Visual walkthrough — Double integrals in polar coordinates — Jacobian r

4.4.19 · D2 · Maths › Multivariable Calculus › Double integrals in polar coordinates — Jacobian r

Har step ko order mein padho. Har step batata hai KYA draw kiya, KYUN draw kiya, aur us PICTURE ki taraf point karta hai jo idea carry karti hai.

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Step 1 — Polar language mein ek point kya hota hai?

KYA. "$x$ right jao, $y$ upar jao" ki jagah hum kehte hain "centre se $r$ distance niklo, us direction mein jo rightward axis se $\theta$ angle banaye."

KYUN. Hum abhi chhote tiles radii aur angles se banane waale hain, toh pehle yeh jaanna zaroori hai ki do numbers — ek length, ek angle — kisi bhi point ko fix kar dete hain. Yahi poora polar idea hai.

PICTURE. Neeche, point $P$ distance $r$ (yellow ruler) par ek ray ke along baithe hai jo positive $x$-axis (white) se $\theta$ angle (red wedge) par tilted hai.

$x$ ke liye $\cos$ aur $y$ ke liye $\sin$ kyun, ulta kyun nahi? Kyunki $\cos \theta$ define hota hai adjacent-over-hypotenuse ratio ke roop mein — arrow ka horizontal fraction — aur $\sin \theta$ vertical fraction. Horizontal shadow horizontal hai, isliye woh $\cos$ pahnata hai.

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Step 2 — "$r$ hold karo" aur "$\theta$ hold karo" — dono alag alag kya draw karte hain?

KYA. Hum ek baar mein ek coordinate freeze karte hain aur trace karte hain ki doosra kya sweep karta hai.

KYUN. Cartesian tile bounded hota hai grid lines "$x=$ const" aur "$y=$ const" se. Polar tile banane ke liye hume uske do families of boundary curves chahiye. Toh hum poochhhte hain: "$r=$ const" kaun sa curve hai? "$\theta =$ const" kaun sa curve hai?

PICTURE. $r$ fix karna (yellow) ek circle trace karta hai — har point same distance par. $\theta$ fix karna (red) ek ray trace karta hai — har point same direction mein.

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Step 3 — Ek chhota polar tile banao

KYA. Ek starting distance $r$ aur starting angle $\theta$ chuno. Distance ko thoda $dr$ badhaao aur angle ko thoda $d\theta$ badhaao. Chaar boundary curves — do circles, do rays — ek chhota tile fence kar dete hain.

KYUN. Double integral laakhon chhote tiles par $f(\text{point})\times (\text{tile area})$ ka sum hota hai. Integral ko polar mein convert karne ke liye hume ek polar tile ka area $dr$ aur $d\theta$ ke terms mein jaanna chahiye. Isliye hum exactly ek ko isolate karte hain.

PICTURE. Shaded tile inner circle (radius $r$), outer circle (radius $r+dr$), aur $\theta$ aur $\theta +d\theta$ angles par do rays se bounded hai.

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Step 4 — Do sides measure karo: seedhi wali aur curved wali

KYA. Hum tile ke do tarah ke sides measure karte hain.

• Radial side (bahar ki taraf) do circles ke beech ka gap hai: length $dr$.
• Arc side (circle ke along) circle ki circumference ka ek tukda hai.

KYUN. Almost-rectangular tile ka area hota hai (ek side) $\times$ (adjacent side). $dr$ toh hai; hume arc length chahiye. Yahan ek bahari tool aata hai — arc-length rule — toh hum ise seedha state karte hain.

PICTURE. Straight side $dr$ yellow hai; curved bottom side, length $rd\theta$, blue hai. Labels seedha tile par padho.

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Step 5 — Sides multiply karo tile ka area paane ke liye

KYA. Itne chhote tile ke liye ki do arcs almost same length hain, yeh essentially ek rectangle hai sides $dr$ aur $rd\theta$ ke saath. Iska area unka product hai.

KYUN. Yahi woh answer hai jo hum dhundhne aaye the: polar coordinates mein area element $dA$.

$\underset{\text{tile area}}{\underbrace{dA}}\approx \underset{\text{straight side, length}}{\underbrace{(dr)}}\times \underset{\text{curved side, arc}}{\underbrace{(rd\theta )}}=rdrd\theta .$

Term by term:

• $dA$ — area ka woh chhota tukda jis par hum integrate kar rahe hain.
• $dr$ — tile ki radial thickness.
• $rd\theta$ — tile ki arc width; $r$ Step 4 ka arc-length magnifier hai.
• Product $rdrd\theta$ — thickness times width = area.

