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

Outer variable $x$ hai (uske limits $0$ aur $2$ pure numbers hain). Inner variable $y$, $x$ ki do functions ke beech run karta hai: bottom curve $g_1(x)=x^2$, top curve $g_2(x)=2x$. Kyunki $y$ ko $x$ ki curves ke beech squeeze kiya gaya hai, yeh ek Type I region hai — hum vertical strips sweep karte hain.

Figure dekho: ek fixed $x$ par (maano $x=1$), strip parabola $y=x^2=1$ se shuru hoti hai aur line $y=2x=2$ par khatam hoti hai. Har vertical strip ke pair $y=x^2$ par hain aur sir $y=2x$ par.

(Sanity check karo ki $2x\ge x^2$ on $[0,2]$: $x=1$ par, $2\ge 1$ ✓; dono $x=0$ aur $x=2$ par milte hain.)

Teen vertices teen edges dete hain. Slanted edge $(4,0)$ aur $(0,2)$ ko join karta hai; uski line hai $\frac{x}{4}+\frac{y}{2}=1$, yaani $y=2-\frac{x}{2}$.

Vertical strips ke liye: $x$, triangle ki $x$-axis par shadow ke upar range karta hai, $0$ se $4$ tak (numbers ✓). Fixed $x$ ke liye, strip ka pair $x$-axis $y=0$ par hai aur uska sir slant $y=2-\frac{x}{2}$ par. ${\int }_0^4{\int }_0^{2-x/2}xydydx.$ Hum yahaan ruk te hain — L1 sirf setup maangta hai.

Slanted edge $(0,0)$ aur $(1,2)$ ko join karta hai: line $y=2x$. Right edge $x=1$ hai; bottom $y=0$ hai.

Vertical strip draw karo (neeche figure): fixed $x\in [0,1]$ ke liye, $y$ $x$-axis $y=0$ se slant $y=2x$ tak chadta hai.

${\int }_0^1{\int }_0^{2x}(x+y)dydx.$ Inner ($x$ ko constant mano): ${\int }_0^{2x}(x+y)dy=[xy+\frac{y^2}{2}{]}_0^{2x}=x(2x)+\frac{(2x{)}^2}{2}=2x^2+2x^2=4x^2.$ Outer: ${\int }_0^{1}4x^{2}dx=\frac{4}{3}.$ Toh integral $\boxed{\frac{4}{3}}$ ke barabar hai.

Ab horizontal strips sweep karo. Triangle ka $y$-shadow $0\le y\le 2$ hai (numbers ✓). Fixed height $y$ ke liye, strip ki left boundary slant $y=2x\Rightarrow x=\frac{y}{2}$ hai, aur right boundary $x=1$ hai. ${\int }_0^2{\int }_{y/2}^1(x+y)dxdy.$ Inner: ${\int }_{y/2}^1(x+y)dx=[\frac{x^2}{2}+yx{]}_{y/2}^1=(\frac{1}{2}+y)-(\frac{y^2}{8}+\frac{y^2}{2})=\frac{1}{2}+y-\frac{5y^2}{8}.$ Outer: ${\int }_0^2(\frac{1}{2}+y-\frac{5y^2}{8})dy=[\frac{y}{2}+\frac{y^2}{2}-\frac{5y^3}{24}{]}_0^2=1+2-\frac{40}{24}=3-\frac{5}{3}=\frac{4}{3}.$ ✅ Wahi value $\frac{4}{3}$ — value order-independent hai; sirf algebra alag hai.

Curves wahaan milti hain jahan $\sqrt{x}=x\Rightarrow x=x^2\Rightarrow x=0,1$. Test $x=\frac{1}{4}$: $\sqrt{x}=\frac{1}{2}>\frac{1}{4}=x$, toh $\sqrt{x}$ upar hai. ${\int }_0^1{\int }_x^{\sqrt{x}}2ydydx.$ Inner: ${\int }_x^{\sqrt{x}}2ydy=[y^2{]}_x^{\sqrt{x}}=x-x^2.$ Outer: ${\int }_0^1(x-x^2)dx=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}.$ Result $\boxed{\frac{1}{6}}$.

$\int \sin (y^3)dy$ ka koi elementary antiderivative nahi hai, toh diya gaya ($dy$-first) order dead end hai. Hum order switch karenge.

Region padho. Diya hai: $0\le x\le 1$ aur $\sqrt{x}\le y\le 1$. Kyunki $y\ge \sqrt{x}$ ka matlab hai $x\le y^2$, aur $x\ge 0$, toh fixed $y$ ke liye variable $x$, $0$ se $y^2$ tak run karta hai. $y$-shadow $0\le y\le 1$ hai.

Type II ke roop mein rewrite karo: ${\int }_0^1{\int }_0^{y^2}\sin (y^3)dxdy.$ Inner ($\sin (y^3)$, $x$ mein constant hai): ${\int }_0^{y^2}\sin (y^3)dx=y^2\sin (y^3).$ Outer: $u=y^3,du=3y^{2}dy$ let karo: ${\int }_0^1{y}^2\sin (y^3)dy=\frac{1}{3}{\int }_0^1\sin udu=\frac{1}{3}(1-\cos 1).$ Result $\boxed{\frac{1-\cos 1}{3}}\approx \mathrm{0.1533.}$

Yeh kyun kaam kiya: switching ne impossible inner integral ko trivial bana diya, aur bacha hua $y^2$ exactly woh derivative factor hai jo $u=y^3$ substitution ko chahiye.

