Worked examples — Change of variables — general Jacobian

4.4.21 · D3 · Maths › Multivariable Calculus › Change of variables — general Jacobian

Yeh deep dive general Jacobian ke liye ek case gym hai. Parent note ne tumhe machine dikhayi thi; yahan hum usme har tarah ka input dalte hain aur dekhte hain kya hota hai: acche invertible maps, orientation flip, ek Jacobian jo zero ho jaata hai (degenerate!), ek limiting-radius sanity check, ek real-world word problem, aur ek exam-style trap.

Shuru karne se pehle, ek reminder seedhe shabdon mein taaki koi bhi symbol bina wajah na lage:

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Scenario matrix

Is topic ke har problem ka ek cell mein aata hai. Neeche ke examples mein har ek ke saath cell(s) tag hain, taaki milake poora table cover ho jaaye.

Cell	Kya special hai	Khatre ki baat	Kaun cover karta hai
A. Curvilinear standard	polar / smooth nonlinear map	region change bhool jaana	Ex 1, Ex 6
B. Linear, $J>0$	constant Jacobian, orientation preserved	kuch nahi, bas factor out karo	Ex 2
C. Linear, $J<0$	map plane ko flip kar deta hai, $\det <0$	absolute value chhod dena	Ex 3
D. Inverse shortcut	$u,v$ ko $x,y$ ke terms mein likhna aasaan	galat direction mein ratio	Ex 4
E. Degenerate / $J=0$	Jacobian kahin zero ho jaata hai	map wahan injective nahi	Ex 5
F. Limiting / sanity	ek variable ko $0$ par le jao, units check karo	area sahi tarike se $\to 0$ hona chahiye	Ex 6
G. Word problem	real quantity (mass/paint)	units aur setup	Ex 7
H. Exam twist	woh jaanboojh kar ulta Jacobian dete hain	reciprocal + $\Vert \cdot \Vert$	Ex 8

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Example 1 — Polar, aur region actually kahan jaata hai (Cell A)

Steps

1. Map: $x=r\cos \theta ,y=r\sin \theta$. Yeh step kyun? Integrand $x^2+y^2$ wahi $r^2$ hai — polar ise ek variable bana deta hai, aur round region ek rectangle ban jaata hai.
2. Jacobian: $J=\det \left(\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right)=r({\cos }^2\theta +{\sin }^2\theta )=r$. Yeh step kyun? Humhe old-area per new-cell chahiye; ${\cos }^2+{\sin }^2=1$ ise clean $r$ mein collapse kar deta hai.
3. Naya region $S$: disk $x^2+y^2\le 1$ ban jaata hai $0\le r\le 1,0\le \theta \le 2\pi$. Yeh step kyun? Yahi woh step hai jo log skip karte hain. Poori disk ek baar sweep hoti hai jab $r$ $0\to 1$ jaata hai aur $\theta$ poora chakkar lagataa hai.
4. Integrate karo: ${\int }_0^{2\pi }{\int }_0^1{e}^{-r^2}rdrd\theta =2\pi \cdot {\left[-\frac{1}{2}{e}^{-r^2}\right]}_0^1=2\pi \cdot \frac{1}{2}\left(1-e^{-1}\right)=\pi \left(1-e^{-1}\right).$ Yeh step kyun? Jacobian se aaya extra $r$ bilkul wahi hai jo $e^{-r^2}r$ ko closed form mein integrable banata hai — iske bina tum atak jaate.

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Example 2 — Linear map, $J>0$, factor out karo (Cell B)

Steps

1. Choose $u=x+y,v=x-y$, toh $x=\frac{u+v}{2},y=\frac{u-v}{2}$. Yeh step kyun? Parallelogram ke edges $x+y=\text{const}$ aur $x-y=\text{const}$ ke along hain (figure dekho), toh yeh coordinates $R$ ko ek square mein seedha kar dete hain.
2. Jacobian: $\frac{\partial (x,y)}{\partial (u,v)}=\det \left(\begin{array}{cc}1/2& 1/2\\ 1/2& -1/2\end{array}\right)=-\frac{1}{2}$, toh $|J|=\frac{1}{2}$. Yeh step kyun? Ek linear map ka constant Jacobian hota hai — yeh integral se seedha bahar aa jaata hai. Minus sign ka matlab sirf yeh hai ki yeh map orientation flip karta hai (koi problem nahi, Ex 3 dekho).
3. Region: vertices se milta hai $u=x+y\in [0,2]$ aur $v=x-y\in [0,2]$; $S$ square $[0,2]\times [0,2]$ hai. Yeh step kyun? Har vertex check karo: $(0,0)\to (0,0)$, $(1,1)\to (2,0)$, $(2,0)\to (2,2)$, $(1,-1)\to (0,2)$. Poora square.
4. Integrate karo: ${\int }_0^2{\int }_0^{2}u\cdot \frac{1}{2}dvdu=\frac{1}{2}{\int }_0^{2}2udu=\frac{1}{2}\cdot 4=2.$ Yeh step kyun? $u=x+y$ wahi integrand hai, toh yeh ek single-variable integral times ek constant ban jaata hai.

