Exercises — Changing order of integration

Limits padho. Outer: $0\le x\le 2$ (constants — good). Inner: $0\le y\le x$. Iska matlab: har fixed $x$ ke liye, ek vertical strip wall $y=0$ se slant $y=x$ tak upar jaata hai. Region woh triangle hai jo neeche $y=0$ se, upar line $y=x$ se, aur right mein $x=2$ se bounded hai. Corners: $(0,0),(2,0),(2,2)$. Check a point: $(1,0.5)$ — kya $0\le 1\le 2$? haan. Kya $0\le 0.5\le 1$? haan. Inside. ✓

Inner variable $x$ hai, toh hum pehle across integrate karte hain ⇒ ek horizontal strip. Kyunki $dxdy$ ka matlab hai "$x$ inner, $y$ outer", toh outer limits ($0$ se $3$ tak) pure numbers hone chahiye, aur inner limits $h_1(y),h_2(y)$ curves hain (strip ki left wall se right wall tak). Yeh ek Type II (horizontally simple) description hai — exactly woh form jo upar Type I / Type II box mein define ki gayi hai (dekho Double Integrals over General Regions).

Sketch: $0\le x\le 1$, $x\le y\le 1$ — triangle $y=x$ ke upar, $y=1$ ke neeche, $x=0$ ke right mein; corners $(0,0),(0,1),(1,1)$ (yeh exactly woh triangle hai jo figure s01 mein hai). Naya outer $y$: overall $0\le y\le 1$. Naya inner $x$: ek height $y$ fix karo. Horizontal strip wall $x=0$ pe enter karta hai aur slant $y=x$ pe exit karta hai. Inner variable ke liye slant solve karo: $y=x\Rightarrow x=y$. Toh $0\le x\le y$. $\boxed{{\int }_0^1{\int }_0^{y}f(x,y)dxdy.}$ Check corner $(0.2,0.5)$: original mein chahiye $0.2\le 0.5\le 1$ ✓; new mein chahiye $0\le 0.2\le 0.5$ ✓. Same point, dono descriptions ise accept karte hain.

Sketch: $0\le x\le 2$, $0\le y\le x^2$ — parabola $y=x^2$ ke neeche, $y=0$ ke upar, $x=2$ ke left mein. Corners $(0,0),(2,0),(2,4)$. Naya outer $y$: parabola $y=x^2=2^2=4$ tak upar jaati hai, toh $0\le y\le 4$. Naya inner $x$: $y$ fix karo. Strip parabola pe enter karta hai aur wall $x=2$ pe exit karta hai. Inner variable ke liye $y=x^2$ solve karo, positive branch (region $x\ge 0$ wala hai): $x=\sqrt{y}$. Toh $\sqrt{y}\le x\le 2$. $\boxed{{\int }_0^4{\int }_{\sqrt{y}}^{2}fdxdy.}$

$\int e^{y^2}dy$ ka koi elementary antiderivative nahi hai — yeh us order mein genuinely ek non-elementary integral hai, toh hum zaroor flip karein. Region (L2.1 ke triangle se): $0\le y\le 1$, $0\le x\le y$. $I={\int }_0^1{\int }_0^y{e}^{y^2}dxdy={\int }_0^1{e}^{y^2}[x{]}_0^{y}dy={\int }_0^{1}y{e}^{y^2}dy.$ $u=y^2,du=2ydy$ substitute karo: $I=\frac{1}{2}{\int }_0^1{e}^{u}du=\frac{1}{2}(e-1)$. $\boxed{I=\frac{e-1}{2}\approx \mathrm{0.8591.}}$ Kyun flipping ne kaam kiya: naye order mein strip ki length exactly $y$ hai, aur $y{e}^{y^2}$ ka ek antiderivative hai kyunki extra $y$ woh derivative-factor hai jo substitution ko chahiye.