PICTURE. Tile ko ek seedhe rectangle mein flatten kiya gaya hai taaki multiplication $\text{area}=(\text{height }dr)\times (\text{width }rd\theta )$ dikh sake.

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Step 6 — Lekin do arcs EQUAL nahi hain. Kya difference matter karta hai?

KYA. Hum apni khud ki laziness ko steel-man karte hain. Outer arc hai $(r+dr)d\theta$, jo inner arc $rd\theta$ se lambi hai. Asli tile ek patla curved trapezoid hai, rectangle nahi. Aao iska exact area compute karein aur dekhen kitna cheated kiya.

KYUN. Agar "leftover" $rdrd\theta$ ke barabar hota, toh humara formula galat hota. Hume prove karna hai ki leftover negligibly chhota hai.

Exact tile do circular sectors ka difference hai:

$$d{A}_{\text{exact}}=\underset{\text{bada sector}}{\underbrace{\frac{1}{2}(r+dr{)}^{2}d\theta }}-\underset{\text{chhota sector}}{\underbrace{\frac{1}{2}{r}^{2}d\theta }}=\frac{1}{2}d\theta [(r+dr{)}^2-r^2].$$

$(r+dr{)}^2-r^2=2rdr+(dr{)}^2$ expand karo:

$$d{A}_{\text{exact}}=\underset{\text{woh term jo hamne rakha}}{\underbrace{rdrd\theta }}+\underset{\text{leftover}}{\underbrace{\frac{1}{2}(dr{)}^{2}d\theta }}.$$

Leftover kyun mar jaata hai. Jo term hamne rakha usme $dr$ ka ek factor hai; leftover mein $dr$ ke do factors hain $(dr{)}^2$. Jab tiles shrink hote hain, $dr\to 0$, aur $(dr{)}^2$ bahut zyada tezi se shrink hota hai $dr$ se — jaise ek haathi ke saamne cheenti. Limiting sum mein yeh kuch contribute nahi karta.

PICTURE. Rakha gaya rectangle (green) versus outer corner se chipka tiny leftover sliver (red). Red piece clearly ek "second-order crumb" dikhta hai.

Recall Exact formula mein har symbol kahan se aaya

$\frac{1}{2}{R}^{2}d\theta$ radius $R$ aur angle $d\theta$ ke pie sector ka area hai ::: kyunki sector full disk $\pi R^2$ ka fraction $\frac{d\theta }{2\pi }$ hai, jo deta hai $\frac{d\theta }{2\pi }\cdot \pi R^2=\frac{1}{2}{R}^{2}d\theta$. Term $rdrd\theta$ ::: pehla-order piece jo hum rakhte hain — humara area element. Term $\frac{1}{2}(dr{)}^{2}d\theta$ ::: second-order crumb jo hum discard karte hain jab $dr\to 0$.

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Step 7 — Algebra check: Jacobian determinant bhi yahi kehta hai

KYA. Picture ne $r$ diya. Ek machine — Jacobian Determinant — ko agree karna chahiye, warna hum mein se koi galat hai. Hum ise compute karte hain.

KYUN. General Change of Variables Theorem kehta hai ki jab tum $(x,y)$ ko naye variables $(r,\theta )$ mein rewrite karte ho, toh area $\left|\det J\right|$ se stretch hota hai, jahan $J$ collect karta hai ki $x$ aur $y$ $r$ aur $\theta$ mein nudge par kaise respond karte hain. Yeh "sides measure karo aur multiply karo" ka algebraic twin hai.

$$J=\left|\begin{array}{cc}\frac{\partial x}{\partial r}& \frac{\partial x}{\partial \theta }\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial \theta }\end{array}\right|=\left|\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right|.$$

Har entry padho:

• $\partial x/\partial r=\cos \theta$ — distance nudge karo, $x$-shadow $\cos \theta$ se move karta hai.
• $\partial x/\partial \theta =-r\sin \theta$ — angle nudge karo, $x$-shadow $r\sin \theta$ se swing back karta hai.
• $\partial y/\partial r=\sin \theta$ — distance nudge karo, $y$-shadow $\sin \theta$ se move karta hai.
• $\partial y/\partial \theta =r\cos \theta$ — angle nudge karo, $y$-shadow $r\cos \theta$ se rise karta hai.

Ab determinant (cross-multiply, subtract):

$$\det J=\cos \theta (r\cos \theta )-(-r\sin \theta )(\sin \theta )=r{\cos }^2\theta +r{\sin }^2\theta =r\underset{=1}{\underbrace{({\cos }^2\theta +{\sin }^2\theta )}}=r.$$

Kyunki $r\ge 0$, $\left|\det J\right|=r$. Wahi $r$. Do alag roads, ek answer.