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Example 3 — Orientation flip: $J<0$ aur absolute value (Cell C)

Steps

1. Jacobian: $x_u=0,x_v=2,y_u=1,y_v=0$, toh $J=x_u{y}_v-x_v{y}_u=0\cdot 0-2\cdot 1=-2$. Yeh step kyun? Negative sign yeh signal hai ki $T$ orientation reverse karta hai — figure dekho: $u\to v$ corner order image mein clockwise run karta hai jabki square mein counter-clockwise tha.
2. Absolute value lo: $|J|=2$. Yeh step kyun? Area length-times-length hai, hamesha $\ge 0$. Agar hum naively $-2$ use karte toh negative area milta — jo nonsense hai.
3. Area $={\iint }_{[0,1{]}^2}1\cdot |J|dudv=2\cdot (\text{unit square ka area})=2\cdot 1=2.$ Yeh step kyun? $f=1$ ke saath integral sirf scaled area total karta hai.

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Example 4 — Inverse-Jacobian shortcut (Cell D)

Steps

1. Aasaan direction compute karo: $\frac{\partial (u,v)}{\partial (x,y)}=\det \left(\begin{array}{cc}{u}_x& u_y\\ v_x& v_y\end{array}\right)=\det \left(\begin{array}{cc}2x& -2y\\ 2y& 2x\end{array}\right)=4x^2+4y^2$. Yeh step kyun? $u,v$ $x,y$ ke functions ke roop mein diye gaye hain, toh yeh partials one-liners hain — doosri direction mein messy square roots chahiye.
2. Reciprocate karo: kyunki $J_T$ aur $J_{T^{-1}}$ inverse matrices hain, unke determinants multiply karke $1$ dete hain (dekho Inverse Function Theorem): $\frac{\partial (x,y)}{\partial (u,v)}=\frac{1}{4(x^2+y^2)}.$ Yeh step kyun? Integral ke liye humhe old-area per new-cell chahiye, yaani $\partial (x,y)/\partial (u,v)$; reciprocal use free mein deta hai.
3. $(1,1)$ par: value $=\frac{1}{4(1+1)}=\frac{1}{8}$. Yeh step kyun? Plug in karke concrete number milta hai.

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Example 5 — Degenerate case: Jacobian ZERO hai (Cell E)

Steps

1. Jacobian: $x_u=2u,x_v=0,y_u=0,y_v=1$, toh $J=2u\cdot 1-0\cdot 0=2u$. Yeh step kyun? Humhe local stretch factor position ka function chahiye.
2. Zero locus: $J=0⟺u=0$. Yeh step kyun? $u=0$ par map ki $x$-derivative khatam ho jaati hai — wahan ek tiny cell ek line par flatten ho jaata hai, area $\to 0$ (figure mein collapsed cell dekho).
3. Injectivity fail hoti hai: $u=+a$ aur $u=-a$ dono $x=a^2$ dete hain. Yeh step kyun? Theorem require karta hai ki $T$ one-to-one ho. Yahan do alag $u$ ek hi $x$ par land karte hain, toh $T$ plane ko $u=0$ ke saath fold karta hai. Tumhe $u\ge 0$ aur $u\le 0$ mein split karna hoga aur alag alag handle karna hoga.
4. Consequence: sirf $u>0$ par, $|J|=2u$ aur $T$ injective hai, toh $dxdy=2ududv$ wahan valid hai. Yeh step kyun? Aise piece tak restrict karna jahan $J\ne 0$ saari hypotheses restore kar deta hai.

