is the two-dimensional curl of the vector field F = P, Q. Suppose D is described by a x b and f(x) y g(x) (aka a type I region).

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1 Math 55 - Vector alculus II Notes 4.4 Green s Theorem We begin with Green s Theorem: Let be a positivel oriented (parameterized counterclockwise) piecewise smooth closed simple curve in R and be the region enclosed b. Suppose P (, ) and Q(, ) have continuous first order partial derivatives on. Then ( Q P d + Q d P ) da. Note: Q P is the two-dimensional curl of the vector field F P, Q. We will give a proof of half of Green s Theorem in the case that the region can be described b the inequalities a b and f() g() (aka a tpe I region). The proof of the other half of Green s Theorem in the case that the region can be described b inequalities a b and f() g() (aka a tpe II region) is easil created b adjusting this proof. Suppose is described b a b and f() g() (aka a tpe I region). Then P da P d. Proof: Let be the region bounded b a b and f() g(). Let be the boundar of, specificall have begin at (a, f(a)) and go along f() to (b, f(b)), then verticall along b to (b, g(b)), then along g() to (a, g(a)), and finall verticall along a to return to (a, f(a)). P da b g() a f() P d d b a P (, ) g() f() d b a P (, f()) d b a P (, g()) d P d + P d, 3 where is the curve f() traversed from a to b and 3 is the curve g() traversed from b to a. Let be the vertical line from (b, f(b)) to (b, g(b)) and 4 be the vertical line from (a, g(a)) to (a, f(a)). Since d dt for an parametrization r(t) (t), (t) along either and 4, we get P (, ) d P (, ) d dt dt and P (, ) d P (, ) d dt. 4 4 dt

2 Putting this all together, P d P d+ P d+ P d+ P d P d+ P d P da 3 g() (b,g(b)) (a,g(a)) f() 4 (a,f(a)) (b,f(b)) From a similar argument (ommitted here) we also get a proof that if a region can be described b inequalities a b and f() g() (aka a tpe II region) then Q da Q d. These two facts together prove Green s Theorem in the case that is a region that can be described both was (simultaneousl of tpe I and II).

3 An region can be carved b vertical and horizontal lines into subregions that can be described both was. Here is the region on the previous page divided into 8 such subregions: For each of these such subregion we have established that Green s Theorem holds. Now let s see how that is enough to impl Green s Theorem for an arbitrar region. Let be a region that is chopped into subregions,,..., k that are simultaneousl of tpe I and II. Let E be an internal horizontal or vertical edge created. E belongs to eactl two of the subregions i and j. In one of the regions E is either the top (or right) boundar. In the other it is the bottom (or left) boundar. onsequentl, E is traversed in one direction using a positive orientation for the boundar of i and then traversed in the opposite direction using a positive orientation for the boundar of j. Appling Green s Theorem to each subregion and adding the path integrals over the boundaries of the subregions results in a sum of path integrals that can be rewritten so the path integrals along all internal edges E pair off and cancel, leaving onl those along the outer boundar of the entire region (in the positive orientation). Therefore, ( Q P ) da k i i ( Q P ) da k P d+q d i i P d+q d.

4 To illustrate, here is an internal edge traversed in opposite directions in this region divided into two regions that are simultaneousl of tpe I and II: E The sum of all the path integrals along the boundaries of the two regions is equal to the path integral around the boundar of the overall region. Eample: Let s evaluate the following path integral (a) with and (b) without Green s theorem. d d where is the triangle whose vertices are (, ), (, ), and (, 3). (a) We can break into three parts,, and 3 where is parameterized b r (t) t, for t, r (t) t, + t for t, and r 3 (t), 3 t for t. So d d d d + d d + d d 3 t dt + [( t) ( + t)( ) dt ( t)( + t)() dt] + dt 3 + [ (t t + )( + t) ( + t t )] dt 3 + [ t 3 + 7t t 3] dt 6.

5 (b) Now let s use Green s Theorem! d d region bounded b, and 3. Then ( [ ( ) ( )] da ( ) da, where is the 3 d d ( ) da ( ) d d ) 3 ( d ) (3 ) + (3 ) d ( ) ( + 3 d ) d 6. B chopping up the region (as in our justification) Green s Theorem even works on regions that have holes in them (boundar around the hole must be oriented negativel). Evaluate the origin). B Green s Theorem, 3 d + 3 d where is the union of the circles of radii and (centered at 3 d + 3 d between the circles of radii and (centered at the origin). Using polar coordinates, da da where is the annulus region π 3r r dr dθ π 3 4 (6 ) 45 π.

