Green's theorem — proof sketch, both forms: Exercises

4.4.29 · D4 · Maths › Multivariable Calculus › Green's theorem — proof sketch, both forms

Reminders jo hum baar baar use karenge (dono parent se): $\oint_{C} P d x + Q d y = \iint_{D} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d A (circulation)$ $\oint_{C} F \cdot n d s = \iint_{D} (\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}) d A (flux)$

Yahaan $\oint_{C}$ ka matlab hai "poore closed loop $C$ ke saath-saath add karo", aur $\iint_{D}$ ka matlab hai "poore flat region $D$ ke upar add karo jo loop ne ghera hua hai". Chhota sa $\partial Q / \partial x$ partial derivative hai: $Q$ kitni tezi se badalta hai agar tum sirf $x$ ko thoda hila do aur $y$ ko freeze rakho.

Related toolkit: Line Integrals, Double Integrals, Curl and Divergence, Conservative Vector Fields.

Figure — Green's theorem — proof sketch, both forms

Level 1 — Recognition

L1.1

Batao ki Green's theorem ka kaun sa form use karoge, aur uska integrand likho, is ke liye: $\oint_{C} 3 x^{2} y d x + x^{3} d y .$

Recall Solution

Yeh ek line integral hai jo $P d x + Q d y$ ke roop mein likha hai, isliye yeh circulation form hai. $P = 3 x^{2} y$ , $Q = x^{3}$ padhlo. Integrand hai $Q_{x} - P_{y}$ : $Q_{x} = \frac{\partial}{\partial x} (x^{3}) = 3 x^{2}, P_{y} = \frac{\partial}{\partial y} (3 x^{2} y) = 3 x^{2} .$ Toh integrand hai $3 x^{2} - 3 x^{2} = 0$ . Loop integral $0$ hai kisi bhi closed curve ke liye (field conservative hai — dekho Conservative Vector Fields).

L1.2

$F = (P, Q) = (x^{2}, y^{2})$ ke liye, flux-form integrand $\nabla \cdot F$ likho.

Recall Solution

Flux sum $P_{x} + Q_{y}$ use karta hai: $P_{x} = \frac{\partial}{\partial x} (x^{2}) = 2 x, Q_{y} = \frac{\partial}{\partial y} (y^{2}) = 2 y .$ $\nabla \cdot F = 2 x + 2 y .$

Level 2 — Application

L2.1

$\oint_{C} (y - x) d x + (2 x) d y$ compute karo jahan $C$ square $0 \leq x \leq 1$ , $0 \leq y \leq 1$ ki boundary hai, counterclockwise.

Recall Solution

Circulation form: $P = y - x$ , $Q = 2 x$ . $Q_{x} - P_{y} = \frac{\partial}{\partial x} (2 x) - \frac{\partial}{\partial y} (y - x) = 2 - 1 = 1.$ Toh loop integral seedha square ke plain area mein collapse ho jaata hai: $\iint_{D} 1 d A = Area = 1 \times 1 = 1.$ Answer: $1$ .

L2.2

Boundary–area formula $Area = \frac{1}{2} \oint_{C} (x d y - y d x)$ use karke ellipse $x = 3 cos t, y = 2 sin t$ , $0 \leq t \leq 2 π$ se ghiree area nikalo.

Recall Solution

Yeh formula kyun? $P = - y /2, Q = x /2$ choose karne se $Q_{x} - P_{y} = 1$ ho jaata hai, isliye double integral raw area ban jaata hai — parent mein prove kiya gaya hai. Hum sirf parametrisation plug in karte hain. $d x = - 3 sin t d t, d y = 2 cos t d t .$ $x d y - y d x = (3 cos t) (2 cos t d t) - (2 sin t) (- 3 sin t d t) = 6 cos^{2} t d t + 6 sin^{2} t d t = 6 d t .$ $Area = \frac{1}{2} \int_{0}^{2 π} 6 d t = \frac{1}{2} \cdot 12 π = 6 π .$ Answer: $6 π$ . (Sanity check: semi-axes $a = 3, b = 2$ wali ellipse ka area $π ab = 6 π$ hota hai. ✓)

L2.3

$F = (x^{2}, y^{2})$ ka outward flux $\oint_{C} F \cdot n d s$ unit square $0 \leq x \leq 1, 0 \leq y \leq 1$ ki boundary ke across compute karo.

