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

4.4.33 · D3 · Maths › Multivariable Calculus › Divergence theorem (Gauss's theorem) — statement, flux-diver

Yeh page ek case zoo hai. Parent note ne theorem prove kiya aur teen examples dikhaye. Yahan hum har tarah ki situation dhundte hain jo theorem tumhare saamne rakh sakta hai — positive aur negative divergence, zero divergence, ek degenerate (flat) region, ek limiting shrink-to-a-point case, ek singularity jo rule todta hai, ek physics word problem, aur ek exam trap. Har ek line by line work kiya gaya hai, aur har ek ka jawab deta hai Kya / Kyun / Kaisa dikhta hai.

Shuru karne se pehle, ek reminder plain words mein, taaki koi bhi symbol bina samjhe andar na aa jaye.

Definition Teen characters

$F = (P, Q, R)$ ek vector field hai: space ke har point par yeh tumhe ek arrow deta hai. Socho "us jagah paani ki velocity."
$\nabla \cdot F = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}$ divergence hai: har point par ek single number jo batata hai ki arrows wahan kitni tezi se alag ho rahe hain (positive ho toh tap, negative ho toh drain, zero ho toh balanced).
$\iint_{S} F \cdot \hat{n} d S$ flux hai: total arrow-flow jo region $V$ ki skin $S$ se bahar ki taraf nikal raha hai — jahan "skin" plain-English shorthand hai us closed boundary surface ke liye jo solid $V$ ko poori tarah wrap karti hai (jaise andey ka shell uske andar ko wrap karta hai).

Symbol $∭_{V} (\dots) d V$ ka matlab hai "solid $V$ ko bharne wale har tiny cube of volume $d V$ par little number $(\dots)$ ko add karo." Theorem kehta hai ki yeh dono totals same number hain.

Scenario matrix

Har problem jo tum miloge, in cells mein se kisi ek mein aayega. Neeche ke examples mein unke cover kiye gaye cell ka label hai.

Cell	Kya special hai	$\nabla \cdot F$ ka sign	Example
A	Constant positive divergence (uniform source)	$> 0$	Ex 1
B	Positive divergence, position ke saath varies karta hai	$> 0$	Ex 2
C	Negative divergence (net sink / inflow)	$< 0$	Ex 3
D	Exactly zero divergence (solenoidal)	$= 0$	Ex 4
E	Degenerate region — flat/zero-volume	any	Ex 5
F	Limiting case — shrink volume to a point	any	Ex 6
G	Singularity andar — theorem naively fail karta hai	ek point par undefined	Ex 7
H	Real-world word problem (physics)	$> 0$	Ex 8
I	Exam twist — inward normal / hidden orientation	$> 0$	Ex 9

Nau cells, nau examples. Milke yeh positive/negative/zero divergence, degenerate aur limiting inputs, failure mode, ek application, aur ek trap cover karte hain.

Cell A — constant positive divergence

Worked example Ex 1 ·

$F = (2 x, 2 y, 2 z)$ radius $R$ ki ball se bahar Radius $R$ wale sphere se, jo origin par centred hai, outward flux nikalo.

Forecast: field har jagah seedha bahar ki taraf point karti hai aur distance ke saath badhti hai — guess: ek bada positive number jo $R^{3}$ ki tarah scale karta hai.

Divergence compute karo. $\nabla \cdot F = \frac{\partial ( 2 x )}{\partial x} + \frac{\partial ( 2 y )}{\partial y} + \frac{\partial ( 2 z )}{\partial z} = 2 + 2 + 2 = 6.$ Yeh step kyun? Theorem humein mushkil surface integral ko is number ke volume integral se swap karne deta hai, aur yahan yeh number constant hai — sabse aasaan possible integrand.
Ball par integrate karo. Constant bahar aa jaata hai: $∭_{V} 6 d V = 6 \cdot Vol (V) = 6 \cdot \frac{4}{3} π R^{3} = 8 π R^{3} .$ Yeh step kyun? $∭_{V} 1 d V$ by definition volume hai, aur radius $R$ ki ball ka volume $\frac{4}{3} π R^{3}$ hota hai.

Answer: $Φ = 8 π R^{3}$ .

