Surface integrals — scalar and vector (flux): Question bank

4.4.31 · D5 · Maths › Multivariable Calculus › Surface integrals — scalar and vector (flux)

Jo core objects par tumhari testing hogi:

Stretch factor $∥ r_{u} \times r_{v} ∥$ — ek tiny parameter square curved surface par map hone par kitna area gain karta hai (built from the Cross product).
Scalar element $d S = ∥ r_{u} \times r_{v} ∥ d u d v$ (hamesha positive).
Vector element $d S = (r_{u} \times r_{v}) d u d v$ (sign / direction carry karta hai).

True or false — justify

Answers cover karo. Har reveal ek full reason hai.

Flux

\iint_{S} F \cdot d S

hamesha

\geq 0

hota hai jab

F

ke components positive hoon.

False. Flux chosen normal ki direction mein flow measure karta hai; agar field surface ke kisi hisse par normal ke opposite point kare, toh woh hissa negatively contribute karta hai, isliye total koi bhi sign ho sakta hai.

Ek scalar surface integral

\iint_{S} f d S

jahan

f \geq 0

ho, hamesha

\geq 0

hoti hai.

True. Yahan

d S = ∥ r_{u} \times r_{v} ∥ d u d v

ek magnitude hai (hamesha

\geq 0

), aur koi normal direction nahi hai jo sign carry kare, isliye nonnegative integrand nonnegative result deta hai.

Parametrization reverse karna (

u

aur

v

swap karna) scalar surface integral ki value badal deta hai.

False.

u, v

swap karne par

r_{u} \times r_{v}

ki jagah

r_{v} \times r_{u} = - (r_{u} \times r_{v})

aa jaata hai, lekin

d S

magnitude use karta hai, jo unchanged rehti hai; sirf flux (jo sign rakhta hai) flip hota hai.

Stretch factor

∥ r_{u} \times r_{v} ∥

Jacobian ka 3D analogue hai jo Double integrals mein use hota hai.

True. Dono ek tiny parameter-region area ko map ke baad actual area mein convert karte hain; Jacobian yeh ek plane ke andar karta hai,

∥ r_{u} \times r_{v} ∥

yeh ek plane ko 3D space mein map karne ke liye karta hai.

Flux integral ke liye tumhe hamesha square root

∥ r_{u} \times r_{v} ∥

compute karna padta hai.

False.

\iint (F \cdot \hat{n}) d S

mein

\hat{n}

∥ \dots ∥

aur

d S

∥ \dots ∥

cancel ho jaate hain, aur bacha rehta hai

F \cdot (r_{u} \times r_{v}) d u d v

— koi root nahi chahiye.

Har smooth closed surface orientable hoti hai.

True. Ek smooth closed surface (jaise sphere) ka consistent inside/outside hota hai, isliye tum "outward" har jagah choose kar sakte ho; Möbius strip jaisi non-orientable surfaces exceptions hain aur closed nahi hoti.

Agar do alag parametrizations ek hi oriented surface dete hain, toh dono same flux denge.

True. Flux sirf surface, field, aur chosen normal direction par depend karta hai — kaise tum parameters sweep karte ho usp par nahi — jab tak orientation (normal direction) agree kare.

Vector

r_{u} \times r_{v}

hamesha unit vector hota hai.

False. Iski length stretch factor

∥ r_{u} \times r_{v} ∥

hai, jo generally

1

nahi hoti; unit normal

\hat{n}

paane ke liye tumhe us length se divide karna padega.

Graph

z = g (x, y)

ke liye, element

r_{x} \times r_{y} = (- g_{x}, - g_{y}, 1)

hamesha upward point karta hai.

True. Iska

z

-component

+ 1 > 0

hota hai

g

chahe kuch bhi ho, isliye yeh specific parametrization hamesha upward-pointing normal deta hai.

Spot the error

Har line mein ek flawed move bataya gaya hai; reveal galti ka naam aur fix batata hai.

"Kyunki

F = (x, y, z)

hai aur sphere par

r

outward normal hai, main flux ke liye

F \cdot r

use karunga."

Vector area element

r_{ϕ} \times r_{θ}

hai, $r$ nahi. Yeh sirf sphere par

r

ke parallel hota hai; correct object hamesha tangent vectors ka cross product hota hai.

\iint_{S} f d S = \iint_{D} f (x, y, z) d u d v

Do errors hain:

∥ r_{u} \times r_{v} ∥

stretch factor missing hai, aur

f

f (r (u, v))

ke roop mein rewrite karna hoga — integrate karne se pehle sab kuch

u, v

ka function banana padega.

"Density

f

wali curved sheet ki mass nikalne ke liye,

f

ko flat shadow region

D

par integrate karo."

Tum stretch bhool gaye: slanted patches apne shadow se zyada real area cover karte hain, isliye mass

= \iint_{D} f ∥ r_{u} \times r_{v} ∥ d u d v

hai,

\iint_{D} f d u d v

nahi.

"Flux negative aaya, toh maine arithmetic mein galti ki."

Zaruri nahi — negative flux ka matlab sirf yeh hai ki net flow tumhare chosen normal ke opposite hai. Yeh tabhi galat hai jab yeh problem ki maangi gayi orientation se contradict kare.

z = g (x, y)

se flux ke liye main

\iint_{D} (P + Q + R) d x d y

use karunga."

Tum slope weighting bhool gaye.

F = (P, Q, R)

(- g_{x}, - g_{y}, 1)

se dot karne par

\iint_{D} (- P g_{x} - Q g_{y} + R) d x d y

milta hai.

