Green's theorem — proof sketch, both forms

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WHAT it says

The two forms are the same theorem written for two different field arrangements — proving one gives the other for free (shown below).

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WHY orientation matters

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HOW to prove it (derivation from scratch)

We prove the circulation form. The trick: split it into two halves and prove each. $\oint_C P\,dx = -\iint_D \frac{\partial P}{\partial y}\,dA \qquad(\star)$ $\oint_C Q\,dy = +\iint_D \frac{\partial Q}{\partial x}\,dA \qquad(\star\star)$ Add them and you get the full theorem.

Step 1: Prove $(\star)$ on a "Type I" region

A Type I region is one that can be written as $D = \{(x,y) : a \le x \le b,\; g_1(x) \le y \le g_2(x)\}.$ Why this shape? Because then for each $x$ the region is a vertical strip from a bottom curve $y=g_1(x)$ to a top curve $y=g_2(x)$ — easy to integrate in $y$.

Right-hand side. Integrate $-P_y$ over $D$, doing the inner $y$-integral first: $-\iint_D \frac{\partial P}{\partial y}\,dA = -\int_a^b\!\!\int_{g_1(x)}^{g_2(x)} \frac{\partial P}{\partial y}\,dy\,dx.$ Why this step? The inner integral is a perfect derivative in $y$, so the Fundamental Theorem of Calculus collapses it: $\int_{g_1(x)}^{g_2(x)}\frac{\partial P}{\partial y}\,dy = P(x,g_2(x)) - P(x,g_1(x)).$ Hence -\iint_D P_y\,dA = \int_a^b \big[P(x,g_1(x)) - P(x,g_2(x))\big]\,dx. \tag{RHS}

Left-hand side. Split the boundary $C$ into bottom curve $C_1$ ($y=g_1$, left→right) and top curve $C_2$ ($y=g_2$, right→left), plus the two vertical sides. Why split? Because $\oint P\,dx$ only "sees" horizontal motion — on vertical sides $dx=0$, so they contribute nothing.