Wahi triangle phir: $0\le y\le 1,0\le x\le y$. $K={\int }_0^1{\int }_0^y\frac{\sin y}{y}dxdy={\int }_0^1\frac{\sin y}{y}\cdot ydy={\int }_0^1\sin ydy=1-\cos 1.$ $\boxed{K=1-\cos 1\approx \mathrm{0.4597.}}$ Strip ki length $y$, $1/y$ ko cancel kar deti hai — wahi cancellation poora reason hai reorder karne ka.

Sketch: parabola $y=x^2$ (neeche) aur line $y=x$ (upar) ke beech; yeh $(0,0)$ aur $(1,1)$ pe milte hain (ek lens — figure s03 dekho). Kyun $x^2\le x$ yahan: $0\le x\le 1$ pe, squaring shrink karta hai, toh $x^2\le x$ — parabola line ke neeche baithti hai. ✓ Naya outer $y$: $0\le y\le 1$. Naya inner $x$: $y$ fix karo. Strip line $y=x\Rightarrow x=y$ pe enter karta hai (left edge) aur parabola $y=x^2\Rightarrow x=\sqrt{y}$ pe exit karta hai (right edge). $0<y<1$ ke liye, $y<\sqrt{y}$, toh $y\le x\le \sqrt{y}$. $\boxed{{\int }_0^1{\int }_y^{\sqrt{y}}fdxdy.}$

Jaise diya gaya hai (Type II, horizontal strips): overall $y$-range $0\le y\le 2$ hai. $y$ fix karo; strip slant $x=y$ pe enter karta hai aur wall $x=2$ pe exit karta hai. Ek clean strip, koi splitting nahi: ${\int }_0^2{\int }_y^{2}fdxdy.$ Type I mein flip karo (vertical strips): overall $x$-range $0\le x\le 2$ hai. $x$ fix karo; strip wall $y=0$ pe enter karta hai aur slant $y=x$ pe exit karta hai (inner variable ke liye $x=y$ solve karo ⇒ $y=x$). Ek clean strip phir se: ${\int }_0^2{\int }_0^{x}fdydx.$ Kyunki $R$ ek single triangle hai, kisi bhi order ko splitting ki zaroorat nahi; sirf inner limits alag hain (slant $x=y$ ek order mein ceiling ki tarah act karta hai, doosre mein floor ki tarah). Same tiles, alag bookkeeping.

Inner $\int x^3{e}^{x^5}dx$ resist karta hai (tumhe $e^{x^5}$ ke saath $x^4$ chahiye hoga, $x^3$ nahi) — toh flip karo. Region: $0\le y\le 4$, $\sqrt{y}\le x\le 2$ ⇒ yeh L2.2 ka region hai ($0\le x\le 2$, $0\le y\le x^2$). $L={\int }_0^2{\int }_0^{x^2}{x}^3{e}^{x^5}dydx={\int }_0^2{x}^3{e}^{x^5}[y{]}_0^{x^2}dx={\int }_0^2{x}^5{e}^{x^5}dx.$ Strip ki length $x^2$ hai, jo $x^3\cdot x^2=x^5$ deta hai. Ek clean substitution chahti hai $u=x^5,du=5x^{4}dx$ — lekin hamare paas $x^5$ hai, $x^4$ nahi, toh yeh integrand elementary terms mein close nahi hota. Reorder legal hai lekin sufficient nahi; hum numerically finish karte hain. $\boxed{L={\int }_0^2{x}^5{e}^{x^5}dx\approx 8.09\times 10^{12}.}$ (Value bahut badi hai kyunki $e^{x^5}$ at $x=2$ is $e^{32}\approx 7.9\times 10^{13}$ hai.)

Boundaries padho: fixed $y\in [0,1]$ ke liye, $x$ line $x=y$ se $x=\sqrt{2-y^2}$ tak run karta hai, yaani $x^2+y^2=2$ — radius $\sqrt{2}$ ka ek circle. Region: circle $r=\sqrt{2}$ ke andar, line $y=x$ aur $x$-axis ke beech. Line $x=y$ circle se milti hai jahan $2y^2=2\Rightarrow y=1$, point $(1,1)$. $y=0$ pe: $x$ $0$ se $\sqrt{2}$ tak.

