Changing order of integration: Visual walkthrough — Hinglish deep dive

4.4.18 · D2 · Maths › Multivariable Calculus › Changing order of integration

Step 1 — Double integral actually kya add karta hai

KYA HAI. Hum $R$ ko chhote rectangles ("tiles") ke grid mein chop karte hain, har tile ki width $Δ x$ aur height $Δ y$ hai, toh har tile ka area $Δ x Δ y$ hai. Tile number $(i, j)$ point $(x_{i}, y_{j})$ par baitha hai aur $f (x_{i}, y_{j}) Δ x Δ y$ amount ki dust carry karta hai.

KYUN. Integral ka poora idea yeh hai: ek mushkil curved cheez ko bahut saare chhote easy rectangles se replace karo, add karo, phir rectangles ko zero par shrink karo. Total hai

$\iint_{R} f d A = lim_{Δ x, Δ y \to 0} \sum_{i, j} dust density f (x_{i}, y_{j}) tile area Δ x Δ y .$

PICTURE. Neeche grid dekho: har chhota square ek tile hai, uska colour dikhata hai kitni dust $f$ hai usme. Jo number chahiye hum use hain woh hai sabhi coloured tiles mein total dust.

Yahan $\sum_{i, j}$ ka matlab hai "har tile par jaao." Yahi ek object hai jise hum manipulate karenge. Jo bhi aage hoga woh is baare mein hai ki hum in tiles ko kis order mein add karte hain.

Step 2 — Tiles add karna: pehle columns, ya pehle rows

KYA HAI. Sum $\sum_{i, j} f (x_{i}, y_{j}) Δ x Δ y$ lo aur tiles ke through ek route choose karo:

Pehle columns: ek column freeze karo (fix $x_{i}$ ), us column mein upar har tile add karo, phir agli column par jaao.
Pehle rows: ek row freeze karo (fix $y_{j}$ ), us row mein across har tile add karo, phir agli row par jaao.

KYUN. Dono routes har tile ko exactly ek baar visit karte hain, aur har tile usi amount contribute karta hai pile mein chahe tum use kab count karo. Toh do grand totals forced hain equal hone ke liye — yahi hai Fubini's theorem ka arithmetic dil.

$pehle columns i \sum (j \sum f Δ y) Δ x = pehle rows j \sum (i \sum f Δ x) Δ y .$

Inner bracket hai "ek strip ke along add karo"; outer sum hai "strips ko across sweep karo."

PICTURE. Left panel: green vertical strips (tiles ka ek column) left→right sweep. Right panel: pink horizontal strips (tiles ki ek row) bottom→top sweep. Same tiles, do walking orders.

Definition Woh hypothesis jo swap ko legal banata hai

Tiles ko zero par shrink karne se har strip-sum ek inner integral ban jaata hai aur sweep ek outer integral ban jaati hai. Equality limit survive karti hai provided ek yeh ho:

$f \geq 0$ on $R$ (nonnegative — dust pile sirf badh sakta hai), ya
$f$ absolutely integrable hai, yaani $\iint_{R} ∣ f ∣ d A < \infty$ (khaaskar koi bhi $f$ jo closed, bounded $R$ par continuous ho woh qualify karta hai).

Kisi bhi condition ke under, $\iint_{R} f d A = pehle columns: d y d x \int_{x = a}^{b} (\int_{y = g_{1} (x)}^{g_{2} (x)} f d y) d x = pehle rows: d x d y \int_{y = c}^{d} (\int_{x = h_{1} (y)}^{h_{2} (y)} f d x) d y .$ Term by term: outer bounds $a, b$ (ya $c, d$ ) constants hain — swept variable ki poori range; inner bounds $g_{1}, g_{2}$ (ya $h_{1}, h_{2}$ ) curves hain jo outer variable par depend karte hain. $d y d x$ inside-out padha jaata hai: pehle $y$ add karo vertical strip ke along, phir $x$ sweep karo. Neeche jo bhi $f$ integrate karte hain woh sab closed bounded triangle par continuous hai, toh swap hamesha licensed hai.

Step 3 — Ek concrete region: triangle, aur uski vertical strips

KYA HAI. Ek region lo taaki abstract sums real limits ban jayein. Triangle lo $R : 0 \leq x \leq 1, 0 \leq y \leq x .$ Words mein padho: $x$ (outer, swept variable) numbers $0$ aur $1$ ke beech rehta hai; har aise $x$ ke liye, height $y$ floor $y = 0$ se slanted line $y = x$ tak jaati hai.

