Unless a vector eld F is conservative computing the line integral 16 4 Green s Theorem cid 90 cid 90 F dr C C P dx Q dy is often dif cult and time consuming For a given integral one must 1 Split C into separate smooth subcurves C1 C2 C3 2 Parameterize each curve Ci by a vector valued function ri t ai t bi 3 Evaluate each integral cid 82 F dr cid 82 bi ai F ri t r cid 48 i t dt Ci Thankfully there is a short cut available for line integrals over particularly simple curves C De nition A curve C is the planes is 1 Closed if it starts and nishes at the same point 2 Simple if it has no self intersections it does not cross itself 3 Positively oriented if the direction of travel around C is such that the inside of C is on one s left Alternatively a simple closed curve is positively oriented if one traverses it counter clockwise This categorization can easily lead you astray however so it is better to think about the inside cid 73 De nition The notation oriented simple closed curve C C F dr denotes a line integral around a positively If D is a region then its boundary curve is denoted D Observe that D is simply connected iff its boundary D is simple and closed Theorem Green s Theorem Let D be a simply connected region of the plane with positively oriented simple closed piecewise smooth boundary C D Suppose that P Q have continuous partial derivatives on some open region containing D and its boundary Then cid 73 C P x y dx Q x y dy Q x P y D dA Qx Py dA D cid 90 cid 90 cid 90 cid 90 1 Anon simpleclosedcurveApositivelyorientedsimpleclosedcurveAnegativelyorientedsimpleclosedcurveC DD cid 73 Green s theorem is often useful in examples since double integrals are typically easier to evaluate than line integrals Example Find 1 0 1 1 0 1 and F dr where C is the square with corners 0 0 C Proof of Green s Theorem The proof has three stages First prove half each of the theorem when the region D is either Type 1 or Type 2 Putting these together proves the theorem when D is both type 1 and 2 The proof is completed by cutting up a general region into regions of both types By Green s Theorem F x y x3 1 i xy2 2 j cid 73 F dr C x cid 90 1 R xy2 2 y 1 y2 dx dy 3 0 0 cid 90 cid 90 cid 90 1 x3 1 dA We could check this by evaluating the line integral directly First suppose that R is a region of Type 1 C consists of two three or four curves y f x and y g x between x a and b and possibly two vertical edges at x a b On the straight edges x is constant and so dx 0 Therefore P x y dx P x g x dx cid 73 C However cid 90 cid 90 cid 73 D Hence a b cid 90 a P x f x dx cid 90 b cid 90 b P x f x P x g x dx cid 90 g x cid 90 b cid 90 b cid 90 cid 90 Py x y dy dx f x a a a Py dA cid 73 C C P x y dx cid 90 c cid 90 d cid 90 d d c Q x y dy Py dA which is half of the theorem D Now suppose that R is a region of Type 2 Analogously to before we compute cid 90 d c Q j y y dy Q k y y dy cid 90 k y cid 90 cid 90 Q k y y Q j y y dy Qx x y dx dy c j y Qx dA R 2 cid 73 P x g x P x f x dx C P x y dx 01y01xDCyxabDCy f x y g x yxcdDx j y Cx k y This is the other half of the theorem It follows that if D is a region of both Types 1 2 we have the result cid 73 cid 90 cid 90 P dx Q dy C Q x P y D dA For general regions cut D into pieces of both Types 1 2 line integrals along common edges are counted in both directions and thus cancel 1 For example D1 D4 are of Types 1 2 so Green s Theorem holds for each piece Notice however that all boundaries are traversed twice in opposite directions For example the curves C1 and C2 run opposite to each other on the common boundary of D1 and D2 It follows that the line integrals along these pieces cancel each other out Therefore cid 73 C cid 20 cid 73 cid 20 cid 90 cid 90 C1 cid 73 cid 21 cid 90 cid 90 C4 P dx Q dy P dx Q dy cid 21 Q cid 90 cid 90 D1 D4 x P y dA Q x P y D dA Examples 1 Let C be the perimeter of the ellipse x2 4 y2 1 and Then cid 73 F x y C F dr cid 19 4x cos y2 cid 18 sin ex y cid 90 cid 90 cid 90 cid 90 x D 4x cos y2 y sin ex y dA cid 90 cid 90 D 4 1 dA 5 dA 10 D area of ellipse formula This is contrived but it would have been much harder to evaluate directly as a line integral 1We will have to leave it as an article of faith that any simply connected region D may be so decomposed 3 DCD1C1D2C2D3C3D4C4 1 0 500 51y 2 1012xDC cid 90 cid 19 cid 18 cid 90 C ex2 y e2x y 2 Calculate ex2 C y dx e2x y dy where C is formed from the parabola y 1 x2 and the x axis as shown The orientation of C is negative so Green s Theorem gets a minus sign cid 90 cid 90 cid 90 1 x2 cid 90 1 dr cid 18 x R 0 1 e2x x2 x 3 1 e 2 e2 2 2 4 3 y dA y ex2 e2x y cid 19 1 2e2x dy dx 1 x3 3 x cid 90 1 cid 12 cid 12 cid 12 cid 12 1 1 2 1 1 1 x2 1 2e2x dx integration by parts Simple connectedness revisited We are now in a position to prove our simple formula if F has continuous partial derivatives on a simply connected region D then Q x P y By Green s Theorem F Pi Qj conservative Proof Recall that F is conservative on D if and only if cid 72 D simply connected interior cid 101 D of every simple closed curve C in D is also simply connected cid 90 cid 90 cid 101 D for …