= \int_a^b P(x,g_1(x))\,dx + \int_b^a P(x,g_2(x))\,dx.$$ *Why the flipped limits on $C_2$?* Counterclockwise orientation traverses the top curve **right to left**, i.e. $x$ from $b$ to $a$. Flip it to standard order: $$\oint_C P\,dx = \int_a^b \big[P(x,g_1(x)) - P(x,g_2(x))\big]\,dx. \tag{LHS}$$ **(LHS) = (RHS).** Done: $(\star)$ holds. ✅ ### Step 2: Prove $(\star\star)$ on a "Type II" region By the *mirror argument* — write $D = \{c\le y\le d,\ h_1(y)\le x\le h_2(y)\}$, integrate $Q_x$ first in $x$, use FTC, and match the left/right boundary pieces (now vertical sides carry the integral since $dy\ne 0$). Same logic with the roles of $x,y$ swapped. This gives $(\star\star)$. ### Step 3: General regions by decomposition > [!intuition] Cancelling interior cuts > A region that isn't simultaneously Type I and Type II is **chopped into sub-pieces** that are. Apply the theorem to each piece. When you add the line integrals, every **internal cut is traversed twice in opposite directions**, so those contributions cancel pairwise. Only the outer boundary survives — leaving Green's theorem on the whole region. ### Step 4: Get the flux form for free Take the circulation form and apply it to the rotated field $\mathbf{G} = (-Q', P')$... cleaner: start from $\oint \mathbf{F}\cdot\mathbf{n}\,ds$. On a curve with unit tangent $\mathbf{T}=(dx/ds, dy/ds)$, the outward normal is $\mathbf{n}=(dy/ds, -dx/ds)$. *Why?* Rotating the tangent **clockwise by 90°** points outward when the region is on your left. So $$\mathbf{F}\cdot\mathbf{n}\,ds = (P,Q)\cdot(dy,-dx) = P\,dy - Q\,dx.$$ Now apply the **circulation form** to the field $(\tilde P,\tilde Q)=(-Q,P)$: $$\oint -Q\,dx + P\,dy = \iint_D\Big(\frac{\partial P}{\partial x} - \frac{\partial(-Q)}{\partial y}\Big)dA = \iint_D (P_x + Q_y)\,dA.$$ The left side is exactly $\oint \mathbf{F}\cdot\mathbf{n}\,ds$. Flux form proved. ✅ --- ## Worked Examples > [!example] 1 — Area via the boundary > Show $\text{Area}(D) = \tfrac12\oint_C (x\,dy - y\,dx)$ and use it on the unit disk. > > Choose $P=-y/2,\ Q=x/2$. Then $Q_x - P_y = \tfrac12-(-\tfrac12)=1$. > *Why this choice?* We engineer the integrand to equal $1$, so the double integral becomes $\iint_D 1\,dA = \text{Area}$. > $$\oint_C \tfrac12(x\,dy - y\,dx) = \iint_D 1\,dA = \text{Area}(D).$$ > Unit circle $x=\cos t,\ y=\sin t$: $x\,dy - y\,dx = \cos^2 t + \sin^2 t = 1$, so > $$\text{Area}=\tfrac12\int_0^{2\pi}1\,dt = \pi.\ ✅$$ > [!example] 2 — Direct verification > $\mathbf{F}=(P,Q)=(y^2, 3xy)$ around the unit circle. > $$Q_x - P_y = 3y - 2y = y.$$ > *Why compute this first?* The double integral is usually easier than the line integral, and it tells us the answer to expect. > $$\iint_D y\,dA = 0$$ > *Why zero?* The disk is symmetric about $y=0$ and $y$ is odd in $y$ — positive and negative halves cancel. So $\oint_C \mathbf{F}\cdot d\mathbf{r}=0$ without computing the line integral at all. > [!example] 3 — Flux form > $\mathbf{F}=(x,y)$ across the boundary of the unit disk. > $\nabla\cdot\mathbf{F}=P_x+Q_y = 1+1 = 2$. > $$\oint_C \mathbf{F}\cdot\mathbf{n}\,ds = \iint_D 2\,dA = 2\pi.$$ > *Why this is intuitive:* $\mathbf F$ points radially outward; the total outward flux through the circle should grow with the area — and it does, at rate $2$ per unit area. --- ## Common Mistakes > [!mistake] "It works for any closed curve I like." > **Why it feels right:** the formula looks self-contained. **The catch:** $C$ must be **simple** (no self-crossings) and $P,Q$ must have continuous partials *everywhere inside* $D$. A singularity like $\mathbf{F}=(-y,x)/(x^2+y^2)$ at the origin **breaks the hypothesis**, giving $\oint = 2\pi$ but $Q_x-P_y=0$. **Fix:** check for singularities; exclude them with a small hole. > [!mistake] Forgetting orientation. > **Why it feels right:** "a loop is a loop." **Fix:** clockwise flips the sign. Keep the region on your **left**; if a problem gives clockwise, prepend a minus sign. > [!mistake] Confusing the two forms' integrands. > **Why it feels right:** both are sums/differences of partials. **Fix:** circulation = $Q_x - P_y$ (**curl**); flux = $P_x + Q_y$ (**divergence**). Mnemonic below. --- > [!recall]- Feynman: explain to a 12-year-old > Imagine a flat field with little spinning tops scattered all over it. If you want to know the **total spin** of all of them, you don't have to count each top — you can just walk all the way around the fence at the edge of the field, measuring how much the wind pushes you *along* the fence. The push along the boundary adds up to exactly the total spin inside! And if instead you measure how much wind blows *out through* the fence