Verify: $R = 1$ rakho: $Φ = 8 π$ . Unit sphere par directly $\hat{n} = (x, y, z)$ toh $F \cdot \hat{n} = 2 (x^{2} + y^{2} + z^{2}) = 2$ , aur $\iint_{S} 2 d S = 2 \cdot (4 π) = 8 π$ . ✓ Dono sides agree karte hain, aur $R^{3}$ growth forecast se match karti hai.

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

Figure mein sphere dikhaya gaya hai jisme field arrows bahar jaate jaate lambe hote hain — yeh "spreading" bilkul wahi hai jo positive divergence measure karta hai.

Cell B — positive but position-dependent divergence

Worked example Ex 2 ·

$F = (x^{2}, 0, 0)$ box $[0, 2] \times [0, 1] \times [0, 1]$ par Total outward flux nikalo.

Forecast: field sirf $x$ direction mein flow karti hai aur $x$ badhne ke saath tez hoti hai, toh far face ( $x = 2$ ) se zyada niklega near face ( $x = 0$ ) se ghusne se. Positive number expect karo.

Divergence. $\nabla \cdot F = \frac{\partial ( x ^{2} )}{\partial x} + 0 + 0 = 2 x .$ Yeh step kyun? Divergence ab constant nahi hai — yeh $x$ ke saath badhta hai. Theorem phir bhi apply hota hai; bas ab genuinely integrate karna hoga.
Integrate karo. $∭_{V} 2 x d V = \int_{0}^{1} \int_{0}^{1} \int_{0}^{2} 2 x d x d y d z .$ Yeh step kyun? Box ek product region hai, toh hum ek variable ko ek time par integrate kar sakte hain.
Pehle $x$ -integral karo. $\int_{0}^{2} 2 x d x = [x^{2}]_{0}^{2} = 4.$ Bache hue $y$ aur $z$ integrals mein se har ek length $1$ cover karta hai: $4 \cdot 1 \cdot 1 = 4.$ Yeh step kyun? Innermost variable integrate karne se triple integral ek simple product mein collapse ho jaata hai kyunki kuch bhi $y$ ya $z$ par depend nahi karta.

Answer: $Φ = 4$ .

Verify (surface side). Sirf woh do faces matter karte hain jo $x$ ke perpendicular hain (chaar side faces par $\hat{n}$ $F$ ke perpendicular hai, toh $F \cdot \hat{n} = 0$ ).

Far face $x = 2$ : $\hat{n} = (+ 1, 0, 0)$ , $F \cdot \hat{n} = x^{2} = 4$ , area $= 1$ ⟹ $+ 4$ .
Near face $x = 0$ : $\hat{n} = (- 1, 0, 0)$ , $F \cdot \hat{n} = - x^{2} = 0$ , area $= 1$ ⟹ $0$ . Total $= 4$ . ✓ Match karta hai, aur far face dominant hai jaise forecast tha.

Dono shaded faces par arrows dekho: left par ( $x = 0$ ) chhote, right par ( $x = 2$ ) lambe. Yeh mismatch hi flux hai, aur box ke across bhadte arrow length hi positive divergence $2 x$ hai.

Cell C — negative divergence (ek net sink)

Ex 1 ki figure se compare karo: wahan arrows bahar bhag rahe the (source, positive flux); yahan har arrow skin se andar ghus raha hai (sink, negative flux). Same shape, opposite sign.

Cell D — zero divergence (solenoidal)

Worked example Ex 4 ·

$F = (y, z, x)$ kisi bhi closed surface se Ek arbitrary closed surface $S$ se flux nikalo.

Forecast: components coordinates mix karte hain lekin unme se koi bhi apne variable ka function nahi hai — divergence shayad zero ho jaye, har region ke liye $0$ de.

Divergence. $\nabla \cdot F = \frac{\partial y}{\partial x} + \frac{\partial z}{\partial y} + \frac{\partial x}{\partial z} = 0 + 0 + 0 = 0.$ Yeh step kyun? Har partial derivative ek aisi variable ka hai jo usme appear nahi karta, toh har ek zero hai. Field divergence-free hai — solenoidal.
Integrate karo. $∭_{V} 0 d V = 0$ kisi bhi $V$ ke liye. Yeh step kyun? Kisi bhi cheez par zero integrate karne se zero milta hai — $S$ ki shape jaanne ki zarurat nahi.