Step A — Cartesian $dx\leftrightarrow dy$ swap (jo title maangta hai). Region sketch karo: yeh left mein line $x=y$ se, right/top mein circular arc $x^2+y^2=2$ se, aur neeche $y=0$ se bounded hai. $dydx$ (vertical strips) mein flip karne ke liye, wahan split karo jahan strip ki top curve change hoti hai:

• $0\le x\le 1$ ke liye: ek vertical strip $y=0$ pe enter karta hai aur line $y=x$ pe exit karta hai (line ke neeche wala region hai). Toh $0\le y\le x$.
• $1\le x\le \sqrt{2}$ ke liye: line corner $(1,1)$ se aage nikal chuki hai; strip ab circle pe exit karta hai $y=\sqrt{2-x^2}$. Toh $0\le y\le \sqrt{2-x^2}$. $M={\int }_0^1{\int }_0^x\frac{dydx}{x^2+y^2}+{\int }_1^{\sqrt{2}}{\int }_0^{\sqrt{2-x^2}}\frac{dydx}{x^2+y^2}.$ Dhyaan do ki Cartesian swap do pieces mein split karne pe majboor karta hai — ek sign ki region ek alag coordinate system maangta hai.

Step B — polar mein switch karo (clean route). Polar coordinates mein ek point $(x,y)=(r\cos \theta ,r\sin \theta )$ hai, toh line $y=x$ ka matlab hai $r\sin \theta =r\cos \theta$, yaani $\tan \theta =1$, jiska first-quadrant solution $\theta =\pi /4$ hai. Origin se jaane wali line exactly "ek fixed angle" hoti hai, isliye woh ek constant $\theta$-limit ban jaati hai. Toh $R$ sector $0\le \theta \le \pi /4$, $0\le r\le \sqrt{2}$ hai — dekho Polar Coordinates in Double Integrals. Integrand $\frac{1}{x^2+y^2}=\frac{1}{r^2}$ aur $dA=rdrd\theta$: $M={\int }_0^{\pi /4}{\int }_0^{\sqrt{2}}\frac{1}{r^2}rdrd\theta ={\int }_0^{\pi /4}{\int }_0^{\sqrt{2}}\frac{1}{r}drd\theta .$ Lekin ${\int }_0^{\sqrt{2}}\frac{1}{r}dr=[\ln r{]}_0^{\sqrt{2}}$ at $r=0$ pe diverge karta hai — integrand origin pe blow up karta hai, jo $R$ ka ek corner hai. Toh $\boxed{M\text{ diverges }(=+\infty ).}$ Mastery point: Cartesian swap ne pehle hi warn kar diya tha (pieces mein split hua); polar (via Jacobian) ne phir asli difficulty expose ki — origin pe ek singularity — jo dono Cartesian forms ne chhupayi thi. Order/coordinates change karna ek diagnostic tool hai, sirf ek computation trick nahi.

Region: $0\le x\le 1$, $0\le y\le 1-x$ ⇒ line $x+y=1$ ke neeche triangle; corners $(0,0),(1,0),(0,1)$. Flip: naya outer $y\in [0,1]$; $y$ fix karo, strip $x=0$ se $x=1-y$ tak ($x$ ke liye $x+y=1$ solve karo). $\boxed{{\int }_0^1{\int }_0^{1-y}fdxdy.}$ 3D generalisation: ek solid ke upar triple integral $dx,dy,dz$ ke $3!=6$ orders mein se kisi mein bhi sum ki ja sakti hai; har order usi solid ko re-describe karta hai yeh choose karke ki kis variable ka overall range sabse bahar constant hai aur baaki ko outer variables ke functions ki tarah express karo — identical logic, ek extra layer.

Har problem: $R$ sketch karo, outer variable choose karo (uska range = outer constants), use fix karo aur padho ki strip kahan enter/exit karta hai (woh curves, inner variable ke liye re-solved, = inner limits) — aur non-elementary inners aur singularities ke liye alert raho.