PICTURE. $J$ ke do column-vectors tile ki actual edges ke roop mein draw kiye gaye: "$\partial /\partial r$" arrow length $1$ ka aur "$\partial /\partial \theta$" arrow length $r$ ka. Jo parallelogram unse banta hai uska area hi determinant hai — aur woh $r$ ke barabar hai.

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Step 8 — Edge aur degenerate cases (skip mat karo)

KYA. Hum formula ko wahan check karte hain jahan cheezein weird ho jaati hain: origin par, aur full-circle sweep ke liye.

KYUN. Jo rule apni boundaries par survive nahi kar sakta woh ek trap hai. Har case cover karo.

Case A — origin par, $r=0$. Yahan $dA=0\cdot drd\theta =0$. Tile ka zero area hai. Kya yeh sense karta hai? Haan: bilkul centre par, saari rays ek point par milti hain, isliye "arc width" $rd\theta =0$. Tile ek degenerate sliver hai jiska koi area nahi. $r$ ise sahi se khatam karta hai.

Case B — full angular sweep, $\theta :0\to 2\pi$. Angle ko full turn par integrate karna familiar circle facts reproduce karna chahiye. Ring of radii $r$ se $r+dr$ tak: iska area hona chahiye circumference $\times$ thickness $=2\pi rdr$. Hamare formula se check karo:

$${\int }_0^{2\pi }rdrd\theta =rdr{\int }_0^{2\pi }d\theta =2\pi rdr.✓$$

Case C — radius $a$ ki poori disk. Saari rings $0$ se $a$ tak sum karo:

$${\int }_0^{2\pi }{\int }_0^{a}rdrd\theta =2\pi {\left[\frac{r^2}{2}\right]}_0^a=2\pi \cdot \frac{a^2}{2}=\pi a^2.✓$$

Disk ka area nikal aata hai — ultimate sanity check.

PICTURE. Teen panels: (A) shrinking tile jo centre par vanish ho raha hai; (B) area $2\pi rdr$ ki ek patli ring; (C) rings nested hoke disk fill kar rahi hain. Tiles ko centre se rim tak badhte dekho.

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Ek-picture summary

Upar sab kuch, compressed: centre se rim tak polar tiles ka ek fan. Har jagah same $dr$, same $d\theta$ — lekin tiles clearly bahar jaate jaate mote hote hain, unki arc width $rd\theta$ distance $r$ ke saath lockstep mein badhti hai. Har tile par label uska apna area padhta hai, $rdrd\theta$.

Recall Feynman: poora walkthrough seedhe words mein

Hum ek plane mein ek chhote tile ka size jaanna chahte the jise "distance out" ($r$) aur "angle around" ($\theta$) se describe kiya jaata hai. Humne ek chhota tile isolate kiya: woh do circles aur do rays se boxed hai. Iska up-down side bas itna hai ki distance kitni badi, $dr$. Iska along-the-circle side ek arc hai, aur arcs on a circle hote hain radius times angle, toh woh hai $rd\theta$ — aur yahi trick hai, side badhti hai kyunki bade circles same angle ke liye lambe arcs rakhte hain. Do sides multiply karo: area $=dr\times rd\theta =rdrd\theta$. Humne sochaa ki tile sach mein ek curved trapezoid hai, lekin "extra bit" mein $(dr{)}^2$ hai, jo ek crumb hai jo tiles chhote hone par vanish ho jaata hai. Confirm karne ke liye humne algebra machine (Jacobian determinant) chalai aur usne wahi same $r$ diya. Bilkul centre par tiles kuch bhi nahi reh jaate ($r=0$ unhe khatam karta hai), ek full turn ek ring of area $2\pi rdr$ rebuild karta hai, aur rings stack karne se disk ka $\pi a^2$ rebuild hota hai. Yaad rakhne ka ek word: arc grows with radius, aur woh growth hi $r$ hai.

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Flashcards

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Connections

• Double integrals in polar coordinates — Jacobian r — parent; yeh page uska picture-book proof hai.
• Arc Length and Radian Measure — crucial $rd\theta$ arc provide karta hai.
• Jacobian Determinant — algebraic twin jo $r$ bhi deta hai.
• Change of Variables Theorem — woh general law jiska humara tile paalан karta hai.
• Area of Regions Bounded by Polar Curves — $rdrd\theta$ tile action mein.