Double integrals over rectangles — Fubini's theorem: Question bank

4.4.16 · D5 · Maths › Multivariable Calculus › Double integrals over rectangles — Fubini's theorem

True or false — justify

Har answer reasoning ko restate karta hai, sirf label nahi deta.

Ek rectangle

R = [a, b] \times [c, d]

par, tum

d x d y

d y d x

mein swap kar sakte ho aur limit slots mein same numbers rakh sakte ho.

True — rectangle par har limit ek pure constant hoti hai, isliye numbers

a, b

hamesha

x

ke saath aur

c, d

hamesha

y

ke saath pair karte hain, chahe order kuch bhi ho. Ye sirf isliye safe hai kyunki region ek box hai.

Yahi limit-swapping ki aadat non-rectangular region par bhi safe hai.

False — wahan inner limits outer variable par depend karti hain, isliye unhe blindly doosre slot mein daalne se ek variable apni hi bound mein mix ho jaata hai; tumhe region ko dobara padhna aur limits ko dobara derive karna hoga.

\iint_{R} f d A

hamesha ek positive number ke barabar hota hai.

False — ye ek signed volume hai. Jahan

f < 0

hota hai, surface

x y

-plane ke neeche hoti hai aur negatively contribute karti hai, isliye total zero ya negative bhi ho sakta hai.

Agar

f (x, y) = 0

kisi partition ke har sample point par hai, toh double integral zaroor

0

hoga.

Aam taur par False — ek partition kuch bhi prove nahi karta;

f

sample points ke beech nonzero ho sakta hai. Integral sabhi shrinking partitions par ek limit hai, kisi ek lucky choice par nahi.

Fubini's theorem essentially single-variable volume-by-slicing formula hai jo do baar apply hota hai.

True — fixed

x

par slice karne se cross-sectional area

A (x) = \int_{c}^{d} f d y

milta hai, aur

\int_{a}^{b} A (x) d x

ordinary "cross-section integrate karo" formula hai; doosri taraf se slice karna sirf usi idea ko

x

aur

y

swap karke repeat karta hai.

Separable integrand

g (x) h (y)

ke liye, double integral kisi bhi region

R

par do single integrals ke product ke barabar hota hai.

False — split sirf rectangle par kaam karta hai, jahan inner limits constants hoti hain isliye

g (x)

(aur numeric inner result) ko bahar cleanly nikala ja sakta hai; general region par inner limits outer variable par depend karti hain aur product form toot jaata hai.

Agar rectangle par

f

ke dono iterated integrals exist karte hain aur finite hain, toh wo automatically equal hain.

Sirf iss baat se guarantee nahi hoti — genuinely non-integrable

f

ke liye dono orders differ kar sakte hain chahe dono exist karein; equality tab guaranteed hoti hai jab

f

integrable ho (jaise continuous, ya

\iint ∣ f ∣ < \infty

by Fubini–Tonelli theorem).

Pehle kaun sa variable integrate karna hai ye change karna ek continuous rectangle problem par answer badal sakta hai.

False — continuous

f

ke liye rectangle par answer fixed hai; sirf mehnat badlti hai. Aasaan order chunna Fubini ka poora practical payoff hai.

Spot the error

Har reveal galat step ko name karta hai aur use repair karta hai.

"Inner integral

\int_{c}^{d} f (x, y) d y

karne ke liye, main

x

aur

y

dono integrate karta hoon kyunki dono appear karte hain."

Galat — inner integral sirf apne variable

y

ko integrate karta hai; yahan

x

ek frozen constant hai.

x

ko variable treat karne se galat antiderivative milta hai (jaise

x y

term drop ho jaata hai).

\int_{0}^{2} \int_{1}^{3} (x + 2 y) d y d x

: inner se

\int_{1}^{3} 2 y d y = [y^{2}]

milta hai aur maine

x

term bhool gaya, isliye inner

= 8

hai."

Error —

x

ek constant hai, zero nahi, isliye

y

mein iska antiderivative

x y

hai, kuch nahi. Inner hai

[x y + y^{2}]_{1}^{3} = 2 x + 8

, aur

x

term outer step mein matter karta hai.

"Kyunki Fubini kehta hai order free hai, main outer limits ko andar le ja sakta hoon bina kuch badlaaw ke, yahan bhi is triangle par."

Error — Fubini ki order swap karne ki freedom non-rectangle par limit slots ko preserve nahi karti. Order-swap aur limit values do alag cheezein hain; triangle par tumhe dono bounds recompute karni padenge.

"Integrand

R

par kahin blow up karta hai, lekin Fubini ek theorem hai isliye main equal orders par trust karunga."

Error — Fubini ko

f

integrable chahiye (continuous, ya absolutely integrable). Agar

f

unbounded/non-integrable hai, toh hypothesis fail hoti hai aur dono orders legitimately disagree kar sakte hain.

f = x e^{y}

separable hai, isliye

R = {y \leq x} \cap [0, 1]^{2}

par main ise

(\int x d x) (\int e^{y} d y)

likhta hoon."

