Double integrals over general regions — Type I and II: Visual walkthrough

4.4.17 · D2 · Maths › Multivariable Calculus › Double integrals over general regions — Type I and II

Pehli line se pehle, teen saada-alfaaz waade:

Step 1 — "Volume under a surface" ka matlab actually kya hai

KYA. Hum floor ko chhote chhote square patches mein kaat dete hain. Ek patch jo $(x, y)$ par baitha hai aur jiska area $Δ A$ hai, us par tent ki height roughly constant $f (x, y)$ hai, toh uske upar ka chhota box $f (x, y) Δ A$ volume rakhta hai. Sab boxes add karo: total volume $\approx \sum f (x, y) Δ A$ .

KYUN. Hum double integral ko sirf is sum ki limit ke roop mein define kar sakte hain jab patches chhote hote jaate hain: $\iint_{D} f d A = lim_{patches \to 0} \sum_{patches in D} box height f (x, y) box floor Δ A .$ Yahan $\sum$ matlab "add up karo", aur subscript "patches in $D$ " sabse crucial part hai — hum sirf wahi patches count karte hain jo $D$ ke andar hain.

PICTURE. Laal boxes sirf shaded region $D$ par khade hain; har box ki height us jagah par $f$ ki tent value hai.

Figure — Double integrals over general regions — Type I and II

Is page ka problem yeh hai: jab " $\sum$ over patches in $D$ " ho aur $D$ ki edges tedhi-medhi hoon toh mushkil ho jaata hai. Hume ise clear limits wale ordinary one-variable integrals mein convert karna hai.

Step 2 — Ek cheez jo hum jaante hain kaise karni hai: rectangles

KYA. Ek rectangle $R = [a, b] \times [c, d]$ par (notation ka matlab hai $a \leq x \leq b$ AUR $c \leq y \leq d$ — ek box jiske vertical sides $x = a, b$ par hain aur horizontal sides $y = c, d$ par hain), sum aasaan hai kyunki dono variables constants ke beech run karte hain. Hume yeh pehle se pata hai ise kaise handle karein.

KYUN. Constant limits ke saath hum boxes ko ek organised tarike se add kar sakte hain: pehle ek column stack karo, phir slide karo across. Woh organised adding hi iterated integral hai.

PICTURE. Rectangle $R$ ki seedhi sides hain; $a$ se $b$ tak har $x$ same top ( $d$ ) aur bottom ( $c$ ) se milta hai. Kuch bhi bend nahi hota.

Toh hamari poori strategy yeh hogi: curved region $D$ ko ek rectangle mein chhupa lo.

Step 3 — $D$ ko ek rectangle mein band karo

KYA. Sabse chhota rectangle $R = [a, b] \times [c, d]$ banao jo $D$ ko poora contain kare. Choose karo:

$a, b$ = wo leftmost aur rightmost $x$ -values jo $D$ tak pahunchti hain,
$c, d$ = wo sabse neeche aur sabse upar wali $y$ -values jo $D$ tak pahunchti hain,

taaki $c \leq g_{1} (x)$ aur $g_{2} (x) \leq d$ har jagah hold hote rahen (curves box ke andar rahein).

KYUN. Hum sirf rectangles par integrate karna jaante hain (Step 2). $D$ ko $R$ ke andar trap karke, hum $R$ par kaam kar sakte hain — bas ek condition ke saath ke $R$ ke woh corners jo $D$ se bahar nikle hain unhe zero count karwaana hoga.

PICTURE. Teal rectangle plum region $D$ se chipka hua hai. Chaar corner slivers (abhi bhi cream) $R$ ke andar hain lekin $D$ ke bahar — yahi woh parts hain jinhe hume silence karna hai.