KYUN pehle columns. $x$ ko outer variable choose karna matlab hai hum vertical strips sweep karte hain. Ek vertical strip ek fixed $x$ par baitha hai; humein jaanna hai woh $R$ mein kahan enter karta hai (bottom) aur kahan exit karta hai (top).

PICTURE. Neeche, ek fixed $x$ par red vertical strip region mein $y = 0$ (floor) par enter karta hai aur $y = x$ (slanted roof) par exit karta hai. Jab $x$ $0$ se $1$ tak slide karta hai, yeh strips poora triangle fill kar dete hain. Yeh Type I / vertically simple description hai.

$\iint_{R} f d A = \int_{x = 0}^{1} \int_{y = 0}^{x} f (x, y) d y d x .$

Term by term:

outer $\int_{x = 0}^{1}$ ::: strip ko saare $x$ se $0$ se $1$ tak sweep karo — constants, kyunki poora region us $x$ -range mein rehta hai.
inner $\int_{y = 0}^{x}$ ::: frozen $x$ ke liye, strip floor $y = 0$ se roof $y = x$ tak climb karo — top limit $x$ ek curve hai, yeh outer variable par depend karta hai.

Step 4 — Wahi triangle horizontal strips ke saath

KYA HAI. Ab doosre taraf chalo: pehle rows, toh $y$ outer swept variable ban jaata hai. Humein exact wahi triangle ko horizontal strips use karke re-describe karna hai.

KYUN. Region ke baare mein kuch nahi badla — sirf uske through hamara route. Ek horizontal strip ek fixed $y$ par baitha hai; humein jaanna hai woh kahan enter karta hai (left) aur kahan exit karta hai (right).

PICTURE. Ek fixed $y$ par blue horizontal strip slanted line par enter karta hai aur right wall $x = 1$ par exit karta hai. Dhyan do left edge wahi same line $y = x$ hai Step 3 se — lekin ab yeh ek horizontal strip ki left boundary hai, toh humein ise $x$ -value ki tarah padhna hai.

Step 5 — Ek algebraic move: boundary ko naye inner variable ke liye solve karo

KYA HAI. Shared boundary $y = x$ ko $x$ ke liye solve karo: $y = x ⟹ x = y .$ Bas yahi hai. Woh ek line ek top-limit ko left-limit mein convert karti hai.

KYUN. Ek horizontal strip ke inner limits $y$ ke functions ke roop mein likhe jaane chahiye (outer variable), kyunki har frozen $y$ ke liye humein woh $x$ chahiye jahan strip enter aur exit karti hai. Toh har boundary curve ko " $x = \dots$ " form mein express karna padta hai.

PICTURE. Neeche, wahi slanted line do baar label hai: green " $y = x$ (roof)" vertical strip ke liye, aur blue " $x = y$ (left wall)" horizontal strip ke liye. Step 3 ka floor $y = 0$ corner ban gaya hai; right wall ab constant $x = 1$ hai.

Ek fixed $y$ ke liye (jab $0 \leq y \leq 1$ ), horizontal strip run karta hai $x = y (left) to x = 1 (right),$ aur $y$ khud triangle ke lowest point ( $y = 0$ ) se highest ( $y = 1$ ) tak sweep karta hai. Isliye

$\iint_{R} f d A = \int_{y = 0}^{1} \int_{x = y}^{1} f (x, y) d x d y .$

Step 3 se compare karo: $\int_{0}^{1} \int_{0}^{x} f d y d x = \int_{0}^{1} \int_{y}^{1} f d x d y .$ Dono identical set of tiles describe karte hain. Flip mein exactly ek algebra step lagi: $y = x \Rightarrow x = y$ .

Step 6 — Kyun bother karte hain: ek inner integral jo possible ho jaata hai

KYA HAI. $f (x, y) = e^{y^{2}}$ lo Steps 3–5 ke exact triangle par ( $0 \leq x \leq 1, 0 \leq y \leq x$ ). Pehle columns mein likha (vertical strips, $d y d x$ ), integral hai $I = \iint_{R} e^{y^{2}} d A = \int_{0}^{1} \int_{0}^{x} e^{y^{2}} d y d x .$ Yeh starting order hai — inner limits $y : 0 \to x$ exactly wahi floor-to-roof limits hain jo humne Step 3 mein padhey. Ab hum ise evaluate karne ki koshish karte hain aur wall se takraate hain.