Recall Solution

Flux form loop ko divergence ke double integral mein badal deta hai (L1.2 se, $\nabla \cdot F = 2 x + 2 y$ ): $\iint_{D} (2 x + 2 y) d A = \int_{0}^{1} \int_{0}^{1} (2 x + 2 y) d y d x .$ $y$ mein inner integral: $\int_{0}^{1} (2 x + 2 y) d y = [2 x y + y^{2}]_{0}^{1} = 2 x + 1.$ $x$ mein outer integral: $\int_{0}^{1} (2 x + 1) d x = [x^{2} + x]_{0}^{1} = 1 + 1 = 2.$ Answer: flux $= 2$ .

Level 3 — Analysis

L3.1

$\oint_{C} (e^{x} - y^{3}) d x + (cos y + x^{3}) d y$ compute karo jahan $C$ unit circle $x^{2} + y^{2} = 1$ hai, counterclockwise.

Recall Solution

Line integral bahut mushkil lagti hai ( $e^{x}$ , $cos y$ ...). Green's theorem hume bachata hai — woh "apne variable" wale pieces $Q_{x} - P_{y}$ ke neeche gayab ho jaate hain. $P = e^{x} - y^{3}, Q = cos y + x^{3} .$ $Q_{x} = 3 x^{2}, P_{y} = - 3 y^{2}, Q_{x} - P_{y} = 3 x^{2} + 3 y^{2} = 3 (x^{2} + y^{2}) .$ Polar coordinates $x = r cos θ, y = r sin θ$ mein convert karo, jahan $x^{2} + y^{2} = r^{2}$ aur $d A = r d r d θ$ . Polar kyun? Integrand aur region dono circular hain, isliye polar se limits trivial ho jaati hain. $\iint_{D} 3 r^{2} d A = \int_{0}^{2 π} \int_{0}^{1} 3 r^{2} \cdot r d r d θ = \int_{0}^{2 π} \int_{0}^{1} 3 r^{3} d r d θ .$ $\int_{0}^{1} 3 r^{3} d r = [\frac{3}{4} r^{4}]_{0}^{1} = \frac{3}{4}, \int_{0}^{2 π} \frac{3}{4} d θ = \frac{3}{4} \cdot 2 π = \frac{3 π}{2} .$ Answer: $\frac{3 π}{2}$ .

L3.2

Vector field $F = \frac{( - y , x )}{x ^{2} + y ^{2}}$ ka curl jahan bhi defined hai wahan zero hai, phir bhi unit circle ke around $\oint_{C} F \cdot d r = 2 π$ hai. Apparent contradiction explain karo aur $(x, y) \neq = (0, 0)$ ke liye $Q_{x} - P_{y}$ compute karo.

Recall Solution

Likho $P = \frac{- y}{x ^{2} + y ^{2}}$ , $Q = \frac{x}{x ^{2} + y ^{2}}$ . Dhyan se differentiate karo: $Q_{x} = \frac{( x ^{2} + y ^{2} ) - x ( 2 x )}{( x ^{2} + y ^{2} ) ^{2}} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}},$ $P_{y} = \frac{- ( x ^{2} + y ^{2} ) + y ( 2 y )}{( x ^{2} + y ^{2} ) ^{2}} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}} .$ Toh $Q_{x} - P_{y} = 0$ jahan bhi $F$ defined hai. Agar Green's theorem apply hota, toh loop integral $0$ hoti — lekin woh $2 π$ hai. Resolution: Green's theorem ke liye zaroori hai ki $P, Q$ ke continuous partials $D$ ke andar har jagah hon. Yahaan field origin pe blow up karti hai, jo unit disk ke andar hai. Hypothesis fail ho jaati hai, isliye hum theorem apply nahi kar sakte. $2 π$ real hai; "contradiction" sirf apparent hai kyunki theorem kabhi licensed hi nahi tha.

