Exercises — Change of variables — general Jacobian

Chaar partials hain: $x_u=3,x_v=0,y_u=0,y_v=5$. $J=\frac{\partial (x,y)}{\partial (u,v)}=\det \left(\begin{array}{cc}3& 0\\ 0& 5\end{array}\right)=3\cdot 5-0\cdot 0=15.$ Yeh kaisa dikhta hai: $uv$ mein ek unit square, $xy$ mein ek $3\times 5$ rectangle ban jaata hai — area $15$ guna bada. Scaling factor hai $|J|=|15|=\boxed{15}$.

Polar Jacobian $r$ hota hai (parent note mein derive kiya gaya hai). $r=2$ pe yeh $\boxed{2}$ ke barabar hai: radius $2$ pe ek polar cell mein radius $1$ par same $drd\theta$ box ki tulna mein double area hoti hai. Jitna door, utna mota wedge — Polar Coordinates picture dekho.

(a) $J=\frac{\partial (x,y)}{\partial (u,v)}=\det \left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)=-2\ne 0$ har jagah — valid. (b) $J=\frac{\partial (x,y)}{\partial (u,v)}=\det \left(\begin{array}{cc}2u& 0\\ 0& 1\end{array}\right)=2u$, jo $u=0$ pe $0$ hai. Map wahan one-to-one bhi nahi hai ($u$ aur $-u$ same $x$ dete hain). $u=0$ pe invalid. Zero Jacobian matlab chhota cell flat crush ho gaya — Inverse Function Theorem dekho.

$J=\frac{\partial (x,y)}{\partial (u,v)}=\det \left(\begin{array}{cc}\cos v& -u\sin v\\ \sin v& u\cos v\end{array}\right)=u{\cos }^{2}v+u{\sin }^{2}v=u.$ Kyun collapse hota hai: ${\cos }^{2}v+{\sin }^{2}v=1$, toh sab kuch $u$ mein fold ho jaata hai. Isliye $dxdy=ududv$, exactly parent note ka $rdrd\theta$.

Pehle forward Jacobian (easy, kyunki $u,v$ diye hain $x,y$ ke functions ke roop mein): $\frac{\partial (u,v)}{\partial (x,y)}=\det \left(\begin{array}{cc}2x& -2y\\ 2y& 2x\end{array}\right)=4x^2+4y^2.$ Reciprocate kyun karein: theorem ko old area per new cell chahiye, yaani $\partial (x,y)/\partial (u,v)$, aur dono Jacobians inverse matrices hain toh unke determinants $1$ tak multiply hote hain. Isliye $J=\frac{\partial (x,y)}{\partial (u,v)}=\frac{1}{4(x^2+y^2)}.$ $(1,1)$ pe: $\frac{1}{4(1+1)}=\frac{1}{8}=\boxed{0.125}$.

$J=\frac{\partial (x,y)}{\partial (u,v)}=\det \left(\begin{array}{cc}2& 1\\ 1& 3\end{array}\right)=6-1=5.$ Unit $uv$-square $R$ pe map hota hai, toh $\text{Area}={\int }_0^1{\int }_0^1|J|dudv={\int }_0^1{\int }_0^{1}5dudv=5.$ Yeh exactly Determinant as Volume Scaling statement hai: parallelogram ka area edge vectors ke $|\det |$ ke barabar hota hai. Answer: $\boxed{5}$.

Chaar corners map hote hain $(0,0),(1,1),(2,0),(1,-1)$ pe — neeche figure mein parallelogram. $J=\frac{\partial (x,y)}{\partial (u,v)}=\det \left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)=-2,|J|=2.$ Parallelogram ke edge vectors hain $\mathbf{a}=(1,1)$ ($u$-edge ka image) aur $\mathbf{b}=(1,-1)$ ($v$-edge ka image). Cross Product and Area rule se iska area $|a_x{b}_y-a_y{b}_x|=|1\cdot (-1)-1\cdot 1|=2$ hai. Yeh exactly $|J|$ se match karta hai, jaise hona chahiye.