KYUN flip karna zaroori hai. Inner integral $\int e^{y^{2}} d y$ ka koi elementary antiderivative nahi hai — $exp, sin, lo g, \dots$ ka koi combination $e^{y^{2}}$ tak differentiate nahi hota (dekho Improper and Non-elementary Integrals). Toh columns-first order mein hum bilkul pehle step par hi stuck hain. Rows-first mein flip karna change karta hai kaunsa variable andar integrate hota hai.

PICTURE. Same triangle Steps 3–5 jaisa; shading dikhata hai $e^{y^{2}}$ grow karta hai jab $y$ badha. Ise horizontal strips mein padhna har strip ko $x$ mein trivially integrate karne deta hai.

Step 5 se rows-first limits use karke flip karo ( $0 \leq y \leq 1$ , $y \leq x \leq 1$ ):

=\int_0^1 e^{y^2}\underbrace{[x]_{y}^{1}}_{=\,1-y}\,dy =\int_0^1 (1-y)\,e^{y^2}\,dy.$$ Inner $x$-integral ne sirf **strip length** $1-y$ se multiply kiya (right wall $x=1$ minus left wall $x=y$). Pieces split karo: $$I=\underbrace{\int_0^1 e^{y^2}\,dy}_{\text{abhi bhi koi formula nahi!}}-\int_0^1 y\,e^{y^2}\,dy.$$ Pehla piece *abhi bhi* non-elementary wala hai — toh **is** triangle par naive flip se poori tarah rescue nahi hoti. Clean rescue **complementary** triangle par hoti hai jahan strip length exactly $y$ hai. > [!example] Woh version jo fully collapse ho jaata hai — same idea, complementary triangle > Triangle $R':\ 0\le x\le1,\ x\le y\le1$ lo (**line $y=x$ ke upar**). Columns-first mein yeh hai > $$I'=\int_0^1\!\int_x^1 e^{y^2}\,dy\,dx,$$ > jiska inner $\int e^{y^2}dy$ phir se non-elementary hai. Iske horizontal strips (fix $y$, $x$ runs $0\to y$) strip length exactly $y$ dete hain: > $$I'=\int_0^1\!\int_0^{y} e^{y^2}\,dx\,dy=\int_0^1 e^{y^2}\,[x]_0^{y}\,dy=\int_0^1 y\,e^{y^2}\,dy.$$ > Substitute karo $u=y^2,\ du=2y\,dy$: > $$I'=\tfrac12\int_0^1 e^u\,du=\tfrac12\big(e^1-e^0\big)=\boxed{\dfrac{e-1}{2}}\approx 0.859.$$ > Factor $y$ (strip length) exactly wahi hai jo $\int y\,e^{y^2}dy$ ko solvable banata hai. Yahi payoff hai — aur yeh bhi dikhata hai **flip tab hi help karta hai jab strip length missing factor supply kare**. Hamesha pehle sketch karo dekhne ke liye ki tum kis triangle par ho. --- ## Step 7 — Woh edge cases jo tumhe skip nahi karne chahiye > [!mistake] Jahan naive flip galat ho jaata hai > **Case (a) — corner par degenerate strip.** Steps 3–5 ke triangle par, $x=0$ par vertical strip ki length $0$ hai (floor aur roof dono $y=0$ par); $y=1$ par horizontal strip ki length $0$ hai ($x$ $1$ se $1$ tak). Yeh zero-length strips kuch contribute nahi karte — accha hai — lekin yeh *endpoints* $x=0$ aur $y=1$ set karte hain. Inhe kabhi drop mat karo. > > **Case (b) — flip tab hi help karta hai jab strip length missing factor carry kare.** Step 6 ne dikhaya: lower triangle par strip length $1-y$ thi aur flip ke baad $\int e^{y^2}dy$ chhoot gaya; upper triangle par strip length $y$ thi aur sab collapse ho gaya. Sketch tumhe batata hai tum kis case mein ho. > > **Case (c) — ek curve jo apna role flip kar leta hai.** Ek region ke liye jo *neeche* ek parabola se bound hai aur *upar* ek line se, e.g. $0\le x\le1,\ x^2\le y\le x$, do boundaries **swap karti hain kaun left/right hai** jab tum flip karte ho. Har ek ko $x$ ke liye solve karo: line $y=x\Rightarrow x=y$; parabola $y=x^2\Rightarrow x=+\sqrt{y}$ (**sirf positive branch**, kyunki region $x\ge0$ par baitha hai). $0<y<1$ ke liye hamein $y<\sqrt y$ milta hai, toh strip run karta hai $y\le x\le\sqrt y$. > > **Case (d) — ek orientation ko splitting chahiye.