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

4.4.33 · D5 · Maths › Multivariable Calculus › Divergence theorem (Gauss's theorem) — statement, flux-diver

Ye page sirf woh vocabulary assume karta hai jo parent note ne build ki: flux $Φ = \iint_{S} F \cdot \hat{n} d S$ (flow across the skin), divergence $\nabla \cdot F = \partial_{x} P + \partial_{y} Q + \partial_{z} R$ (source strength per unit volume), aur theorem $\iint_{S} F \cdot \hat{n} d S = ∭_{V} (\nabla \cdot F) d V$ . Baki sab jaise-jaise aata hai waise define hota hai.

True or false — justify karo

Divergence theorem ko

S

ek closed surface honi chahiye.

True — ek "leaky" open surface ka koi enclosed volume

V

nahi hota, isliye right-hand side

∭_{V}

undefined hai; sirf woh boundary

S = \partial V

qualify karti hai jo ek solid ko poori tarah seal kare.

Agar

\nabla \cdot F = 0

everywhere inside

V

ho, toh

S

se bahar flux zero hona chahiye.

True provided

F

V

mein koi singularity na ho: tab

∭_{V} 0 d V = 0

. Agar andar koi singularity chhupa ho, toh integrand wahan

0

nahi hai aur flux nonzero ho sakta hai (jaise

r /∣ r ∣^{3}

gives

4 π

Do alag closed surfaces jo same region ko enclose karti hain hamesha same flux deti hain.

True by definition — "same region" ka matlab hai same

V

, aur flux

∭_{V} (\nabla \cdot F) d V

ke barabar hai, jo sirf

V

aur

F

pe depend karta hai, na ki aap surface ko kaise describe karte ho.

Do closed surfaces jo alag regions ko enclose karti hain phir bhi equal flux de sakti hain.

True — agar unke beech ke shell mein

\nabla \cdot F = 0

ho, toh extra volume zero contribute karta hai, isliye dono surfaces same flux report karti hain (yahi trick Gauss's law ke peeche hai).

Ek field jo

S

pe everywhere large ho uska flux bhi large hona chahiye.

False — sirf

\hat{n}

ke along component matter karta hai. Ek bahut bada field jo surface ke tangent flow kar raha ho (jaise

(y, - x, 0)

swirl karta hua)

F \cdot \hat{n}

mein kuch contribute nahi karta.

Ek point pe positive divergence guarantee karta hai ki uske around kisi bhi surface se total flux bahar positive ho.

False — total flux poore volume pe divergence sum karta hai;

V

mein kahin aur strong sinks (negative divergence) ek positive point ko outweigh kar sakti hain, net flux negative bana sakti hain.

Ye theorem sirf FTC ka 3D copy hai.

True in spirit — dono ek region ke integral ko uske boundary pe evaluation ke liye trade karte hain. FTC:

\int_{a}^{b} f^{'} = f (b) - f (a)

; Gauss:

∭_{V} \nabla \cdot F = \iint_{\partial V} F \cdot \hat{n}

. Ek interval ki boundary do points hoti hai; ek solid ki boundary uski skin hoti hai.

Aap cube

[0, 1]^{3}

F = (1/ x, 0, 0)

ke liye divergence theorem apply kar sakte ho.

False —

1/ x

aur uske partial face

x = 0

pe blow up karte hain, isliye

F

mein

V

throughout continuous first partials nahi hain; theorem ki hypothesis violate hoti hai.

Divergence theorem aur Stokes' theorem ko same integral pe interchangeably apply kiya ja sakta hai.

False — Gauss ek flux (

F \cdot \hat{n}

, across) ko volume integral mein convert karta hai; Stokes circulation (

F \cdot d r

, ek curve ke along) ko curl ke surface integral mein convert karta hai. Alag dot products, alag objects.

Inward normal ki taraf reverse karne se sirf flux ka sign badalta hai, size nahi.

True —

\hat{n} \to - \hat{n}

everywhere

F \cdot \hat{n}

flip karta hai,

- ∭_{V} \nabla \cdot F d V

deta hai; magnitude unchanged rehta hai lekin theorem jaise stated hai outward choice require karta hai.

Error dhundho

\nabla \cdot F = 0

toh

F

ki koi field lines

S

cross nahi karti."

Galat conclusion — divergence-free ka matlab hai net flux zero hai (lines in = lines out), nahi ki koi lines cross nahi karti. Bahut saari lines

S

pierce kar sakti hain aur phir bhi zero pe balance ho sakti hain.

F = (y, - x, 0)

ka zero flux hai, toh ye zero field hai."

Nahi — field nonzero hai aur swirl karti hai (iska curl hai); ye solenoidal (divergence-free) hai, jo sirf net flux ko zero force karta hai, field ko vanish nahi karta.

"Unit sphere ke liye

F = (x, y, z)

ka flux

4 π

hai, toh radius-2 sphere ka flux bhi

4 π

hai."

Error —

\nabla \cdot F = 3

, isliye flux

= 3 \cdot Vol

. Radius-2 sphere ka volume

\frac{4}{3} π (2)^{3}

hai, flux

32 π

deta hai,

4 π

nahi. Constant divergence ka matlab constant flux nahi hota.

"Kyunki divergence ek derivative hai, divergence theorem ko

F

continuous chahiye lekin differentiable nahi."