Figure — aise padho: left panel $uv$ mein plain unit square hai (area $1$); uska bottom yellow arrow $u$-edge hai, left pink arrow $v$-edge hai. Right panel $xy$ mein image hai: yellow arrow $\mathbf{a}=(1,1)$ wahan hai jahan $u$-edge land kiya, pink arrow $\mathbf{b}=(1,-1)$ wahan jahan $v$-edge land kiya. Jo tilted blue parallelogram woh span karte hain uska area $2$ hai — exactly $|J|$, woh number jo uske andar likha hai.

Maano $J_T$, $T:(u,v)\mapsto (x,y)$ ka Jacobian matrix hai aur $J_{T^{-1}}$ inverse map ka. $T^{-1}\circ T$ compose karne se identity milti hai, toh multivariable chain rule se (jo Jacobian matrices multiply karta hai) $J_{T^{-1}}{J}_T=I$. Determinants lete hain aur $\det (AB)=\det A\det B$ use karte hain: $\det J_{T^{-1}}\cdot \det J_T=\det I=1.$ Kyunki ye determinants exactly $\partial (u,v)/\partial (x,y)$ aur $\partial (x,y)/\partial (u,v)$ hain, unka product $1$ hai. Geometrically kya matlab hai: agar $T$ areas triple karta hai, toh $T^{-1}$ unhe tihaayi tak shrink karta hai — stretches ek doosre ko undo karte hain.

$r=0$ pe map injective nahi hai: har angle $\theta$ same point $(0,0)$ pe map hota hai. Jacobian ka vanish hona signal karta hai ki poori $\theta$-line ek point pe collapse ho rahi hai — cell zero area tak squash ho gayi. Theorem ki hypotheses (injective, nonzero Jacobian) is single line/point pe fail hoti hain. Lekin origin ek area-zero set hai, toh integral mein kuch contribute nahi karta; hum ise simply exclude kar sakte hain. Yeh standard "measure-zero exception" hai — Double Integrals over General Regions dekho.

Invert karo: $x=\frac{u+v}{2},y=\frac{u-v}{2}$. Jacobian: $J=\frac{\partial (x,y)}{\partial (u,v)}=\det \left(\begin{array}{cc}1/2& 1/2\\ 1/2& -1/2\end{array}\right)=-\frac{1}{2},|J|=\frac{1}{2}.$ Naya region: vertices dete hain $u=x+y\in [0,2]$ aur $v=x-y\in [0,2]$ — ek clean square $0\le u\le 2,0\le v\le 2$. Integrand $x+y=u$. Toh ${\iint }_R(x+y)dA={\int }_0^2{\int }_0^{2}u\cdot \frac{1}{2}dvdu=\frac{1}{2}{\int }_0^{2}2udu=\frac{1}{2}\cdot 4=2.$ Answer: $\boxed{2}$.

Invert karo: add/subtract karke, $y=\frac{u+v}{2},x=\frac{v-u}{2}$. $J=\frac{\partial (x,y)}{\partial (u,v)}=\det \left(\begin{array}{cc}-1/2& 1/2\\ 1/2& 1/2\end{array}\right)=\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=-\frac{1}{2},|J|=\frac{1}{2}.$ Naya region: triangle ki edges $x=0,y=0,x+y=1$ ban jaati hain $v=u$ ($x=0$ se: $v-u=0$), $v=-u$ ($y=0$ se), aur $v=1$ ($x+y=1$ se). Milaakar: $0\le v\le 1$ aur $-v\le u\le v$. Integrand $=e^{u/v}$. ${\iint }_R{e}^{(y-x)/(y+x)}dA={\int }_0^1{\int }_{-v}^v{e}^{u/v}\frac{1}{2}dudv.$ Inner: ${\int }_{-v}^v{e}^{u/v}du=v(e^1-e^{-1})$. Phir $\frac{1}{2}(e-e^{-1}){\int }_0^{1}vdv=\frac{1}{2}(e-e^{-1})\cdot \frac{1}{2}=\frac{e-e^{-1}}{4}.$ Numerically $\frac{e-e^{-1}}{4}\approx 0.5875$. Answer: $\boxed{\frac{e-e^{-1}}{4}}$.