Level 4 — Synthesis

L4.1

$F = (y - sin x, cos x - y)$ ke liye $\oint_{C} F \cdot d r$ compute karo jahan $C$ vertices $(0, 0)$ , $(2, 0)$ , $(0, 2)$ wala triangle hai, counterclockwise traverse kiya gaya.

Recall Solution

$P = y - sin x, Q = cos x - y$ . $Q_{x} = - sin x, P_{y} = 1, Q_{x} - P_{y} = - sin x - 1.$ Triangle Type I region hai $0 \leq x \leq 2$ , $0 \leq y \leq 2 - x$ (bottom edge $x$ -axis ke saath, hypotenuse $y = 2 - x$ ). Yeh limits kyun? Fixed $x$ ke liye vertical strip $x$ -axis se upar slanted top tak jaati hai. $\iint_{D} (- sin x - 1) d A = \int_{0}^{2} \int_{0}^{2 - x} (- sin x - 1) d y d x .$ Inner integral (integrand $y$ mein constant hai): $\int_{0}^{2 - x} (- sin x - 1) d y = (- sin x - 1) (2 - x) .$ Ab expand karo aur $x$ mein integrate karo: $\int_{0}^{2} (- sin x - 1) (2 - x) d x = \int_{0}^{2} [- 2 sin x + x sin x - 2 + x] d x .$ $[0, 2]$ pe term by term:

$\int_{0}^{2} - 2 sin x d x = 2 cos x_{0}^{2} = 2 cos 2 - 2.$
$\int_{0}^{2} x sin x d x = [sin x - x cos x]_{0}^{2} = sin 2 - 2 cos 2.$
$\int_{0}^{2} - 2 d x = - 4.$
$\int_{0}^{2} x d x = 2.$

Sab jodo: $(2 cos 2 - 2) + (sin 2 - 2 cos 2) + (- 4) + 2 = sin 2 - 4.$ Answer: $sin 2 - 4 \approx 0.9093 - 4 = - 3.0907$ .

L4.2

Ek closed curve $C$ (counterclockwise) ki property hai ki $\frac{1}{2} \oint_{C} (x d y - y d x) = 10$ . Agar tum poori curve ko origin ki taraf factor $\frac{1}{2}$ se shrink karo (har point $(x, y) \mapsto (\frac{1}{2} x, \frac{1}{2} y)$ map karo), toh $\frac{1}{2} \oint (x d y - y d x)$ ki nayi value kya hogi?

Recall Solution

Woh integral hi enclosed area hai (parent Example 1). Har coordinate ko factor $k$ se scale karne se saari lengths $k$ se aur isliye areas $k^{2}$ se scale hoti hain. Yahaan $k = \frac{1}{2}$ hai, toh area $\to (\frac{1}{2})^{2} \times 10 = \frac{1}{4} \times 10 = 2.5$ . Answer: $2.5$ .

Level 5 — Mastery

L5.1 (Full argument)

Maano $D$ ek annulus hai: circle $x^{2} + y^{2} = 1$ (inner) aur $x^{2} + y^{2} = 4$ (outer) ke beech ka region. Maano $F = (- y, x)$ . Hole wale region ke liye Green's theorem sahi se use karke, total circulation compute karo, yaani (outer boundary counterclockwise) $+$ (inner boundary clockwise).

Recall Solution

Annulus ko dhyan se kyun handle karna hai. Hole wale region ke liye Green's theorem kehta hai: boundary integrals ko is tarah jodo ki region hamesha tumhari left pe rahe. Matlab outer circle counterclockwise aur inner circle clockwise. Phir $\oint_{\partial D} F \cdot d r = \iint_{D} (Q_{x} - P_{y}) d A .$ Yahaan $P = - y, Q = x$ , toh $Q_{x} - P_{y} = 1 - (- 1) = 2$ (annulus mein har jagah well-defined hai — koi singularity nahi, kyunki origin exclude hai). Annulus ka area hai $π (2^{2}) - π (1^{2}) = 4 π - π = 3 π$ . $\iint_{D} 2 d A = 2 \cdot 3 π = 6 π .$ Answer: total circulation $= 6 π$ .