Answer: $Φ = 0$ har closed surface ke liye.

Verify: answer $0$ hona chahiye regardless of $V$ ; ek solenoidal field kuch create nahi karta, toh jo andar jaata hai woh wapas bahar aa jaata hai. ✓ (Parent ka Ex 3 $F = (y, - x, 0)$ compare karo — same principle.)

Figure ek circle se solenoidal flow sketch karta hai: har arrow jo andar jaata hai (region mein) uss ke liye ek bahar jaane wala bhi hai. Ins aur outs exactly balance karte hain — yeh visual cancellation hi $Φ = 0$ kaisa dikhta hai.

Cell E — degenerate region (zero volume)

Worked example Ex 5 ·

$F = (x, y, z)$ ek flat region par (ek plane mein disk) Koi kehta hai: "divergence theorem ko flat disk $z = 0$ , $x^{2} + y^{2} \leq 1$ par apply karo." Kya hota hai?

Forecast: disk 2D hai — iska area hai lekin koi volume nahi. Agar theorem phir bhi kuch bolta hai, toh volume integral $0$ hona chahiye, toh flux bhi $0$ hona chahiye.

Hypothesis check karo. Divergence theorem ko $V$ ek solid (3D) region chahiye. Ek flat disk ka $Vol = 0$ hai; yeh ek degenerate input hai. Yeh step kyun? Theorem ka right-hand side $∭_{V} (\dots) d V$ hai. Agar $V$ ka volume zero hai, toh yeh integral $0$ hai chahe $F$ kuch bhi ho.
Honestly interpret karo. Ek genuine 3D region ke liye theorem flux deta hai. Flat disk ke liye koi "inside" nahi hai, toh sahi reading yeh hai: divergence theorem apply nahi hota — tumhe instead disk se seedha ordinary surface flux compute karna hoga (jo generally zero nahi hoga). Yeh step kyun? Degenerate cases woh jagah hai jahan careless students plug in karke $0$ paa lete hain; fix yeh hai ki recognize karo ki $V$ solid hona chahiye.

Answer: divergence theorem zero-volume region ke liye $0$ return karta hai, lekin woh number ek one-sided disk se flux ke roop mein meaningless hai — theorem simply sahi tool nahi hai. Direct surface integral use karo.

Verify: $F = (x, y, z)$ ka flux disk se seedha (normal $\hat{n} = (0, 0, 1)$ ) $\iint_{disk} z d S$ hai, aur $z = 0$ par yeh $\iint 0 d S = 0$ hai yahan — coincidentally $0$ , lekin ek general field ke liye nahi hota. Lesson: tool ko real solid ke saath match karo. ✓ (volume-integral $0$ ek hypothesis-failure artefact hai, physics nahi.)

Cell F — limiting case (ek point tak shrink karo)

Worked example Ex 6 ·

$F = (x^{2}, y^{2}, z^{2})$ origin par side $ε$ ke tiny cube par Box $[0, ε]^{3}$ lo aur $ε \to 0$ ke roop mein flux-per-volume dekho.

Forecast: parent note kehta hai "divergence flux per box hai." Toh $\frac{Φ}{Vol}$ ko $\nabla \cdot F$ origin par approach karna chahiye, jo wahan $2 x + 2 y + 2 z = 0$ hai. Guess: ratio $\to 0$ .

Divergence. $\nabla \cdot F = 2 x + 2 y + 2 z .$ Volume side: $∭_{[0, ε]^{3}} (2 x + 2 y + 2 z) d V .$ Yeh step kyun? Hum dekhna chahte hain ki divergence ka local meaning ek shrinking region se kaise emerge hota hai.
Integrate karo. $\int_{0}^{ε} 2 x d x = ε^{2}$ , aur baaki dono variables mein se har ek ka factor $ε$ contribute karta hai: toh $2 x$ term deta hai $ε^{2} \cdot ε \cdot ε = ε^{4}$ . Symmetry se teeno terms match karte hain: $Φ = 3 ε^{4} .$ Yeh step kyun? Har term symmetric box par same tarah integrate hota hai.
Volume $ε^{3}$ se divide karo aur limit lo. $\frac{Φ}{ε ^{3}} = \frac{3 ε ^{4}}{ε ^{3}} = 3 ε ε \to 0 0.$ Yeh step kyun? Yeh ratio hi box par average divergence hai; iska limit point par divergence hai — yahan $0$ , jo $2 (0) + 2 (0) + 2 (0)$ se match karta hai.

