Green s Theorem. MATH 311, Calculus III. J. Robert Buchanan. Fall Department of Mathematics. J. Robert Buchanan Green s Theorem

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1 Green s Theorem MATH 311, alculus III J. obert Buchanan Department of Mathematics Fall 2011

2 Main Idea Main idea: the line integral around a positively oriented, simple closed curve is related to a double integral over the region enclosed by the curve.

3 Main Idea Main idea: the line integral around a positively oriented, simple closed curve is related to a double integral over the region enclosed by the curve. Definition A curve in the plane = {(x, y) x = f (t), y = g(t), a t b} is closed if its two endpoints are the same, i.e. if (f (a), g(a)) = (f (b), g(b)). A closed curve is simple if it does not intersect itself, except at the endpoints.

4 Illustration x x losed (not simple) losed and simple

5 Orientation y y Definition A simple closed curve has positive orientation if the region enclosed by the curve stays to the left of as the curve is traversed. A curve has negative orientation if the region stays to the right of. Positive orientation Negative orientation

6 Green s Theorem Notation: a line integral of a vector field around a positively oriented simple closed curve will be denoted F(x, y) dr.

7 Green s Theorem Notation: a line integral of a vector field around a positively oriented simple closed curve will be denoted F(x, y) dr. Theorem Let is a piecewise-smooth, simple closed curve in the plane with positive orientation and let be the region enclosed by, together with. Suppose that M(x, y) and N(x, y) are continuous and have continuous first partial derivatives in some open region D, with D. Then ( N M(x, y) dx + N(x, y) dy = x M ) da. y

8 Proof of Green s Theorem (1 of 3) Suppose = {(x, y) a x b and g 1 (x) y g 2 (x)}. 2 1 y x

9 Proof of Green s Theorem (2 of 3) onsider the path and components: = = {(x, y) a x b and y = g 1 (x)} 2 = {(x, y) a x b and y = g 2 (x)}. Then M(x, y) dx = = = = M(x, y) dx + M(x, y) dx 1 2 b a b M(x, g 1 (x)) dx a b g1 (x) a b a M(x, g 2 (x)) dx [M(x, g 1 (x)) M(x, g 2 (x))] dx g 2 (x) M y M dy dx = y da.

10 Proof of Green s Theorem (3 of 3) If we can also describe = {(x, y) c y d and h 1 (y) x h 2 (y)}, then in a similar way we can show N(x, y) dy = Putting these two results together gives M(x, y) dx + N(x, y) dy = N x da. ( N x M ) da. y

11 Examples (1 of 6) Evaluate the line integral x 4 dx + xy dy where is the boundary of the triangle with vertices at (0, 0), (1, 0), and (0, 1).

12 Graph of and 1.0 y

13 Examples (2 of 6) x 4 dx + xy dy = = = = x (xy) y (x 4 ) da y da 1 1 y y y 2 dy y dx dy = 1 6

14 Examples (3 of 6) Evaluate the line integral (3y e sin x ) dx + (7x + y ) dy where is the boundary of the disk of radius 3 centered at the origin.

15 Examples (4 of 6) (3y e sin x ) dx + (7x + y ) dy = x (7x + 4 y 4 + 1) y (3y esin x ) da = 7 3 da = 4 1 da = 4(π3 2 ) = 36π

16 Examples (5 of 6) Evaluate the line integral (y 2 tan 1 x) dx + (3x + sin y) dy where is the boundary of the region enclosed by y = x 2 and y = 4.

17 Graph of and 4 y

18 Examples (6 of 6) (y 2 tan 1 x) dx + (3x + sin y) dy = (3x + sin y) x y (y 2 tan 1 x) da = 3 2y da = = = 96 5 x 2 (3 2y) dy dx x 4 3x 2 4 dx

19 Finding the Area of a egion If is the region enclosed by the positively oriented simple closed curve then the area of is A = 1 da = 0 dx + x dy = x dy. This can be generalized to the following theorem.

20 Finding the Area of a egion If is the region enclosed by the positively oriented simple closed curve then the area of is A = 1 da = 0 dx + x dy = x dy. This can be generalized to the following theorem. Theorem If is a region in the xy-plane which is bounded by a piecewise-smooth simple closed curve, the area of region is A = x dy = y dx = 1 x dy y dx 2

21 Example (1 of 2) Find the area enclosed by the ellipse x 2 a 2 + y 2 b 2 = 1.

22 Example (2 of 2) If the boundary of the ellipse is parameterized as x = a cos t y = b sin t for 0 y 2π, then A = = 2π x dy 0 2π = ab = ab 2 = abπ. (a cos t)(b cos t) dt 0 2π 0 cos 2 t dt (1 + cos 2t) dt

23 Decomposing a egion (1 of 2) Notation: if is a region in the xy-plane, the boundary curve of the region oriented positively will be denoted.

24 Decomposing a egion (2 of 2) By cutting the region we may apply Green s Theorem to each piece

25 Extensions to Green s Theorem If is the region between two positively oriented curves 1 and 2 then ( N x M ) da = M(x, y) dx + N(x, y) dy y 1 + M(x, y) dx + N(x, y) dy 2

26 Example (1 of 6) Evaluate the line integral y 2 dx + 3xy dy where is the boundary of the shaded region shown below y x

27 Example (2 of 6) is the upper half of an annulus with inner radius 1 and outer radius 2. y 2 dx + 3xy dy = x (3xy) y (y 2 ) da = (3y 2y) da = = π 2 0 π 0 = (r sin θ)r dr dθ 7 sin θ dθ 3

28 Example (3 of 6) 1 Suppose F(x, y) = x 2 y, x and show that + y 2 F(x, y) dr = 2π for every simple closed curve that encloses the origin.

29 Example (4 of 6) If is a simple closed curve containing the origin then there is a circle of radius a > 0 centered at the origin interior to. Let 1 be the positively oriented boundary of the circle y x

30 Example (5 of 6) Thus F(x, y) dr F(x, y) dr 1 = F(x, y) dr [ N = x M ] da y [ x 2 + y 2 2x 2 = (x 2 + y 2 ) 2 x 2 y 2 + 2y 2 ] (x 2 + y 2 ) 2 = 0 da = 0 F(x, y) dr = F(x, y) dr. 1 da

31 Example (6 of 6) Thus we may evaluate the line integral around 1 instead of around. F(x, y) dr 1 = 1 y, x dr 1 a2 = 1 a 2 1 y dx + x dy = 1 2π a 2 ( a sin t)( a sin t) + (a cos t)(a cos t) dt = 0 2π 0 = 2π 1 dt

32 Homework ead Section Exercises: 1 29 odd