t with t, so that r (t) =,4 and F(t) = 4t, 8. Thus, F dr = 4t, 8,4 =, so that elements agree with the direction of. = dt = [t] =. The vector field For, we have r (t) = t, 4 4t with t, so that r (t) =, 4 and F(t) = 4 4t, 8 8t. Thus, F dr = 4 4t, 8 8t, 4 = 4t 4, so that = (4t 4) dt = [t 4t] =. The vector field elements disagree (point against) the direction of. Since = + +, we have = + =. 8

3 Now, let s use Green s Theorem. We find that N x M y = 4 =, so that = da = da = (4) =. In this case, the constant integrand was moved to the front, leaving da, which is the area of region. Using geometry, the area of is that of a triangle with base and height 4, so ()(4) = 4. da = Example 49.: Evaluate, where F(x, y) = xy, x and traverses from (,) to (,) along a semi-circle of radius, centered at the origin, in the counter-clockwise direction, then from (,) back to (,) along a straight line. Solution: Path is a simple closed loop traversed in a counter-clockwise direction. To find, we use Green s Theorem. Since the region is a semicircle of radius, we will evaluate the double integral using polar coordinates. = (N x M y ) = ( x) π 8 da da = ( r cos θ) r dr dθ π = (r r cos θ) dr dθ.

4 The inside integral is evaluated with respect to r: (r r cos θ) dr This is then integrated with respect to θ: = [ r r cos θ] = 6 cos θ. π ( 6 cos θ) dθ = [θ 6 sin θ] = π. π Thus, the line integral along is induced by the vector field. F dr = π. There is positive circulation along this path Example 49.: Evaluate, where F(x, y) = y, x + y and traverses a rectangle from (,) to (,6) to (7,6) to (7,) back to (,). Solution: A sketch of the path shows it to be a simple closed loop traversed in a clockwise direction. In order to use Green s Theorem, we would traverse it in the counter-clockwise direction, which is equivalent to traversing each segment in its opposite direction. This means that we will multiply our result by in order to account for this opposite direction. Using Green s Theorem, we have N x M y = + ( ) = 4: (N x M y ) da = ( 4) da = 4 dy dx The double integral dy dx is the area of the rectangle, which is (6)(5) =. Thus, (N x M y ) da = 4() =. However, since was traversed in the opposite direction, we negate this result. We have =. 84

5 Example 49.4: Evaluate, where F(x, y) = 5x 4 + y, yx and is an ellipse with major axis of along the x-axis, and minor axis of 8 along the y-axis, in a counter-clockwise direction. Solution: Using Green s Theorem, we have (N x M y ) da = (y y) da = da =. Note that F is conservative, since M y = N x. There is no need to parameterize the ellipse. Green s Theorem can be used to find the line integral of a non-loop path. We close off the path forming a loop, as this next example shows: Example 49.5: Evaluate, where F(x, y) = y, x and is a sequence of line segments from (,) to (,) to (,4) to ( 4,4). Solution: The path is not a simple closed loop. Thus, we would have to parametrize each line segment one at a time and determine the value of each line integral individually. Instead, we can add in the final line segment, that from ( 4,4) to (,), thus creating a simple closed loop traversed counter-clockwise. Using Green s Theorem, we have (N x M y ) 4 da = (x ) y dx 4 dy = (x ) y dx dy. 85

6 The inside integral is first evaluated: (x ) y dx = [ x x] y = ( 9 ) ( y + y) This result is then integrated with respect to y: 4 ( y y + ) dy = y y +. = [ 4 y y + y] = 76. We now need to evaluate the line integral from ( 4,4) to (,), the segment that we added in to form the closed loop. We have r(t) = 4 + 4t, 4 4t, where t. Thus, r (t) = 4, 4 and F(t) = (4 4t), ( 4 + 4t), which simplifies to F(t) = 8 8t, 6t t + 6. Along this path segment, we have = 8 8t, 6t t + 6 4, 4 dt = ( 64t + 96t ) dt = [ 64 t + 48t t] = 6. Therefore, the line integral from (,) to (,) to (,4) to ( 4,4) is the value we found from Green s Theorem, 76, subtracted by the value of the line integral along the segment we used to close off the region, 6. We then have = 76 6 ( ) = 6 =. 86

7 As a check, here are the individual line integrals along the three line segments: From (,) to (,): We have r(t) = t, with t. Thus, r (t) =, and F(t) =,9t, and therefore F dr =,9t, dt = dt =. From (,) to (,4): We have r(t) =,4t, with t. Thus, r (t) =,4 and F(t) = 8t, 9, and therefore F dr = 8t, 9,4 dt = 6 dt = [6t] = 6. From (,4) to ( 4,4): We have r(t) = 7t, 4, with t. Thus, r (t) = 7, and F(t) = 8, ( 7t), and therefore, 56 dt = [ 56t] = 56. F dr = 8, ( 7t) 7, dt = The sum is =, which agrees with our earlier answer. If is a simple closed loop, then the region bounded by is simply connected. All of the regions in the preceding examples in this section are simply connected. In a very intuitive sense, a simply-connected region in the plane has no holes. Green s Theorem requires a simply connected region. However, a non-simply connected region can be made into two (or more) simply-connected regions by dividing the region carefully. In the above image, a non-simply connected region is strategically divided into two subregions, A and B, that are each simply connected. Notice that the counter-clockwise circulation is preserved in both cases. The line integrals along the two cuts will cancel, since the flow is in opposite directions depending on whether A or B is being considered. Green s Theorem can then be applied to each subregion, and often combined into one double integral covering the entire region. 87

8 Example 49.6: Evaluate, where F(x, y) = e x + y, x sin y and is the boundary of a region enclosed by two concentric circles, centered at the origin, one of radius 5 and the other of radius. Assume the circulation in the outer circle is counter-clockwise, and that the circulation on the inner circle is clockwise. Solution: The region and its boundary are shown below. Using Green s Theorem, we have N x M y = =. Using polar coordinates, the region can be defined as r 5 and θ π. Thus, π 5 = r dr dθ π 5 = [ r ] dθ π = 8 dθ = 8(π) = 6π. ([ r 5 ] = 8) π 5 Note that r dr dθ is the area of represented as a double integral, so we can verify using geometry. The area inside a circle of radius 5 is 5π, and the area inside a circle of radius is 9π, and their difference is 6π. 9-6 Scott Surgent (surgent@asu.edu) 88