Step 4 — Zero tak extend karo (bahar ko silence karo)

KYA. Ek nayi height $F$ banao jo $D$ par real height ke barabar ho aur baaki har jagah flat-zero ho: $F (x, y) = {f (x, y) 0 (x, y) \in D (region par, real tent rakho) (x, y) \in / D (region ke bahar, floor level)$ Term by term padho: upar wali line tent ko wahan rakhti hai jahan hum chahte hain; neeche wali line tent ko corner slivers mein height $0$ tak flatten kar deti hai.

KYUN. Height $0$ ka box volume $0$ ka hota hai. Toh $F$ ko poore rectangle par integrate karne se exactly wohi volume milta hai jaisa $f$ ko sirf $D$ par integrate karne se milta: $\iint_{D} f d A = \iint_{R} F d A .$ Left side woh hai jo hum chahte hain; right side ek rectangle par hai jahan hum iterate kar sakte hain.

PICTURE. Wohi rectangle, lekin ab corner slivers floor tak dabaaye gaye hain (height $0$ ). Sirf $D$ ke upar wala tent utha hua hai.

Step 5 — Rectangle par iterate karo (Fubini)

KYA. Ek rectangle par, Fubini's Theorem kehta hai ke hum boxes ko column-by-column add kar sakte hain: $\iint_{R} F d A = \int_{a}^{b} (ek poora column, x fixed \int_{c}^{d} F (x, y) d y) d x .$ Term by term:

inner $\int_{c}^{d} F (x, y) d y$ $x$ ko freeze karta hai aur $y$ ko us column mein $c$ se $d$ tak slide karta hai, us ek vertical strip mein box heights add karta hua;
outer $\int_{a}^{b} (\dots) d x$ phir us finished strip ko left ( $a$ ) se right ( $b$ ) tak slide karta hai.

KYUN. Yeh Step 2 ka "ek column add karo, phir across slide karo" wala idea hai, ab symbols mein likha hua. Yeh sirf isliye kaam karta hai kyunki rectangle ki limits $a, b, c, d$ sab constants hain.

PICTURE. Ek fixed $x$ par ek vertical teal strip highlight ki gayi hai; arrow dikhata hai ke $y$ $c$ se $d$ tak sweep kar raha hai. Ek ghosted strip outer sweep ko $a$ se $b$ tak slide karte dikhata hai.

Step 6 — Dead zones ko collapse karo (the punchline)

KYA. Ek fixed $x$ par ek column dekho. Jab $y$ $c$ se $d$ tak chadh raha hota hai, height $F$ zero hoti hai jab tak $y$ bottom curve $g_{1} (x)$ tak nahi pahunchta, phir real $f$ ke barabar hoti hai top curve $g_{2} (x)$ tak, phir $d$ tak zero ho jaati hai. Inner integral ko un do curves par split karo: $\int_{c}^{d} F d y = = 0 \int_{c}^{g_{1} (x)} 0 d y + \int_{g_{1} (x)}^{g_{2} (x)} f d y + = 0 \int_{g_{2} (x)}^{d} 0 d y = \int_{g_{1} (x)}^{g_{2} (x)} f (x, y) d y .$ Dono zero pieces gayab ho jaate hain, sirf tent-part bachta hai.

KYUN. Dead zones ne literally kuch contribute nahi kiya (height $0$ ), isliye column ka sach mein content sirf curves ke beech run karta hai. Isliye inner limits curves $g_{1} (x)$ aur $g_{2} (x)$ hain: yahi woh jagah hai jahan us column par tent shuru aur khatam hota hai.

PICTURE. Wohi column, ab colour-coded: grey dead segments $g_{1}$ ke neeche aur $g_{2}$ ke upar, orange live segment unke beech mein.

Ise Step 5 ke outer integral mein wapas daalo: $\iint_{D} f d A = \int_{a}^{b} \int_{g_{1} (x)}^{g_{2} (x)} f (x, y) d y d x$ Padho: outer = $x$ mein constant-se-constant tak slide karo; inner = har column mein curve-se-curve tak chadho. $■$

Step 7 — Edge case: kya $D$ hamesha Type I hota hai?