"Maine Divergence Theorem se check kiya lekin surface ek hemisphere hai jiska bottom open hai, isliye flux

= ∭ \nabla \cdot F d V

Divergence Theorem ko closed surface chahiye. Volume integral ke saath equate karne se pehle tumhe hemisphere ko cap karne wali flat disk ka flux bhi add karna hoga.

"Maine surface parametrize ki, lekin

F

mein original

x, y, z

daale aur

D

par integrate kiya."

F

r (u, v)

par evaluate karna hoga, yaani

F (x (u, v), y (u, v), z (u, v))

, taaki integrand

D

par

u, v

ka genuine function ban sake.

Why questions

Mapped parallelogram ka area

∥ r_{u} \times r_{v} ∥

kyun hota hai,

∥ r_{u} ∥ \cdot ∥ r_{v} ∥

kyun nahi?

Kyunki tangent vectors perpendicular nahi hote;

∥ r_{u} \times r_{v} ∥ = ∥ r_{u} ∥∥ r_{v} ∥ sin θ

unke beech ka angle account karta hai (base times height), jabki plain product overcount karta.

F

ka sirf

\hat{n}

ke along wala component flux mein kyun contribute karta hai?

Surface ke parallel flow uspe slide karta hai bina through gaye; sirf perpendicular part actually cross karta hai, aur

F \cdot \hat{n}

exactly wahi perpendicular component extract karta hai.

Flux aksar same surface par scalar surface integral se zyada aasaan kyun hota hai compute karna?

Scalar integral mein awkward square root

∥ r_{u} \times r_{v} ∥

rehta hai, jabki flux mein woh norm cancel ho jaata hai, aur sirf plain polynomial-jaisa dot product

F \cdot (r_{u} \times r_{v})

bachta hai.

Hum

d S

banane ke liye Cross product kyun use karte hain, instead of dot product ke?

Humein do vectors se area aur dono ke perpendicular ek direction chahiye; cross product dono deliver karta hai (iski magnitude area hai, direction normal hai), jo dot product nahi kar sakta.

\hat{n}

- \hat{n}

choose karna flux ke liye kyun matter karta hai lekin scalar integral ke liye nahi?

Flux ek signed directional quantity hai ("kis taraf se through?"), isliye normal ki direction sign set karti hai; scalar integral sirf positive area contributions add karta hai aur reverse karne ke liye koi direction nahi hai.

Graph case mein

∥ r_{x} \times r_{y} ∥ = 1 + g_{x}^{2} + g_{y}^{2}

kyun aata hai, aur geometrically extra

\geq 1

ka kya matlab hai?

1

flat shadow contribution hai aur

g_{x}^{2} + g_{y}^{2}

tilt add karta hai;

\geq 1

hona matlab hai ki ek sloped patch apne horizontal shadow se hamesha bada hota hai, kabhi chota nahi.

Möbius strip ka consistent orientation kyun nahi ho sakta?

Normal ko strip ke around ek baar slide karne par woh opposite direction mein laut aata hai, isliye

\hat{n}

ka koi continuous global choice exist nahi karta — surface ka sirf "ek side" hai.

Edge cases

Agar

F

har jagah surface ke tangent ho, toh flux kya hoga?

Bilkul zero, kyunki

F \cdot \hat{n} = 0

har point par hoga — kuch bhi through cross nahi karta, sab kuch along slide karta hai.

\iint_{S} f d S

kya deta hai jab

f \equiv 1

ho?

Surface area khud, kyunki tum sirf har patch ka area bina kisi weighting ke sum kar rahe ho.

Sphere jo

(ϕ, θ)

se parametrize hai, uske equator par

ϕ = π /2

hai; poles

ϕ = 0, π

par

sin ϕ

kya special role play karta hai?

Spherical area factor

sin ϕ

poles par vanish ho jaata hai, yeh correctly reflect karta hai ki parameter grid squares wahan zero area mein collapse ho jaate hain (ek coordinate degeneracy, koi hole nahi).

Agar ek surface sirf piecewise smooth ho (jaise cube with edges), kya tum phir bhi usp par integrate kar sakte ho?

Haan — ise smooth faces mein split karo, har ek ko alag integrate karo, aur add karo; edges measure-zero hain aur kuch contribute nahi karte, lekin flux ke liye har face ki orientation consistent (sab outward) rakhni hogi.

Agar surface ke andar koi net source nahi hai aur

\nabla \cdot F = 0

ho ek closed region mein, toh flux kya hoga?

Divergence Theorem se total outward flux zero hoga — jitna fluid enter karta hai utna hi leave karta hai, isliye field us closed surface ke across "incompressible" hai.

Kya

∥ r_{u} \times r_{v} ∥

kabhi zero ho sakta hai, aur yeh kya signal deta hai?

Haan, jahan bhi

r_{u}

aur

r_{v}

parallel hoon (ya koi ek zero ho); parametrization wahan degenerate ho jaata hai (jaise upar poles), zero area produce karta hai — yeh chart ka defect hai, surface ka nahi.

Surface integral ka line integral se Stokes' Theorem ke zariye kya relation hai?

Stokes' Theorem ek surface se

\nabla \times F

ke flux ko

F

ke line integral ke saath boundary curve ke around equate karta hai, 2D surface flux ko ek 1D loop se link karta hai.

Recall Carry away karne ke liye one-line summaries

Scalar $d S$ : magnitude, hamesha positive, root rakho.
Vector $d S$ : signed, direction $= \hat{n}$ , root cancel.
Orientation reverse karo → flux sign flip, scalar unchanged.
Tangent field → zero flux; $\nabla \cdot F = 0$ wali closed surface → zero net flux.