Requirement zyada strong hai:

F

V

throughout continuous first partial derivatives hone chahiye, kyunki

\nabla \cdot F

unhi partials se bana hai aur integrable hona chahiye.

F = r /∣ r ∣^{3}

\nabla \cdot F = 0

hai, toh kisi bhi sphere se iska flux

0

hai."

Divergence origin ko chhodkar zero hai, jahan field singular hai. Agar sphere origin ko enclose karti hai, toh theorem directly apply nahi hota aur actual flux

4 π

hai — pehle origin ke around ek tiny sphere carve out karo.

"Theorem use karne ke liye mujhe pehle cube ke saare chhe faces parameterize karne padenge."

Ulta hai — theorem ka point hi surface work avoid karna hai: chhe face integrals ko ek volume integral

∭_{V} \nabla \cdot F d V

se replace karo.

Why questions

Jab hum tiny boxes ko glue karte hain toh interior faces kuch contribute kyun nahi karte?

Ek shared wall ek box ki outward face hai aur uske neighbor ki inward face, isliye do flux contributions size mein equal aur sign mein opposite hote hain — ye exactly cancel ho jaate hain, sirf exterior skin bachti hai.

Flux mein sirf normal component

F \cdot \hat{n}

kyun enter karta hai?

Surface ke parallel flow skin ke along slide karta hai bina escape kiye; sirf

S

ke perpendicular motion actually cheezein cross karata hai, aur

\hat{n}

ke saath dot product precisely woh perpendicular part extract karta hai.

Divergence "flux per unit volume" kyun hai na ki total?

Ye ek local quantity hai — (ek tiny box se net flux bahar)/(box volume) ka limit jab box shrink hota hai — isliye ise

V

pe integrate karna (local rates ko unke volumes se multiply karna aur sum karna) total flux recover karta hai.

Theorem humein ek hard surface integral ko easier volume se swap kyun karne deta hai?

Kyunki dono same physical cheez count karte hain (net production = net escape); hum woh description choose karte hain jo integrate karna simpler ho, aksar volume jab

\nabla \cdot F

ek simple polynomial ho.

Outward orientation, sirf "a normal" nahi, kyun essential hai?

Answer ka sign net flow ki direction encode karta hai; "outward positive" woh convention hai jo source strength (

\nabla \cdot F > 0

) ko positive escaping flux se correspond karata hai. Ek inward normal is bookkeeping ko reverse karta hai.

Continuity equation naturally is theorem se kyun follow karti hai?

Mass-flux field pe Gauss apply karne se "mass leaving through

S

" turn hota hai "

∭_{V} \nabla \cdot (ρ v) d V

" mein; use rate se equate karna jis par mass andar girta hai local law

\partial_{t} ρ + \nabla \cdot (ρ v) = 0

deta hai jab volume arbitrary ho.

Edge cases

Kya hota hai agar

V

ek single point tak shrink ho jaaye?

Dono sides

\to 0

, lekin ratio flux/volume us point pe

\nabla \cdot F

tend karta hai — yeh limiting case actually divergence ki definition hai, woh seed jisse poora theorem grow karta hai.

Agar

V

ke andar ek hole (cavity) ho, jaise ek spherical shell?

Boundary

S = \partial V

dono surfaces hain — outer sphere outward normal ke saath aur inner sphere woh normal ke saath jo cavity ke andar point kare (solid se bahar). Flux dono pe sum karna hoga, har ek correctly oriented.

Agar

\nabla \cdot F

V

ke ek part mein positive aur doosre mein negative ho?

Volume integral signed contributions add karta hai, isliye sources aur sinks partially ya poori tarah cancel ho jaate hain; flux sirf net imbalance report karta hai, jo positive, negative, ya zero ho sakta hai.

Agar closed surface

S

exactly

F

ki singularity se hokar guzre?

Wahan theorem undefined hai —

F \cdot \hat{n}

S

pe unbounded hai aur hypotheses fail hote hain. Aapko

S

ko singular point se bachne ke liye deform karna hoga (ya ise improper limit ki tarah handle karna hoga).

Kisi bhi closed surface se ek constant field

F = c

ka flux kya hai?

Zero —

\nabla \cdot c = 0

, isliye

∭_{V} 0 d V = 0

. Geometrically, ek taraf se jo constant flow enter karta hai woh doosri taraf se nikalta hai, koi net gain nahi.

Agar region

V

simply connected nahi hai ya awkward shape ka hai?

Theorem phir bhi hold karta hai jab tak

S = \partial V

ek genuine closed boundary hai aur

F

V

pe smooth hai; hum simply "box-friendly" pieces mein subdivide karte hain jinke shared interior faces cancel ho jaate hain, exactly jaise derivation mein.

Recall One-line survival kit

Gauss apply karne se pehle teen questions poocho: (1) Kya $S$ closed aur outward-oriented hai? (2) Kya $F$ andar everywhere smooth hai — koi hidden singularities nahi? (3) Kya main flux compute kar raha hoon (across, divergence use karo) ya circulation (along, Stokes' theorem / Green's theorem use karo)? Agar koi bhi jawab "no" hai, toh equation trust karne se pehle fix karo.

Connections

Flux integrals — "across" side jise ye traps test karte rehte hain.
Divergence and curl — divergence vs. curl flux-vs-circulation confusion ke neeche hai.
Stokes' theorem · Green's theorem — circulation cousins.
Gauss's law (electromagnetism) · Continuity equation — jahan "same flux, different surface" aur local law aate hain.