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Example 6 — Limiting sanity check ek ring ke saath (Cells A, F)

Steps

1. Setup: $\text{Area}={\int }_0^{2\pi }{\int }_a^{b}rdrd\theta$. Yeh step kyun? $f=1$, aur Jacobian $r$ bilkul wahi hai jo "radial slab" ko true area mein convert karta hai.
2. Evaluate: ${\int }_a^{b}rdr=\frac{1}{2}(b^2-a^2)$, times ${\int }_0^{2\pi }d\theta =2\pi$, giving $\pi (b^2-a^2)$. Yeh step kyun? Seedha integration; note karo ki $r$ ne $rdr$ ko $\frac{1}{2}{r}^2$ integrate karne yogya banaya.
3. Limit $a\to b$: $\pi (b^2-a^2)\to \pi (b^2-b^2)=0$. Yeh step kyun? Confirm karta hai ki ek degenerate (zero-width) ring ka area zero hai — formula validity ki boundary par sensibly behave karta hai.
4. Sanity $a=0$ par: area $=\pi b^2$, poori disk. Yeh step kyun? Known disk area recover hota hai — formula ka doosra extreme.

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Example 7 — Word problem: ek stretched plate ka mass (Cell G)

Steps

1. Mass $={\iint }_R\rho dA={\iint }_{[0,1{]}^2}1\cdot |J|dudv$. Yeh step kyun? Mass = density integrated over area; constant $\rho$ ke saath, mass = $\rho \times (\text{area of }R)$, aur area $|J|$ se aata hai.
2. Jacobian: $x_u=1,x_v=1,y_u=1,y_v=-1\Rightarrow J=1\cdot (-1)-1\cdot 1=-2$, toh $|J|=2$ (units: m² of plate per m² of $uv$-cell). Yeh step kyun? Map areas ko double karta hai; sign phir se sirf orientation flag karta hai.
3. Mass $={\int }_0^1{\int }_0^{1}2dudv=2\cdot 1=2$ kg. Yeh step kyun? Unit square par constant integrand.

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Example 8 — Exam twist: woh galat-direction Jacobian dete hain (Cell H)

Steps

1. Unka determinant check karo: $u_x=\frac{1}{y},u_y=-\frac{x}{y^2},v_x=y,v_y=x$. $\frac{\partial (u,v)}{\partial (x,y)}=\det \left(\begin{array}{cc}1/y& -x/y^2\\ y& x\end{array}\right)=\frac{x}{y}-\left(-\frac{x}{y}\right)=\frac{2x}{y}.$ Yeh step kyun? Value $\frac{2x}{y}$ actually sahi hai — lekin yeh integral ke liye galat direction hai.
2. Error 1 — direction. $dA=dxdy$ ke liye humhe $\frac{\partial (x,y)}{\partial (u,v)}$ chahiye, yaani reciprocal: $\frac{\partial (x,y)}{\partial (u,v)}=\frac{1}{2x/y}=\frac{y}{2x}.$ Yeh step kyun? Integral old-area ($dxdy$) per new-cell ($dudv$) chahta hai; "nayi variables denominator mein jaati hain" (parent note ka JAB rule).
3. Error 2 — $u,v$ mein express karo, aur absolute value lo. Kyunki $u=x/y$ aur $v=xy$, hame milta hai $x/y=u$ toh $\frac{y}{2x}=\frac{1}{2u}$, aur $|J|=\frac{1}{2u}$ (positive kyunki $u>0$). $dA=\left|\frac{\partial (x,y)}{\partial (u,v)}\right|dudv=\frac{1}{2u}dudv.$ Yeh step kyun? Jacobian nayi variables mein likha hona chahiye actually integrate karne ke liye, aur absolute value area ko positive rakhti hai.

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Connections

• Change of variables — general Jacobian — woh parent machine jin par yeh examples practice hain.
• Polar Coordinates — Ex 1, 6 mein Jacobian $r$ action mein hai.
• Determinant as Volume Scaling — kyun $\det J$ hi area factor hai.
• Cross Product and Area — Ex 7 ke verify mein use hua.
• Inverse Function Theorem — reciprocal rule (Ex 4, 8) aur $J\ne 0$ requirement (Ex 5) justify karta hai.
• Double Integrals over General Regions — jahan region-change step (Ex 1) rehta hai.
• Chain Rule (Multivariable) — prove karta hai $\det J_T\cdot \det J_{T^{-1}}=1$.