Pieces ko directly cross-check karo. Outer circle pe (radius $2$ , CCW), $\oint (- y d x + x d y) = 2 \times area = 2 π (2^{2}) = 8 π$ . Inner circle pe (radius $1$ , clockwise = uski CCW value ka minus) yeh $- 2 π (1^{2}) = - 2 π$ hai. Sum $= 8 π - 2 π = 6 π$ . ✓

L5.2 (Full argument)

Prove karo ki agar $F = (P, Q)$ mein $Q_{x} = P_{y}$ poore simply-connected region $D$ mein hai, toh $\oint_{C} F \cdot d r = 0$ har simple closed curve $C$ ke liye jo $D$ ke andar ho. Phir isse use karke $\oint_{C} (2 x y d x + x^{2} d y)$ kisi bhi loop pe evaluate karo.

Recall Solution

Argument. $D$ ke andar koi bhi simple closed $C$ lo jo region $R \subseteq D$ gherta ho. Kyunki $D$ simply connected hai (koi holes nahi), $R$ mein koi forbidden points nahi hain, isliye Green's theorem apply hota hai: $\oint_{C} F \cdot d r = \iint_{R} (Q_{x} - P_{y}) d A .$ Hypothesis se $Q_{x} = P_{y}$ hai, isliye integrand har jagah $0$ hai, toh double integral $0$ hai, toh $\oint_{C} = 0$ . Yeh bilkul wahi statement hai ki $F$ $D$ pe conservative (path-independent) hai — dekho Conservative Vector Fields aur line integrals ke liye Fundamental Theorem of Calculus analogue. Application. $F = (2 x y, x^{2})$ ke liye: $Q_{x} = \frac{\partial}{\partial x} (x^{2}) = 2 x, P_{y} = \frac{\partial}{\partial y} (2 x y) = 2 x .$ Woh plane mein har jagah match karte hain (jo simply connected hai), isliye $\oint_{C} = 0$ har closed loop ke liye. Answer: $0$ . (Waqai $F = \nabla (x^{2} y)$ hai, ek gradient field.)

L5.3 (Synthesis with the flux form)

$F = (x^{3}, y^{3})$ ke liye, circle $x^{2} + y^{2} = a^{2}$ (radius $a$ ) ke across outward flux compute karo, answer $a$ ke terms mein do.

Recall Solution

Flux form: $\nabla \cdot F = P_{x} + Q_{y} = 3 x^{2} + 3 y^{2} = 3 (x^{2} + y^{2})$ . Polar coordinates mein ( $x^{2} + y^{2} = r^{2}$ , $d A = r d r d θ$ ): $\iint_{D} 3 r^{2} d A = \int_{0}^{2 π} \int_{0}^{a} 3 r^{2} \cdot r d r d θ = \int_{0}^{2 π} \int_{0}^{a} 3 r^{3} d r d θ .$ $\int_{0}^{a} 3 r^{3} d r = [\frac{3}{4} r^{4}]_{0}^{a} = \frac{3}{4} a^{4}, \int_{0}^{2 π} \frac{3}{4} a^{4} d θ = \frac{3}{4} a^{4} \cdot 2 π = \frac{3 π a ^{4}}{2} .$ Answer: $\frac{3 π a ^{4}}{2}$ . ( $a = 1$ se check karo: $\frac{3 π}{2}$ milta hai, jo L3.1 ke radial computation ke shape se match karta hai.)

Recall One-line self-test

$\oint_{C} F \cdot d r = 0$ sabhi loops ke liye kab guaranteed hai? ::: Jab $Q_{x} = P_{y}$ simply-connected region pe ho (andar koi singular points nahi) — field conservative hai. Hole wale region ke liye, inner boundary kis direction mein jaati hai? ::: Clockwise (taaki region tumhari left pe rahe).