4.4.32
Intuition Ek sentence mein ideaEk surface par phaila hua total swirl (curl) uski boundary ke around net circulation ke barabar hota hai.
Agar tum ek surface ke andar ho rahi saari choti rotations ko jodo, toh interior contributions apne neighbors ke saath cancel ho jaati hain aur sirf edge bachti hai — jo boundary ke around line integral deti hai.
Maano S R 3 ∂ S = C F S ∮ C F ⋅ d r = ∬ S ( ∇ × F ) ⋅ d S C C n ^ left mein ho (right-hand rule).
LHS: F circulation boundary C
RHS: surface S ∇ × F flux .
Intuition Kyun "curl = circulation density"Curl ek point par ek infinitesimal loop ki circulation per unit area measure karta hai. Stokes' theorem bas yeh kehta hai: is local circulation-density ko poori surface par integrate karo, aur tumhe edge ke around global circulation mil jaati hai.
Intuition Cancellation ka pictureS dono directions mein ek-ek baar traverse hoti hai → contributions cancel ho jaate hain. Sirf outer boundary C C
Step 1 — Curl ko circulation per area ke roop mein define karo.
Ek chota rectangle lo jiska area Δ A n ^ ∮ ∂ ( patch ) F ⋅ d r ≈ ( ∇ × F ) ⋅ n ^ Δ A .
Yeh step kyun? F hi curl hai. Aaiye isko ek flat x y
Step 2 — Ek chote x y
Rectangle [ x 0 , x 0 + Δ x ] × [ y 0 , y 0 + Δ y ] n ^ = k ^ F = ( P , Q , R ) ∮ F ⋅ d r = y = y 0 ∫ P d x bottom + ∫ Q d y right + y = y 0 + Δ y ∫ P d x top + ∫ Q d y left . P Q ≈ [ − ∂ y ∂ P + ∂ x ∂ Q ] Δ x Δ y = ( ∇ × F ) z Δ A .
Yeh step kyun? Bottom aur top integrals ek hi same x y 0 y 0 + Δ y − ∂ P / ∂ y Δ y Δ x Green's theorem hai — Stokes uska 3D upgrade hai.
Step 3 — Sum karo aur limit lo.
∮ C F ⋅ d r = ∑ patches ∮ ∂ patch F ⋅ d r ≈ ∑ ( ∇ × F ) ⋅ n ^ Δ A Δ A → 0 ∬ S ( ∇ × F ) ⋅ d S .
Worked example Example 1 — Constant curl wala fieldF = ( − y , x , 0 ) ∮ C F ⋅ d r C x y
Curl: ∇ × F = ( 0 , 0 , ∂ x ( x ) − ∂ y ( − y )) = ( 0 , 0 , 2 ) Kyun? Sirf z P = − y , Q = x ⇒ Q x − P y = 1 − ( − 1 ) = 2
Surface: x y S n ^ = k ^ = π ∬ S ( ∇ × F ) ⋅ d S = ∬ S 2 d A = 2 π . Yeh step kyun? ( 0 , 0 , 2 ) ⋅ ( 0 , 0 , 1 ) = 2 2 × Area = 2 π
Directly check karo: C r = ( cos t , sin t , 0 ) d r = ( − sin t , cos t , 0 ) d t F ⋅ d r = ( sin 2 t + cos 2 t ) d t = d t 0 → 2 π 2 π
Worked example Example 2 — Ek hard surface ko easy wale se replace karoF = ( z 2 , x 2 , y 2 ) C ( 1 , 0 , 0 ) , ( 0 , 1 , 0 ) , ( 0 , 0 , 1 ) n ^
Curl: ∇ × F = ( ∂ y y 2 − ∂ z x 2 , ∂ z z 2 − ∂ x y 2 , ∂ x x 2 − ∂ y z 2 ) = ( 2 y , 2 z , 2 x )
Surface: plane x + y + z = 1 n ^ = 3 1 ( 1 , 1 , 1 ) d S = ( 1 , 1 , 1 ) d A x y z = 1 − x − y ( ∇ × F ) ⋅ ( 1 , 1 , 1 ) = 2 y + 2 z + 2 x = 2 ( x + y + z ) = 2 ( 1 ) = 2. Yeh step kyun? Plane x + y + z = 1 x ≥ 0 , y ≥ 0 , x + y ≤ 1 2 1 ∮ C F ⋅ d r = ∬ 2 d A x y = 2 ⋅ 2 1 = 1.