Error — function ka separable hona kaafi nahi; product split ke liye region bhi rectangle honi chahiye. Yahan

y \leq x

limits ko couple karta hai, isliye shortcut invalid hai.

"Maine har sub-rectangle ke corner points sample kiye aur ek value mili, toh wahi double integral hai."

Error — double integral woh limit hai jab sab sub-rectangles zero tak shrink hote hain, chahe koi bhi sample point lo. Ek partition ke saath ek sample choice sirf ek Riemann sum approximation hai, exact value nahi.

Why questions

Har reveal mechanism batata hai, koi slogan nahi.

x

ko freeze karna inner integral ko ordinary single-variable integral mein kyun badal deta hai?

x

x_{0}

par fix karne se, map

y \mapsto f (x_{0}, y)

ek variable ka function ban jaata hai, isliye

[c, d]

par iska integral ek plain 1-D area hai — us

x_{0}

par solid ka cross-section.

A (x) = \int_{c}^{d} f (x, y) d y

"cross-sectional area" naam kyun deserve karta hai?

x = x_{0}

par vertical plane solid ko slice karta hai; sliced face neeche se plane se aur upar se curve

z = f (x_{0}, y)

se bounded hai, aur

A (x_{0})

exactly us face ka area hai.

Separable case mein

g (x)

ko inner

y

-integral ke bahar kyun nikala ja sakta hai?

Inner integration variable

y

ke relative to,

g (x)

ek constant hai, aur constants kisi bhi integral se factor out ho jaate hain —

g (x)

ko ek fixed number se multiply karke chhodke.

"Same solid, two slicings" argument dono orders ko equal hone par kyun force karta hai?

Dono iterated integrals ek hi solid ka volume reconstruct karte hain; ek solid ka ek hi volume hota hai, isliye dono computations same number par land karni chahiye.

Integration order chunna ek real practical tool kyun hai, sirf aesthetics nahi?

Ek order tumhe free mein antiderivative de sakta hai (jaise jab koi factor exactly kisi cheez ka

\frac{d}{d x}

ho), jabki doosra integration by parts force karta hai — same answer, kaafi zyada mehnat, via Change of order of integration.

Hume limit-of-Riemann-sums definition ki zaroorat kyun hai, instead of sirf iterated integral ko answer declare karne ke?

Double integral define hota hai volume-approximating limit ke roop mein; Fubini ek theorem hai jo prove karta hai ki iterated integral us limit ke barabar hai. Definition ke bina Fubini kisi cheez ke baare mein sach hone ke liye kuch nahi hoga.

Edge cases

Boundary aur degenerate scenarios jo theorem quietly cover karta hai.

\iint_{R} f d A

kya hoga jab

R

degenerate ho, jaise

[a, a] \times [c, d]

(zero width)?

0

hai — region ka zero area hai, har sub-rectangle base

Δ A = 0

, isliye har Riemann-sum term vanish ho jaata hai; koi solid nahi hai jiska volume ho.

Agar

f (x, y) = k

constant hai

R = [a, b] \times [c, d]

par, double integral kya deta hai?

k (b - a) (d - c)

— box par height

k

ka ek flat slab, yaani height times base area, exactly wahi jo dono iterated orders return karte hain.

Kya Fubini ek

f

ko handle kar sakta hai jisme

R

ke andar ek line par jump discontinuity ho?

Haan — finitely many discontinuities (ek measure-zero set) wala bounded

f

phir bhi integrable hai, isliye dono orders agree karte hain; continuity sufficient hai lekin strictly necessary nahi.

Agar

f

ek symmetric rectangle ke centre ke baare mein odd hai (jaise

f (x, y) = x

[- 1, 1] \times [c, d]

), toh kya hota hai?

Integral

0

hai — ek half par positive contributions doosre half par equal negative contributions ko cancel karte hain, jo signed-volume interpretation se ek dum seedha dikh jaata hai.

Kya Fubini ka rectangular statement teen variables aur boxes tak extend hota hai?

Haan — box par continuous

f

ke liye tum teen single integrals ko kinhi bhi chhe orders mein iterate kar sakte ho, jo same slicing logic ka Triple integrals generalization hai.

Agar

f

continuous hai lekin rectangle unbounded hai, jaise

[0, \infty) \times [0, 1]

Toh ye ek improper double integral hai; Fubini ka plain form directly apply nahi hota. Orders swap justify karne ke liye tumhe Fubini–Tonelli theorem chahiye jisme

\iint ∣ f ∣ < \infty

ho.

Connections

Riemann sums — "ek partition answer nahi hai" wala har trap yahan tak trace hota hai.
Volume by slicing (single-variable) — "why" answers ke peeche ka engine.
Double integrals over general regions — jahan limit-swap ki aadat dangerous ho jaati hai.
Change of order of integration — "aasaan order chuno" wala payoff.
Fubini–Tonelli theorem — unbounded/improper cases ke liye safety net.
Triple integrals — boxes tak extend karne wala edge case.