Answer: $\frac{Φ}{Vol} \to 0 = (\nabla \cdot F)_{origin} .$

Verify: limit $lim_{ε \to 0} 3 ε = 0$ origin par $\nabla \cdot F$ ke barabar hai. ✓ Yeh theorem se recover ki gayi divergence ki limiting definition hai.

Origin par nested teen cubes of shrinking side $ε$ : jaise box point par collapse hota hai, flux-per-volume number jo har ek ke paas likha hai ( $3 ε$ ) centre par divergence value $0$ ki taraf neeche jaata hai.

Cell G — singularity andar (theorem toot jaata hai)

Worked example Ex 7 ·

$F = \frac{r}{∣ r ∣ ^{3}}$ origin ke andar $S$ ke saath Yahan $r = (x, y, z)$ aur $∣ r ∣ = x^{2} + y^{2} + z^{2}$ . Origin ke around radius $R$ ke sphere se flux nikalo.

Forecast (trap): origin se door yeh field divergence-free hai (tum check kar sakte ho), toh ek naive student kehta hai $∭ 0 d V = 0$ . Lekin field origin par blow up karti hai — toh savdhan raho.

Origin se door divergence. Ek direct computation deta hai $\nabla \cdot F = 0$ har point par except $r = 0$ , jahan $F$ undefined hai (zero se division). Yeh step kyun? Theorem ki hypothesis demand karti hai continuous partials throughout $V$ — bilkul enclosed point par violate hota hai.
Honest surface integral karo. Radius $R$ ke sphere par, $∣ r ∣ = R$ toh $F = \frac{r}{R ^{3}}$ , aur $\hat{n} = \frac{r}{R}$ . Phir $F \cdot \hat{n} = \frac{r \cdot r}{R ^{3} \cdot R} = \frac{R ^{2}}{R ^{4}} = \frac{1}{R ^{2}} .$ Yeh step kyun? Jab theorem par trust nahi kiya ja sakta, flux directly surface par compute karo.
Sphere par integrate karo. $Φ = \iint_{S} \frac{1}{R ^{2}} d S = \frac{1}{R ^{2}} \cdot (4 π R^{2}) = 4 π .$ Yeh step kyun? $F \cdot \hat{n}$ sphere par constant hai, toh hum sirf sphere ki area $4 π R^{2}$ se multiply karte hain.

Answer: $Φ = 4 π$ — nahi $0$ , aur $R$ se independent.

Verify: $\frac{1}{R ^{2}} \cdot 4 π R^{2} = 4 π$ kisi bhi $R$ ke liye. ✓ Naive " $0$ " galat hai bilkul enclosed singularity ki wajah se — fix yeh hai ki origin ko exclude karo (yahi Gauss's law ka mathematical core hai).

Picture mein outer sphere $S$ dikhayi gayi hai jisme origin (spike) pink mein centre par marked hai. Kyunki field wahan explode karti hai, hume theorem par trust karne se pehle ek tiny inner sphere carve out karni padegi — yahi reason hai naive " $0$ " fail kyun hota hai.

Cell H — real-world word problem

Worked example Ex 8 · Heat ek warming metal cube se nikalna

Ek $1 m$ cube $[0, 1]^{3}$ ek heat-flux field $F = (3 x, 3 y, 3 z)$ carry karta hai jo watts per square metre ( $W / m^{2}$ ) mein measure hai. Cube ki surface se kitni total heat power nikalti hai?

Forecast: flux field corner se bahar ki taraf badhta hai — heat andar generate ho rahi hai. Positive wattage expect karo.