Worked example Example 3 — Do surfaces, same boundaryEk hemisphere x 2 + y 2 + z 2 = 1 , z ≥ 0 z = 0 Kisi bhi F ∇ × F same hoga — kyunki common edge ke around line integral same hai.
Kyun? RHS sirf S sabse aasaan surface pe swap karo.
Common mistake "Orientation? Main sign baad mein decide kar lunga."Kyun sahi lagta hai: magnitudes match karte hain, toh lagta hai yeh free choice hai.
Sach yeh hai: ∮ C n ^ compatibly link karna hoga. Galat kiya toh answer sign se off ho jaayega.
Fix: Right-hand thumb n ^
Common mistake "Curl ek number / scalar hai."Kyun sahi lagta hai: 2D mein (Green) "curl" Q x − P y Fix: 3D mein, ∇ × F vector hai. Green's theorem mein scalar bas uska k ^ n ^
F Kyun sahi lagta hai: line integrals ke liye FTC se confusion.
Fix: Stokes kisi bhi C 1 ∇ × F = 0 S tab circulation zero hoti hai — yeh special case batata hai ki F
Common mistake Yeh bhool jaana ki surface koC Fix: S C S ∇ × F
Mnemonic Layout yaad karo"Curl drum-skin se guzarta hai = rim ke around circulation."
Ek drum socho: S C CALC — C irculation A round L oop = C url-flux.
Recall Feynman: ek 12-saal ke bachche ko samjhaoSocho ek shallow tray mein paani hai aur usme ek wire loop flat rakha hua hai. Tum paani ko bahut saare chote-chote spinning whirlpools se hilaate ho. Stokes' theorem kehta hai: agar tum loop ke andar saare chote whirlpools ki spinning ko jodo, tumhe exactly woh mil jaata hai jitna paani loop ke edge ke around flow kar raha hai. Reason kaafi neat hai — jahan do whirlpools touch karte hain, ek paani ko ek taraf push karta hai aur doosra usse wapas push karta hai, toh cancel ho jaate hain. Sirf bilkul edge par ki spinning ko cancel karne waala koi nahi hota. Wahi bacha hua spinning rim ke around flow hai!
Stokes' theorem equation ∮ C F ⋅ d r = ∬ S ( ∇ × F ) ⋅ d S
LHS of Stokes physically kya represent karta hai? Boundary curve C F
RHS physically kya represent karta hai? Surface S ∇ × F
Curl ___ per unit area measure karta hai Circulation (yeh circulation density hai).
C n ^ Right-hand rule: thumb n ^ C
Stokes ek flat x y Green's theorem.
Derivation mein interior patch boundaries kyun cancel ho jaati hain? Shared edges dono directions mein ek-ek baar traverse hoti hain, toh unke contributions opposite hote hain aur cancel ho jaate hain.
( − y , x , 0 ) ( 0 , 0 , 2 )
Agar ∇ × F = 0 S Zero (field locally conservative / irrotational hai).
Kya same boundary wale do alag surfaces alag curl-flux de sakte hain? Nahi — flux sirf shared boundary C
Curl determinant formula ∇ × F = det [ i ^ , j ^ , k ^ ; ∂ x , ∂ y , ∂ z ; P , Q , R ]
Kya Stokes ke liye F Nahi, sirf C 1
Vector field F on surface S
Right-hand rule orientation
Curl = circulation density
