Divergence theorem (Gauss's theorem) — statement, flux-divergence connection

4.4.33 · Maths › Multivariable Calculus

Setup aur notation

• $V$ = 3D mein ek solid region, jiska ek closed, outward-oriented boundary surface $S=\partial V$ hai.
• $\mathbf{F}(x,y,z)=(P,Q,R)$ = ek vector field, $V$ pe continuous partial derivatives ke saath.
• $\hat{\mathbf{n}}$ = $S$ pe outward unit normal.
• $dS$ = scalar surface area element; $d\mathbf{S}=\hat{\mathbf{n}}dS$.

Statement

First principles se derivation

Hum ise ek "box-friendly" region ke liye derive karte hain — ek tiny cube se shuru karke sab add karte hain.

Step 1 — Ek tiny box se flux. Ek box $[x,x+\Delta x]\times [y,y+\Delta y]\times [z,z+\Delta z]$ lo. Sirf $P$-component (flow in $x$) consider karo. $x$ ke perpendicular do faces se bahar flux: $[P(x+\Delta x,\cdot ,\cdot )-P(x,\cdot ,\cdot )]\Delta y\Delta z.$ Yeh step kyun? Right face ka outward normal $+\hat{x}$ hai (contributes $+P$), left face ka normal $-\hat{x}$ hai (contributes $-P$). Unka area $\Delta y\Delta z$ hai.

Step 2 — Difference ko derivative mein badlo. $P(x+\Delta x)-P(x)\approx \frac{\partial P}{\partial x}\Delta x,$ toh $x$-faces contribute karte hain $\frac{\partial P}{\partial x}\Delta x\Delta y\Delta z=\frac{\partial P}{\partial x}\Delta V$. Yeh step kyun? Yahi partial derivative ki definition hai — $P$ mein chhota change ≈ slope × step.

Step 3 — $Q$ (y-faces) aur $R$ (z-faces) ke liye bhi yahi karo. Teeno pairs of faces add karne par, tiny box se net flux hai: $\left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\right)\Delta V=(\nabla \cdot \mathbf{F})\Delta V.$ Yeh theorem ka local version hai: divergence = flux-per-volume.

Step 4 — Boxes ko saath jodhna (telescoping / cancellation). $V$ ko bahut saare tiny boxes se bharo. Jab do boxes ek interior face share karte hain, to ek ka outward normal doosre ka inward normal hota hai, toh woh flux contributions exactly cancel ho jaate hain. Yeh step kyun? Jo box A us shared wall se bahar jaata hai wo box B mein jaata hai — internally barabar hai. Sirf exterior faces bachte hain, aur milke woh $S$ banate hain.

Step 5 — Sum karo aur limit lo. ${\sum }_{\text{boxes}}(\nabla \cdot \mathbf{F})\Delta V={\sum }_{\text{exterior faces}}\mathbf{F}\cdot \hat{\mathbf{n}}\Delta S.$ Boxes ko shrink karne do: left side $\to {\iiint }_V(\nabla \cdot \mathbf{F})dV$, right side $\to {\iint }_S\mathbf{F}\cdot \hat{\mathbf{n}}dS$. ∎

Worked examples

Common mistakes

Flashcards

Connections

• Flux integrals — surface side ${\iint }_S\mathbf{F}\cdot \hat{\mathbf{n}}dS$.
• Divergence and curl — divergence woh scalar hai jo yeh theorem integrate karta hai.
• Stokes' theorem — curl/circulation cousin (flow along boundary curve).
• Green's theorem — 2D special case (flux form).
• Gauss's law (electromagnetism) — physics application: $\oint \mathbf{E}\cdot d\mathbf{S}=Q_{enc}/{\epsilon }_0$.
• Continuity equation — local conservation laws differential (per-box) form se aate hain.

Concept Map

Deep Dive

• D1 · Foundations — har symbol zero se
• D2 · Visual walkthrough — derivation pictures mein
• D3 · Worked examples — har scenario
• D4 · Exercises — graded, full solutions
• D5 · Question bank — concept traps