So, if we are finding the amount of work done over a non-conservative vector field F r, we do that long ur r b ur =

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Peregrine Oscar Day
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1 3.4 Geen s Theoem Geoge Geen: self-taught English scientist, So, if we ae finding the amount of wok done ove a non-consevative vecto field F, we do that long u b u 3. method Wok = F d F( () t ) = () t dt. a If F is a consevative vecto field we can use the 3.3 shotcut (Fundamental Theoem fo Line Integals) whee we must find the potential function f. u f = F Wok = F d = f ( ( b) ) f ( ( a) ) o f potential Function If F is a consevative vecto field and cuve is a closed loop, the wok always equals Befoe we stat 3.4 let s pactice one moe line integal. Pass out woksheet Today: When F is NOT a consevative vecto field but ou cuve is a closed, bounded egion, thee is a much simple method to choose othe than the long 3. method. Geen s Theoem: Let be a positively oiented(counte-clockwise), piecewise-smooth, simple closed cuve in the plane and let be the egion bounded by cuve. If P and Q have continuous patial deivatives on an open egion Q P that contains, then P dx + Q dy = da. Geen s Theoem elates a line integal aound simple closed cuve to a double integal ove egion bounded by. If F ( xy, ) = Pxy (, ), Qxy (, ) Wok = F d = P dx + Q dy = da Note what would happen if F was a consevative vecto field hee: If F is consevative then = x, so the integand,, would equal zeo and you would get zeo wok. y Geen s Theoem convets line integals ove cuve to double integals ove aea (egion). (Flip back to 3.9 elates it to the Fundamental Theoem of alculus) Geen s Theoem is efeed to as the countepat of the Fundamental Theoem alculus fo double integals.

2 Geen s Theoem uses positive oientation to be counte clockwise of closed cuve. If you ae given the opposite oientation (clockwise), we attach a negative in font of the integals. So, if F is not a consevative vecto field, we ely on Geen s Theoem instead of the long 3. method if it meets the equiements. The notation is sometimes used to indicate that the line integal is calculated using positive oientation (being counte-clockwise). I think we should use this notation to emind us to attach a negative if the egion s oientation is clockwise. Geen s Theoem is not easy to pove; we won t be poving it. The book did pove it fo one simple case if you e inteested in exploing it. Note: simple egions vesus simply connected egions simple egions efe to Type I and Type II egions ou dx and dy aows. simply connected egions a egion that doesn t coss ove itself. A figue 8 is made up of two simply connected egions, but itself as a whole is not a simply connected egion. If Geen s Theoem isn t known: Example we aleady did the long way on the woksheet: (simila to a, a, 3a, fom HW) 3 Evaluate this line integal diectly xy dx + 4x y dy, whee is the egion bounded by y = x 3, y = 0 and x= with counteclockwise oientation. So F = xy x y 3 Is F consevative? { 3,4 P Q = 6xy = 8xy No, F is not consevative, so we can t use the FTLI. Region : we need to set up 3 line integals. This will gain you a geat appeciation of Geen s Theoem. a, a, 3a

3 3tdt : x = t, y = 0, 0 t Is oientation coect? dx dt, dy 0dt 0,0 to,0 Yes = = ( ) ( ) : x =, y = t, 0 t Is oientation coect? dx 0 dt, dy dt, 0 to, Yes = = ( ) ( ) 3 3 : x = t, y = t, 0 t Is oientation coect? (, ) back to ( 0, 0 ) NO so the easiest way to deal with this incoect oientation is to switch the 0 and aound in the limits of integation. If you don t like that, when you get to the integal you could thow a negative out in font and switch it back to 0 t. The solutions manual does it the had way and changes the fomulas fo x and y instead of changing t. They ae nuts; it s way moe wok! This is what they did: ( ) 3 3 : x = t, y = t, 0 t, so thei integal looked like: 333-t-t-dt+4-t-t-3-t 0( ) ( ) ( ) (( ) ) ( )( ) ( ) ( ) 3( ) ( ) ( ) ( ) ( ) -3( t) dt3=--t-tdt--tdt404=- = xy dx + 4x y dy = 3-ttdt+4tt t 0 dt+ 4 t dt + t 0dt+ 4 t dt ( ) ( ) ( ) dt = 0 dt + 4 t dt t t 4t -= = = 33 44

