Changing order of integration

4.4.18 · Maths › Multivariable Calculus

WHY karte hain hum order change?

Woh key fact jo hum use karte hain (Fubini's theorem):

$\iint_{R} f (x, y) d A = \int_{x = a}^{b} \int_{y = g_{1} (x)}^{g_{2} (x)} f d y d x = \int_{y = c}^{d} \int_{x = h_{1} (y)}^{h_{2} (y)} f d x d y$

jab $f$ continuous ho (ya sirf integrable ho) $R$ pe. Dono iterated integrals area-weighted total ke barabar hote hain.

WHAT exactly hota hai ek region of integration?

HOW karte hain order change — algorithm

First principles se derivation (limits kyun flip hoti hain)

Concretely, triangle $R : 0 \leq x \leq 1, 0 \leq y \leq x$ lo.

Tiles columns se describe karte hue: $x \in [0, 1]$ pick karo, phir $y$ runs $0 \to x$ . ⇒ $\int_{0}^{1} \int_{0}^{x} f d y d x$ .
Same tiles rows se: $y \in [0, 1]$ pick karo, phir $x$ runs line $x = y$ se $x = 1$ tak. ⇒ $\int_{0}^{1} \int_{y}^{1} f d x d y$ .

Boundary $y = x$ shared hai; humne ise bas doosre variable ke liye solve kiya ( $x = y$ ). Woh single algebraic step hi order change hai.

Worked Example 1 — classic "impossible inner integral"

Evaluate karo $I = \int_{0}^{1} \int_{x}^{1} e^{y^{2}} d y d x$ .

Worked example Step by step

Step 1 — Limits padho. Outer: $0 \leq x \leq 1$ . Inner: $x \leq y \leq 1$ . Kyun? Yeh kehte hain: har $x$ ke liye, $y$ line $y = x$ se $y = 1$ tak jaata hai.

Step 2 — Sketch. Region: $y = x$ ke upar, $y = 1$ ke neeche, $x = 0$ ke right mein. Yeh triangle hai corners $(0, 0), (0, 1), (1, 1)$ ke saath. Yeh step kyun? $\int e^{y^{2}} d y$ ka koi elementary form nahi hai, to hume ZAROOR swap karna hoga.

Step 3 — Naya outer variable $y$ . Overall $y$ ranges $0 \leq y \leq 1$ . Kyun? Lowest point ka $y = 0$ hai, highest ka $y = 1$ hai.

Step 4 — Naya inner $x$ . $y$ fix karo. Horizontal strip $x = 0$ pe enter karti hai aur line $y = x$ pe exit karti hai, yaani $x = y$ . To $0 \leq x \leq y$ . Kyun? Fixed $y$ ke liye, region left wall $x = 0$ aur slanted boundary $x = y$ ke beech lie karta hai.

Step 5 — Rewrite aur evaluate karo. $I = \int_{0}^{1} \int_{0}^{y} e^{y^{2}} d x d y = \int_{0}^{1} e^{y^{2}} [x]_{0}^{y} d y = \int_{0}^{1} y e^{y^{2}} d y .$ $u = y^{2}, d u = 2 y d y$ lo: $I = \frac{1}{2} \int_{0}^{1} e^{u} d u = \frac{1}{2} (e - 1) .$

To $I = \frac{e - 1}{2}$ .

Worked Example 2 — region jo ek way mein splitting maangta hai, doosre mein nahi

Evaluate karo $J = \int_{0}^{1} \int_{x}^{1} \frac{1}{1 + y ^{4}} d y d x$ ... actually ek cleaner wala reorder karte hain:

$K = \int_{0}^{2} \int_{y^{2} /2}^{y} sin (x) d x d y$ mein order reverse karo — ruko, ise concrete aur tidy rakhte hain:

Mujhe ek clean fully worked Example 2 dene do:

Worked Example 3 — region jo ek parabola se describe hota hai

$\int_{0}^{1} \int_{x^{2}}^{x} f d y d x$ ko reverse karo.

Common mistake Classic blunders ko steel-man karna

Blunder A — outer limits mein variable chhodna. $\int_{0}^{x} \int_{0}^{1} d y d x$ likhna sahi lagta hai kyunki " $x$ original mein tha." Kyun sahi lagta hai: tumne ek curve outer slot mein copy kar di. Fix: outermost limits constants hone chahiye; sirf inner limits outer variable pe depend kar sakti hain. Outer ko hamesha numbers tak reduce karo.