6 We can use Green s Theorem to calculate area of a region R enclosed b the closed simple curve : If F, then F dr d da the area of R. R Similarl, if F, then F dr d the area of R. Let s calculate the area of portion of the circle of radius that lies above the line using a line integral. ( ) Let be made up of and, where is the part of the unit circle from, ( ) to, parameterized b r cos (t), sin (t) for π 4 t 3π 4 and is the ( ) ( ) line segment from, back to, parameterized b r t, for t. Then the area is A 3π/4 π/4 cos (t) dt d d + d 3π/4 π/4 + cos (t) dt π 4. 3π/4 π/4 cos (t) cos (t) dt + Note: The clever wa to solve this is to realize it s the area of a quarter disk minus a triangle...

7 Let be a closed simple curve in R enclosing a region R. Let F f, g be a vector field. B Green s Theorem, the flu integral satisfies F n ds g d + f d R ( f + g ) f Note: + g is the two-dimensional divergence of the vector field F f, g. If the divergence of F f, g is throughout a region R the vector field F is said to be source-free. Eample: Use Green s Theorem to compute the flu integral (outward) of the radial vector field F, across the unit circle +. As an eercise left for the reader (ELFY), compute it directl as a line integral. F n ds d + d + da. ( + ) da π. Another eample: onsider the vector field F +, 4. Let s calculate the circulation and flu (outward) where bounds region in the first quadrant between the circles of radii and. irculation: F dr ( + ) d + ( 4) d ( ) da. (which is also implied b the fact that this vector field is conservative.) Flu: F n ds ( 4)d + ( + ) d R R ( 4) da 3π. onsider a vector field F f, g, differentiable on a region R. A stream function for F is a function ψ satisfing ψ f and ψ g. If a vector field F f, g has a stream function ψ (on some region R) then the divergence of the vector field is b the equalit of mied partials. f + g (ψ ) + ( ψ ) ψ ψ, So the eistence of a stream function implies that the field is source-free. The converse turns out to be true (proof skipped) on simpl connected regions in R.

8 Let ψ be a stream function of a source-free vector field F f, g. Let s consider a level curve of ψ determined b ψ(, ) k. Then the gradient ψ(, ) ψ, ψ g, f evaluated a point on is orthogonal to the level curve. B checking that ψ F gf + fg we get that F must be parallel to (at points on ) since it is also orthogonal to the gradient. onclusion: Level curves of stream functions of source-free vector fields are flow curves (aka streamlines) of the vector field. onsider F,. Since ( ) + ( ) this is a source-free vector field. A stream function for this source-free vector field is ψ. Let s compute the flu of this vector field across (a) the unit circle and (b) across the arc αβ, the part of the unit circle parameterized b r(t) cos (t), sin (t) from t α to t β: (a) F n ds ( ) d + d d + d + ( ) da. (b) F n ds ( ) d + d αβ αβ d + d αβ β α cos (t) sin (t)( sin (t) dt) + cos (t)(cos (t) dt) 3 sin3 (t) + sin (t) 3 sin3 (t) β α sin (t) sin 3 (t) β α sin (t) cos (t) β α sin (β) cos (β) sin (α) cos (β) ψ(r(β)) ψ(r(α)). Here we see an important feature of a source-free vector field F with stream function ψ: The flu integral is equal to the value of ψ at the end of the path minus the value of ψ at the start of the path (so is on closed curves). So the flu integral is independent of path for a source-free vector field (similar to the circulation integral being independent of path for a conservative vector field).

9 Finall, it s ver special when we have a vector field that is BOTH conservative and sourcefree. Such a vector field is sometimes called an ideal flow. onsider F,. Since + ( ) + ( ) ( + ), + it follows that this vector field is conservative. Since ( ) ( ) ( ) , + + ( + ) it follows that this vector field is source-free. So it follows that the circulation and flu integral of F on an closed curve around a region where the components of F have continuous partial derivatives (so avoiding the origin) is.