KYA. Nahi. Formula kaam karne ke liye ek column mein sirf ek entry-point aur ek exit-point hona chahiye — matlab har vertical line $D$ ko ek single unbroken segment mein mile. Agar ek vertical line $D$ mein enter kare, exit kare, phir re-enter kare (ek dumbbell ya crescent), toh koi single pair $g_{1}, g_{2}$ us column ko describe nahi kar sakti.

KYUN. Step 6 ne exactly do crossings assume ki theen: zero → live → zero. Teen ya zyada crossings us split ko tod deti hain. Fix: $D$ ko sub-regions $D_{1}, D_{2}, \dots$ mein kato, har ek simple, aur add karo: $\iint_{D} f d A = \iint_{D_{1}} f d A + \iint_{D_{2}} f d A .$

PICTURE. Left: ek achha region — har vertical line ek baar andar, ek baar bahar cross karti hai. Right: ek bura crescent jahan ek vertical line chaar baar cross karti hai, dashed cut se do Type I pieces mein kaata hua.

Step 8 — Degenerate case: curves jo touch karti hain (area collapse ho jaata hai)

KYA. Kisi $x$ par jahan $g_{1} (x) = g_{2} (x)$ ho, column ki height zero hoti hai — inner integral $\int_{g_{1}}^{g_{1}} f d y = 0$ hai. Yeh koi error nahi hai; yeh region ki honest boundary hai.

KYUN. Between-curves area mein do curves endpoints par milti hain. $f \equiv 1$ set karke: $Area (D) = \int_{a}^{b} (g_{2} (x) - g_{1} (x)) d x,$ aur ek touching point par $g_{2} - g_{1} = 0$ , toh woh column kuch add nahi karta — exactly jaisa geometry demand karti hai. Agar curves $[a, b]$ ke andar cross karti hain (top aur bottom swap ho jaate hain), toh tumhe crossing par split karna hoga aur reassign karna hoga ke har piece par kaun si curve $g_{2}$ (top) hai; warna $g_{2} - g_{1}$ negative ho jaata hai aur sign flip ho jaata hai.

PICTURE. $y = x^{2}$ aur $y = x$ ke beech ek region: columns ke beech mein poori height hai aur meeting points $x = 0$ aur $x = 1$ par zero height tak pinch ho jaati hai.

Ek-picture summary

Is page ki poori cheez ek single frame mein: enclose ( $a, b, c, d$ ) → corners ko silence karo ( $F = 0$ ) → ek column $g_{1} \to g_{2}$ tak chadha → $a \to b$ tak slide karo.

Recall Feynman retelling — poora walkthrough saade alfaaz mein

Tum ek ulte-seedhe tent ke neeche ka saamaan measure karna chahte ho, lekin tent ka floor ek ajeeb tedha shape hai, koi neat box nahi (Step 1). Tum sirf ek box handle karna jaante ho, toh tum sabse chhota box banate ho jo poore tent ko andar fit kar le (Step 3). Box ke corners hain jo tent ke real floor se bahar nikalte hain — toh tum un corner-tents ko zameen par press karte ho, unhe kuch nahi count karwaate ho (Step 4). Ab tum poore box ke andar easy tarike se measure karne ki permission rakhte ho: ek patla vertical slice lo, us mein tent height ko neeche se upar tak add karo, phir us slice ko across slide karo (Step 5). Lekin kisi bhi slice mein, flatten wale parts zero hain — tent sirf bottom curve aur top curve ke beech exist karta hai — toh slice actually sirf curve-to-curve run karti hai (Step 6). Yahi formula hai: fixed numbers ke beech left-to-right slide karo, har slice mein bottom-curve to top-curve ke beech climb karo. Agar koi slice kabhi shape ke andar-bahar ek se zyada baar jaaye, toh pehle shape ko simpler pieces mein kato (Step 7). Aur jahan dono curves kiss karti hain, slice ki koi height nahi hoti aur quietly zero add hota hai — jo exactly region ki edge hai (Step 8).