Figure — aise padho: left panel $xy$ mein original triangle hai jismein uska slanted hypotenuse $x+y=1$ yellow mein marked hai. Right panel $uv$-space mein transformed region hai: yeh do pink slanted lines $u=v$ aur $u=-v$ (triangle ke do legs se) aur horizontal yellow line $v=1$ (hypotenuse se) se bounded hai. Hum $u$ ko $-v$ se $v$ tak integrate karte hain, phir $v$ ko $0$ se $1$ tak — woh limits seedhe picture se padho.

Cylindrical: $x=r\cos \theta ,y=r\sin \theta ,z=z$. 3D Jacobian $r$ hai ($z$-row/column ek factor $1$ contribute karta hai; Spherical and Cylindrical Coordinates dekho). Toh $dV=rdrd\theta dz$ aur ${\iiint }_{B}zdV={\int }_0^{2\pi }{\int }_0^1{\int }_0^{2}zrdzdrd\theta .$ Stages mein integrate karo: ${\int }_0^{2}zdz=2$; phir ${\int }_0^{1}2rdr=1$; phir ${\int }_0^{2\pi }1d\theta =2\pi$. Product $=2\pi$. Answer: $\boxed{2\pi }$.

Map linear aur diagonal hai: $J=\frac{\partial (x,y,z)}{\partial (u,v,w)}=\det \left(\begin{array}{ccc}a& 0& 0\\ 0& b& 0\\ 0& 0& c\end{array}\right)=abc.$ Is map ke under ellipsoid ban jaata hai unit ball $u^2+v^2+w^2\le 1$ (volume $\frac{4}{3}\pi$). Isliye $V={\iiint }_{\text{ball}}|abc|dudvdw=abc\cdot \frac{4}{3}\pi =\frac{4}{3}\pi abc.$ Sanity check: $a=b=c=R$ rakhne se $\frac{4}{3}\pi R^3$ milta hai, yaani sphere. Answer: $\boxed{\frac{4}{3}\pi abc}$.

$J=\frac{\partial (x,y)}{\partial (u,v)}=\det \left(\begin{array}{cc}{x}_u& x_v\\ y_u& y_v\end{array}\right)=\det \left(\begin{array}{cc}1& 0\\ 1& 0\end{array}\right)=1\cdot 0-0\cdot 1=0.$ Jacobian har jagah $0$ hai, sirf ek point pe nahi. Yeh kaisa dikhta hai: do image edge vectors ${\mathbf{r}}_u=(1,1)$ aur ${\mathbf{r}}_v=(0,0)$ hain — doosra zero vector hai, toh "parallelogram" line $y=x$ pe flat crush ho jaata hai, jiska area zero hai. Map injective nahi hai aur theorem apply nahi hota; image zero area ka ek curve hai, aur iske upar koi bhi double integral $0$ hai. Yeh polar case (3.3) ke origin ka fully degenerate cousin hai: yahan yeh pure space pe fail hota hai, toh koi rescue nahi.

• L1: Jacobian local area-scaling factor hai (ek number ya function).
• L2: ise compute karo, phir $dxdy=|J|dudv$ — bars hamesha.
• L3: forward aur inverse Jacobians reciprocals hain; measure-zero sets pe zeros harmless hain.
• L4: integrate karne se pehle integrand aur region dono transform karo.
• L5: linear maps constant $|\det |$ dete hain; identically-zero Jacobians genuine degeneracy matlab hain — koi shortcut survive nahi karta.