4 Let s do example again but this time using Geen s Theoem Pdx + Qdy = da. 3 Evaluate this line integal, xy dx + 4x y dy, whee is the egion bounded by y = x 3, y = 0 and x= with counteclockwise oientation. onditions fo using Geen s Theoem: Let be a positively oiented (counte-clockwise), piecewisesmooth, simple closed cuve in the plane and let be the egion bounded by cuve. P and Q must have continuous patial deivatives on an open egion that contains. Geen s Theoem convets a line integal ove cuve to a double integal ove simple egion. Example : u Find the wok done by foce F = x y,x 4y in moving a paticle once counteclockwise aound the cicle ( x ) ( y ) + = 4. Ae the conditions met fo using Geen s Theoem? Is be a positively oiented (counte-clockwise), piecewise-smooth, simple closed cuve in the plane and let be the egion bounded by cuve? o P and Q have continuous patial deivatives on an open egion that contains? W = F d = da (Geen s Theoem)

5 It s difficult to set this double integal up with pola coodinates Just fo fun I did the double integal by conveting to pola coodinates. On second thought, it s not eally all that fun just skip this! I have nothing bette to do. All d aows exit All d aows ente ( x ) ( y ) + = 4 ( θ ) ( θ ) cos + sin = 4 cos θ 4cosθ sin θ 4sinθ + 4 = 4

6 4cosθ 4sinθ + 4= 0 ( θ θ) 4 cos + sin + 4= 0 I plugged this into my solve on my gaphing calculato and got: ( θ θ θ θ) ( θ θ θ θ) = sin cos cos sin OR = sin cos + cos + sin So 4 da = π ( sinθcosθ + cosθ+ sinθ ) 4 d dθ 0 ( sinθcosθ cosθ sinθ) = 6π SO OOL!!! Example 3: Use Geen s Theoem to evaluate F d fo ( ) 3 F xy, = x+ y, x + y whee consists of the ac of the cuve y = sin x fom ( 0, 0 ) to ( π, 0) and the line segment fom (,0) to ( 0,0) π. (heck the oientation of the cuve befoe applying the theoem.) Ae the conditions met fo using Geen s Theoem? Is be a positively oiented (counte-clockwise), piecewise-smooth, simple closed cuve in the plane and let be the egion bounded by cuve? o P and Q have continuous patial deivatives on an open egion that contains? Pdx+ Qdy = da *on t foget to check oientation. If the oientation is clockwise, attach a negative in font of the integal.

7 Example 4: Use Geen s Theoem to evaluate F d fo ( ) y F x, y = tan, ln( x + y ), whee : ing x Pdx+ Qdy = da

8 ( ) da = P dx + Q dy Application of the evese diection of Geen s Theoem is in computing the aea of egion. Since the aea of is da, we would need to choose P and Q such that =. Thee ae many possibilities of choosing P and Q so that =. If I choose P( x, y ) = 0, then what does (, ) If P( x, y ) = 0 and Q( x, y) Since ( ) = x, then Q x y need to be to make =? = 0 = da = P dx + Q dy then you would have 0 dx + x dy = x dy. If I choose Q( x, y ) = 0, then what would (, ) P( x, y) = y and Q( x, y ) = 0, then Since ( ) P x y need to be to make =? = 0 =. da = P dx + Q dy then you would have y dx + 0 dy = y dx The book also chooses a thid, if P( x, y) = y and Q( x, y) = x, then = = Since ( ) da = P dx + Q dy then you would have y dx + x dy = x dy y dx So Geen s Theoem gives the following thee fomulas fo the aea of egion : A = x dy = y dx = x dy y dx Pick any of these 3 fomulas fo #9 on HW. Just follow the fomula exactly. Pick an easy one. I wouldn t choose the thid one.

The Divergence Theorem

The Divergence Theorem 13.8 The ivegence Theoem Back in 13.5 we ewote Geen s Theoem in vecto fom as C F n ds= div F x, y da ( ) whee C is the positively-oiented bounday cuve of the plane egion (in the xy-plane). Notice this