everywhere, that equals how much "new wind" is being created inside. The edge always knows what's happening inside. > [!mnemonic] Curl vs Divergence > **C**irculation has a **C**url → "**Q**uick **x** beats **P**lodding **y**": $Q_x - P_y$ (a **difference**). > **D**ivergence is **D**iagonal/straight-out → $P_x + Q_y$ (a **sum**, both "own" derivatives). --- ## Flashcards #flashcards/maths Green's theorem (circulation form)? ::: $\oint_C P\,dx+Q\,dy=\iint_D (Q_x-P_y)\,dA$, $C$ positively oriented simple closed curve bounding $D$. Green's theorem (flux form)? ::: $\oint_C \mathbf F\cdot\mathbf n\,ds=\iint_D(P_x+Q_y)\,dA=\iint_D \nabla\cdot\mathbf F\,dA$. What scalar does $Q_x-P_y$ represent? ::: The scalar (z-component of the) curl of $\mathbf F=(P,Q)$. Required orientation of $C$? ::: Counterclockwise (positive); region on your left. Why do vertical sides not matter when proving $\oint P\,dx$? ::: On vertical segments $dx=0$, so they contribute nothing to $\int P\,dx$. Which classical theorem powers the inner integral in the proof? ::: The Fundamental Theorem of Calculus (collapses $\int P_y\,dy$). How are general regions handled? ::: Chop into Type I/II pieces; internal cuts traverse twice oppositely and cancel. Area formula from Green's theorem? ::: $\text{Area}=\tfrac12\oint_C(x\,dy-y\,dx)$. Outward normal in terms of arc length? ::: $\mathbf n=(dy/ds,\,-dx/ds)$ (tangent rotated clockwise 90°). Why can a singularity inside $D$ break the theorem? ::: $P,Q$ must have continuous partials on all of $D$; a singularity violates the hypothesis (e.g. $(-y,x)/r^2$ gives $\oint=2\pi$ but integrand $=0$). --- ## Connections - [[Fundamental Theorem of Calculus]] — Green's theorem is its 2D generalization. - [[Stokes' Theorem]] — Green's is Stokes restricted to a flat region in the plane. - [[Divergence Theorem]] — the flux form is the 2D divergence theorem. - [[Curl and Divergence]] — supply the two integrands. - [[Line Integrals]] and [[Double Integrals]] — the two sides being equated. - [[Conservative Vector Fields]] — if $Q_x=P_y$ everywhere, all loops give $0$. ## 🖼️ Concept Map ```mermaid flowchart TD GT[Green's theorem] FTC[Fundamental Thm of Calculus] Circ[Circulation tangential form] Flux[Divergence normal form] Orient[Counterclockwise orientation] Curl[Scalar curl Qx minus Py] Split[Split into two halves] TypeI[Type I region] InnerY[Inner y-integral] Bound[Boundary curves C1 and C2] FTC -->|2D ancestor of| GT GT -->|form A| Circ GT -->|form B| Flux Circ -->|same theorem as| Flux Circ -->|integrand is| Curl Curl -->|requires| Orient Orient -->|region on left| GT Circ -->|proof splits into| Split Split -->|proven on| TypeI TypeI -->|do| InnerY InnerY -->|collapsed by| FTC TypeI -->|split boundary into| Bound Bound -->|matches RHS| InnerY ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Green's theorem ka core idea bahut simple hai: **boundary par jo ho raha hai, woh andar ke total ke barabar hota hai.** Yaani ek closed loop $C$ ke around line integral $\oint P\,dx+Q\,dy$ exactly equal hota hai us region $D$ ke double integral $\iint(Q_x-P_y)\,dA$ ke. Ye Fundamental Theorem of Calculus ka 2D version hai — jaise FTC mein endpoints poora integral yaad rakhte hain, waise hi yahan boundary poore area ka hisaab rakhti hai. > > Do forms hain. **Circulation form** mein integrand $Q_x-P_y$ aata hai — ye **curl** hai, yaani local ghoomna (counterclockwise spin). **Flux form** mein $P_x+Q_y$ aata hai — ye **divergence** hai, yaani kitna bahar nikal raha hai. Dono ek hi theorem hain, bas field ko 90° ghuma do toh ek se doosra mil jaata hai. Yaad rakho: circulation = **difference** (curl), divergence = **sum**. > > Proof ka trick: pehle region ko "Type I" maano (har $x$ ke liye neeche se upar tak strip). Inner $y$-integral ko FTC se collapse karo — woh top minus bottom curve ban jaata hai, aur exactly boundary line integral se match ho jaata hai. Vertical sides par $dx=0$ hota hai isliye woh count nahi hote. Phir Type II ke liye same logic $x,y$ swap karke. Agar region complicated hai toh use chote pieces mein kaato — internal cuts do baar opposite direction mein traverse hote hain, isliye cancel ho jaate hain, sirf outer boundary bachti hai. > > Do warning: (1) **orientation** counterclockwise honi chahiye, region tumhare left mein — clockwise pe sign ulta. (2) Andar koi **singularity** nahi honi chahiye (jaise $(-y,x)/r^2$ origin par), warna theorem fail. Exam mein zyada tar double integral easy hota hai, isliye line integral ko double integral mein convert karke solve karo — kaafi time bachta hai. ![[audio/4.4.29-Green's-theorem-—-proof-sketch,-both-forms.mp3]]