Blunder B — naye variable ke liye boundary re-solve karna bhool jaana. $y = x$ rakhna jab $x$ ab inner hai. Kyun sahi lagta hai: equation "same curve" hai. Fix: tumhe inner variable explicitly express karna hoga — $y = x$ ko $x$ ke liye solve karo to $x = y$ milega; $y = x^{2}$ ko $x$ ke liye solve karo to $x = y$ milega (is region pe positive branch!).

Blunder C — sketch kiye bina limits swallow karna. Mechanically sab chaar numbers ka min/max copy karna. Kyun sahi lagta hai: fast hai. Fix: sketch karo — bahut se regions ko ek orientation mein splitting chahiye aur doosre mein nahi. Sketch batata hai ki kab ek strip doosri curve pe exit karti hai.

Blunder D — square root ki galat branch. $x = - y$ use karna. Fix: check karo ki tum region ke kis side pe ho; woh branch chuno jo $R$ mein lie kare (ek point test karo).

Recall Feynman: ek 12-saal ke bacche ko explain karo

Ek tiled patio imagine karo jo triangle shape ka hai. Tum sab tiles pe total dust chahte ho. Tum column by column count kar sakte ho (bottom ke along jao, har tall stack of tiles add karo) ya row by row (side ke upar jao, har long row add karo). Kisi bhi tarah tum har tile ek baar count karte ho, to total dust same hai! Kabhi kabhi columns count karna nightmare hota hai aur rows easy hoti hain — to hum bas change kar lete hain ki tiles ke through kaise chalte hain. Switch karne ke liye, tum patio shape draw karte ho, phir uske left/right edges describe karte ho instead of top/bottom edges (ya vice versa).

Flashcards

Double integral ka order kab swap kar sakte hain?

Jab

f

integrable ho (continuous kaafi hai)

R

pe — Fubini's theorem guarantee karta hai ki dono iterated integrals

\iint_{R} f d A

ke barabar hain.

Sahi likhe iterated integral mein, OUTER integral ki limits kis type ki honi chahiye?

Constant (numeric) limits; sirf inner limits outer variable pe depend kar sakti hain.

Ek vertical strip order ke terms mein kya correspond karta hai?

Pehle

y

phir

x

integrate karna:

\int_{a}^{b} \int_{g_{1} (x)}^{g_{2} (x)} d y d x

(Type I region).

Order change karne ke liye, har boundary curve pe tum kya algebraic operation karte ho?

Curve ko NEW inner variable ke liye solve karo (jaise

y = x^{2} \to x = y

, correct branch).

\int_{0}^{1} \int_{x}^{1} e^{y^{2}} d y d x

ko reverse kyun karte hain?

e^{y^{2}}

y

mein koi elementary antiderivative nahi hai; swap karne se

\int_{0}^{1} y e^{y^{2}} d y

milta hai, jo easy hai. Answer

= (e - 1) /2

\int_{0}^{1} \int_{x}^{1} \frac{s i n y}{y} d y d x

ko reorder karo aur value do.

\int_{0}^{1} \int_{0}^{y} \frac{s i n y}{y} d x d y = \int_{0}^{1} sin y d y = 1 - cos 1

Triangle

0 \leq x \leq 1, x \leq y \leq 1

ke liye, swapped limits kya hain?

0 \leq y \leq 1, 0 \leq x \leq y

Order change karne se pehle PEHLA kaam kya karna chahiye?

Region

R

ko original inequalities se sketch karo.

\int_{0}^{1} \int_{x^{2}}^{x} f d y d x

ko reorder karo.

\int_{0}^{1} \int_{y}^{y} f d x d y

Ek orientation mein multiple integrals mein splitting kyun zaroori ho sakti hai?

Agar us direction mein ek strip outer variable ki alag sub-ranges pe alag boundary curves pe enter/exit karti hai.

Connections

Fubini's Theorem — woh theorem jo swapping ko license deta hai.
Double Integrals over General Regions — Type I / Type II definitions.
Polar Coordinates in Double Integrals — ek aur "region ko re-describe karna" trick.
Change of Variables (Jacobian) — ek region ko reparametrize karna generalize karta hai.
Improper and Non-elementary Integrals — kyun $e^{x^{2}}, \frac{s i n x}{x}$ reorder force karte hain.
Triple Integrals — Order of Integration — same idea 3D mein (6 possible orders).