** Agar, horizontally sweep karte waqt, ek strip apni height ke hisaab se *do different curves par exit karti hai*, toh tumhe outer $y$-range ko pieces mein split karna hoga. Sketch yahi batata hai — numbers ko blindly shuffle karke limits kabhi flip mat karo. **PICTURE.** Left: $y=x^2$ aur $y=x$ ke beech lens, strip ka left edge line par ($x=y$) aur right edge parabola par ($x=\sqrt y$) dikhata hai. Right: ek region jahan horizontal strip halfway upar apni exit-curve change karti hai, split force karta hai. Lens ke liye, flip deta hai $$\int_0^1\!\int_{x^2}^{x} f\,dy\,dx=\int_0^1\!\int_{y}^{\sqrt y} f\,dx\,dy.$$ > [!recall] Quick check > Parabola ka inner limit $x=\sqrt y$ kyun hai na ki $x=-\sqrt y$? ::: Region $x\ge 0$ par rehta hai; positive branch wahi hai jo $R$ ke andar hai. Hamesha ek point test karo. > Outer limits ($0$ aur $1$) pure numbers kyun hone chahiye? ::: Yeh swept variable ki *full* range mark karte hain; wahan koi variable region ka kuch hissa undescribed chhod dega. > Kis triangle par $\int\!\!\int e^{y^2}$ flip ke baad fully collapse hota hai? ::: Upper wale par ($x\le y\le 1$), jahan strip length exactly $y$ hai. --- ## Ek-picture summary Yeh final figure poori derivation compress karta hai: **ek triangle, do sweeps**. Left par green vertical strips ($dy\,dx$: inner limits $0\to x$); right par pink horizontal strips ($dx\,dy$: inner limits $y\to 1$). Shared slanted boundary left par $y=x$ label hai aur right par uska solved twin $x=y$ — woh single algebra step jo *hi* order change hai. $f=e^{y^2}$ ka upper-triangle version $\dfrac{e-1}{2}$ evaluate hota hai. > [!recall]- Feynman: plain words mein poora walkthrough > Socho ek triangular patio tiles se covered hai, har tile mein kuch dust hai. Tum total dust chahte ho. Tum ise **column by column** walk kar sakte ho — bottom edge par ek jagah khado aur oopar tumhare tile ka tall stack count karo slanted roof tak, phir sideways step karo aur agla column karo. Ya tum ise **row by row** walk kar sakte ho — ek height par side par khado aur slanted line se right wall tak tiles ki long row count karo, phir step up karo aur agla row karo. Har tile exactly ek baar count hoti hai dono taraf se, toh dono totals same number hain — yahi Fubini hai (jab tak pile infinity ki taraf nahi bhaagti, jo hamare finite dusty triangles ke liye kabhi nahi hota). Columns se rows switch karne ke liye, tum sirf slanted edge dekho aur *ise doosre taraf se padho*: ek roof ke taur par yeh kehta hai "$y=x$", lekin ek left wall ke taur par yeh kehta hai "$x=y$" — same line, doosre letter ke liye solve kiya. Woh single re-reading hi poora "change of order" hai. Hum yeh karte hain kyunki kabhi kabhi columns count karna impossible hota hai ($\int e^{y^2}dy$ ka koi formula nahi hai) jabki rows count karna tumhe exact extra factor (strip length $y$) deta hai jo integral ko solvable banata hai. > [!mnemonic] Yeh yaad rakhkar jaao > **"Same tiles, naya walk. Outer = numbers, inner = curves. Strip flip karo, boundary ko naye inner letter ke liye re-solve karo."** Related tools jab tum yeh master kar lo: [[Polar Coordinates in Double Integrals]] aur [[Change of Variables (Jacobian)]] tiles ki *shape* change karte hain; [[Triple Integrals — Order of Integration]] same game teesre stack ke saath khelta hai.