Divergence. $\nabla \cdot F = 3 + 3 + 3 = 9$ (units: $W / m^{3}$ , ek heat-source density). Yeh step kyun? Flux field ka divergence local generation rate per unit volume hai — continuity picture.
Cube par integrate karo. $∭_{[0, 1]^{3}} 9 d V = 9 \cdot (1 m)^{3} = 9 W .$ Yeh step kyun? Constant source density times volume ( $1 m^{3}$ ) total generated power deta hai, jo theorem ke anusaar skin se radiate hone wale total power ke barabar hai.

Answer: $9 W$ heat cube se nikalti hai.

Verify (units & value). $[W / m^{3}] \times [m^{3}] = [W]$ ✓. Numerically $9 \cdot 1 = 9$ . Surface check: far $x = 1$ face contribute karta hai $F \cdot \hat{n} = 3 x = 3$ area $1$ par = $3 W$ ; near $x = 0$ face deta hai $0$ ; same $y$ aur $z$ ke liye ⟹ $3 \cdot 3 = 9 W$ . ✓

Cell I — exam twist (inward normal / hidden orientation)

Worked example Ex 9 · Sign trap

Ek exam poochta hai: " $F = (x, y, z)$ ; unit sphere se $\iint_{S} F \cdot \hat{n} d S$ compute karo, jahan $\hat{n}$ inward normal hai."

Forecast: divergence theorem outward normal ke liye bana hai. Inward mein flip karne se sign flip hona chahiye — $+ 4 π$ nahi, $- 4 π$ expect karo.

Pehle outward answer compute karo. $\nabla \cdot F = 1 + 1 + 1 = 3$ , toh outward flux $= 3 \cdot \frac{4}{3} π (1)^{3} = 4 π .$ Yeh step kyun? Theorem sirf outward orientation mein bolta hai, toh pehle woh number cleanly nikalo kuch aur karne se pehle.
Inward normal ke liye sign flip karo. Inward normal $- \hat{n}_{out}$ hai, toh $\iint_{S} F \cdot \hat{n}_{in} d S = \iint_{S} F \cdot (- \hat{n}_{out}) d S = - \iint_{S} F \cdot \hat{n}_{out} d S = - 4 π .$ Yeh step kyun? Normal reverse karne se har dot product $F \cdot \hat{n}$ ka sign reverse hota hai, isliye pure integral ka bhi. Yahi trap hai jo parent ke mistakes mein warn kiya gaya tha.

Answer: $Φ_{in} = - 4 π$ (jabki outward $Φ_{out} = + 4 π$ ).

Verify: outward $= + 4 π$ (yeh Ex 1 hai $R = 1$ aur aadhi field ke saath: yahan $\nabla \cdot = 3$ , volume $\frac{4}{3} π$ , product $4 π$ ); inward ke liye negate karo $\Rightarrow - 4 π$ . ✓ Theorem ko touch karne se pehle hamesha orientation padho.

Left circle: outward normals (blue) bahar ki taraf pointing, flux $+ 4 π$ . Right circle: same field lekin inward normals (pink) andar ki taraf — har dot product ka sign flip hota hai, toh flux $- 4 π$ padta hai. Ek arrowhead direction hi poora fark hai.

Coverage check

Recall Kya humne har cell cover ki?

Positive constant divergence (A) ::: Ex 1 Position-dependent positive divergence (B) ::: Ex 2 Negative divergence / sink (C) ::: Ex 3 Zero divergence / solenoidal (D) ::: Ex 4 Degenerate zero-volume region (E) ::: Ex 5 Limiting shrink-to-a-point (F) ::: Ex 6 Singularity andar — theorem fail karta hai (G) ::: Ex 7 Real-world word problem with units (H) ::: Ex 8 Exam sign/orientation trap (I) ::: Ex 9

Connections

Divergence theorem (Gauss's theorem) — statement, flux-divergence connection — parent statement aur proof.
Flux integrals — surface side jiske against humne har example mein verify kiya.
Divergence and curl — scalar $\nabla \cdot F$ jo har answer drive karta hai.
Gauss's law (electromagnetism) — singularity case (Ex 7) iska mathematical core hai.
Continuity equation — Ex 8 mein use ki gayi source/sink reading.
Stokes' theorem · Green's theorem — circulation cousins (flow along, not across).