Green s Theorem Calculus III (MATH 2203)

Transcription

1 Green s Theorem Calculus III (MATH 223) S. F. Ellermeyer November 2, 213 Green s Theorem gives an equality between the line integral of a vector field (either a flow integral or a flux integral) around a simple closed curve,, and the double integral of a function over the region,, enclosed by the curve. A simple closed curve is a loop which does not intersect itself (as pictured below). Theorem 1 (Green s Theorem) Suppose that is a simple closed piecewise smooth curve in 2 and that is the region enclosed by. Suppose 1

2 also that is oriented counterclockwise meaning that is parameterized in such a way that it is traced out counterclockwise by the parameterization. Also suppose that F ( ) ( ) i + ( ) j is a vector fieldsuchthat and have continuous first order partial derivatives throughout and on its boundary curve,. Then F r. Example 2 Let be the unit circle oriented counterclockwise and let F ( ) i + j. We will verify that Green s Theorem holds in this case. First we will compute the flow of F around the boundary of (oriented counterclockwise). To do this we parameterize in the usual way by Thus and This gives r () cos() i +sin() j 2. r () sin () i +cos() j F ( () ()) sin () i +cos() j. F r 2 sin 2 ()+cos 2 () 2. Next we will compute the double integral on the right hand side of Green s Theorem using and. Weobtain (1 ( 1) (Area of the unit circle) 2. 2

3 This shows that F r 2. Remark 3 We can also write the conclusion of Green s Theorem as ( + ) and this form is sometimes more convenient to use (as in the following example). Example 4 Let be the triangle with vertices at ( ), (1 ), and(1 1) oriented counterclockwise and let F ( ) i + j. We will verify that Green s Theorem holds here. 3

4 To compute the line integral, we note that 1 and. The curve (broken down into three pieces) is pictured. On 1 we have which gives us On 2 we have which gives us On 3 we have which gives us 1 (+ ) 1 2 (+ ) (1) () 1 2. (+ ) ( 1 + ) We conclude that ( + )

5 Nextweevaluatethedoubleintegralandseethatwegetthesameresult: 1 ( ) Below is a picture of the triangle shown together with the vector field F ( ) i + j. Example 5 As another example, we use Green s Theorem to evaluate 3 3 where is the positively oriented circle of radius 2. 5

6 Note that the region is the interior of the disk of radius 2 centered at the origin and that the functions 3 and 3 have partial derivatives that are continuous on. By Green s Theorem, In order to evaluate this double integral, it will be convenient to use polar coordinates Therefore Exercise 6 For as given in the previous example, show directly (without using Green s Theorem) that You will see that it is easier to use Green s Theorem than to evaluate the line integral directly. Example 7 In this example we will evaluate the flux integral of F ( ) 2 i + 2 j 6

7 around the (pictured) positively oriented curve that consists of the upper half of the circle of radius 2 centered at the origin followed by the line segment connecting the point ( 2 ) to the point (2 ). Sincewearecomputingtheflux of F across, then the line integral we want to compute is 2 2. This problem was given on Practice Exam 7. We can do it directly but it is easier to use Green s Theorem and evaluate the double integral instead. Notice that since this is a flux integral of F, weneedtobecarefultoapply Green s Theorem correctly. In particular we first rewrite the line integral we want to compute as so, in applying Green s Theorem, we need to take 2 and 2.We 7

8 obtain (2 +2) ( cos ()+ sin ()) 2 2 (cos ()+sin()) (cos ()+sin()) 8 2 (sin () cos ()) (( ( 1)) ( 1)) Remark 8 The preceding example suggests how we can write Green s Theorem in another form. The originally stated form of the theorem, called the flow form, is F r ( + ). For a flux integral, we have F n ( ) ( + ) and Green s Theorem, written in flux form, isthus F n ( + ) ( ) or F n ( ) 8 +.

9 1 Application (Finding Area Using Green s Theorem) We know that if 2,then Area of 1. If we take and to be the functions then Thus and by Green s Theorem Area of Area of ( ) 1 2 ( ) 1 2, ( + ) 1 2 ( ) where is the boundary curve of. This remarkable formula allows us to compute the area of a region using a line integral rather than a double integral. Example 9 We will use the formula Area of 1 2 to compute the area of the ellipse ( )

10 To do this first note that the positively oriented boundary of the ellipse is the curve given by For this curve we have r () () i + () j cos () i + sin () j 2. r () i + j sin () i + cos () j. The area of the ellipse is thus Area of 1 ( ) () () ( cos () cos () sin ()( sin ())) Example 1 Suppose that we have a polygon with vertices at the points ( ), ( 1 1 ), ( 2 2 ),...,( ) where ( )( ) as pictured. 1

11 The area of the region,, enclosed by the polygon is Area of 1 ( ) 2 where is the union of the line segments that make up the sides of. Thus Area of 1 X ( ). 2 1 In order to compute R 1 ( ), wenotethat 1 is the line segment joining the point ( ) to the point ( 1 1 ). We can parameterize this segment as +( 1 ) +( 1 ) 1 and thus

12 This gives us 1 ( ) () () (( +( 1 ) )( 1 ) ( +( 1 ) )( 1 )) ( ( 1 ) ( 1 )) ( 1 1 ) 1 1. By similar reasoning we obtain for all 1 2. Therefore ( ) 1 1 Area of 1 2 X ( 1 1 ). 1 This formula allows us to compute the area of the region enclosed by a polygon using only the coordinates of the vertices of the polygon. Example 11 Let us find the area of the region enclosed by the polygon pictured here. 12

13 Using the formula derived in the previous example we obtain Area of 1 2 5X ( 1 1 ) 1 1 [() () (1) () 2 +(1)(4) (3) () +(3)(4) (4) (4) +(4)(6) (3) (4) +(3)() () (6)] 1 [ ] 2 6. Exercise 12 Find the area of the region enclosed by the polygon pictured here. 13

14 2 Proof of Green s Theorem Rather than prove Green s Theorem in its greatest generality, we will just prove the theorem in the special case that is the positively oriented circle of radius centered at some point ( ) and the vector field F ( ) ( ) i + ( ) j is such that and have continuous first partial derivatives on and in the enclosed region. We will actually prove that and. Adding these two equations together then gives Green s Theorem. To prove the first fact, we note that the circle,, has equation ( ) 2 +( )

15 The upper half of the circle has equation q + 2 ( ) 2 andthelowerhalfofthecirclehasequation q 2 ( ) 2. Therefore + q 2 ( ) 2 + q 2 ( ) 2 + q 2 ( ) 2 q 2 ( ) 2 and + 2 ( ) 2 2 ( ) 2 + q 2 ( ) 2 q 2 ( ) 2 q 2 ( ) 2 + q 2 ( ) 2. This shows that as promised. To prove the second fact, we note that the right half of the circle has equation q + 2 ( ) 2 and the left half of the circle has equation q 2 ( ) 2. 15

16 Therefore and q + 2 ( ) 2 + q + 2 ( ) 2 This shows that We conclude that + 2 ( ) 2 2 ( ) 2 + q 2 ( ) 2 ( + ).. q 2 ( ) 2 q 2 ( ) 2 q 2 ( ) 2 Although the above proof only proves Green s Theorem in a special case, it hints at the general idea of the proof. Exercise 13 Prove Green s Theorem in the special case that is the rectangle [ ] [ ]. 16