Wahaan milte hain jahan $y^2=y+2\Rightarrow y^2-y-2=0\Rightarrow (y-2)(y+1)=0\Rightarrow y=-1,2$. Ek Type II region ke roop mein (horizontal strips), har $y\in [-1,2]$ ke liye: kaun si curve left hai, kaun right? Test $y=0$: $x=y^2=0$ (parabola), $x=y+2=2$ (line). Parabola left hai, line right hai. $\text{Area}={\int }_{-1}^2{\int }_{y^2}^{y+2}1dxdy={\int }_{-1}^2((y+2)-y^2)dy.$ $=[\frac{y^2}{2}+2y-\frac{y^3}{3}{]}_{-1}^2=(2+4-\frac{8}{3})-(\frac{1}{2}-2+\frac{1}{3})=\frac{10}{3}-(-\frac{7}{6})=\frac{20}{6}+\frac{7}{6}=\frac{27}{6}=\frac{9}{2}.$ Area $=\boxed{\frac{9}{2}}$.

Type II kyun? Type I mein tumhe $x=1$ par split karna padta (below $x=1$, dono walls parabola hain; above, top wall line ban jaati hai) — do integrals. Type II poori cheez ek hi shot mein sweep karta hai. Yeh wahi area idea hai jaise Area between two curves mein hai.

Curves $x=0,1$ par milti hain; $(0,1)$ par, $x^2<x$ toh $y=x$ top hai. Type I: $m={\int }_0^1{\int }_{x^2}^{x}xdydx={\int }_0^{1}x(x-x^2)dx={\int }_0^1(x^2-x^3)dx=\frac{1}{3}-\frac{1}{4}=\frac{1}{12}.$ Mass $=\boxed{\frac{1}{12}}$. (Yeh wahi setup hai jo Centre of mass and moments mein use hoti hai.)

$M_y={\int }_0^1{\int }_{x^2}^x{x}^{2}dydx={\int }_0^1{x}^2(x-x^2)dx={\int }_0^1(x^3-x^4)dx=\frac{1}{4}-\frac{1}{5}=\frac{1}{20}.$ L4·1 se $m=\frac{1}{12}$ ke saath: $\bar{x}=\frac{M_y}{m}=\frac{1/20}{1/12}=\frac{12}{20}=\frac{3}{5}.$ Toh $M_y=\boxed{\frac{1}{20}}$ aur $\bar{x}=\boxed{\frac{3}{5}}$.

$\int \frac{e^x}{x}dx$ non-elementary hai — $dx$-first order fail karta hai. Switch karo.

Region: $0\le y\le 2$ aur $\frac{y}{2}\le x\le 1$. $x\ge \frac{y}{2}$ se $y\le 2x$ milta hai; $x\le 1$ ke saath $x$-shadow $0\le x\le 1$ hai, aur fixed $x$ ke liye, $y$, $0$ se $2x$ tak run karta hai. ${\int }_0^1{\int }_0^{2x}\frac{e^x}{x}dydx.$ Inner ($\frac{e^x}{x}$, $y$ mein constant hai): ${\int }_0^{2x}\frac{e^x}{x}dy=\frac{e^x}{x}\cdot 2x=2e^x.$ Outer: ${\int }_0^{1}2e^{x}dx=2(e-1).$ Result $\boxed{2(e-1)}\approx \mathrm{3.4366.}$ Yeh kyun kaam kiya: strip length $2x$ ne problematic $\frac{1}{x}$ ko cancel kar diya, ek textbook exponential chhod kar.

Inner: ${\int }_0^{\sqrt{4-x^2}}xdy=x\sqrt{4-x^2}.$ Outer: substitute karo $u=4-x^2,du=-2xdx$. Jab $x=1,u=3$; jab $x=2,u=0$. $I={\int }_1^{2}x\sqrt{4-x^2}dx=-\frac{1}{2}{\int }_3^0\sqrt{u}du=\frac{1}{2}{\int }_0^3\sqrt{u}du=\frac{1}{2}\cdot \frac{2}{3}{u}^{3/2}{|}_0^3=\frac{1}{3}\cdot 3^{3/2}=\sqrt{3}.$ Toh $I=\boxed{\sqrt{3}}\approx \mathrm{1.732.}$ (Poore quarter-disc ke liye $x$, $0$ se $2$ tak, same method se ${\int }_0^{2}x\sqrt{4-x^2}dx=\frac{1}{3}\cdot 8=\frac{8}{3}$ milta hai. Polar mein yeh integral aksar zyada clean hoti hai — dekho Polar coordinates double integrals.)

Region triangle $x\ge 0,y\ge 0,x+y\le 1$ hai, jo $x\leftrightarrow y$ mein symmetric hai. Pehla order: inner ${\int }_0^{1-x}(x+y)dy=[xy+\frac{y^2}{2}{]}_0^{1-x}=x(1-x)+\frac{(1-x{)}^2}{2}=\frac{1-x^2}{2}.$ Phir ${\int }_0^1\frac{1-x^2}{2}dx=\frac{1}{2}(1-\frac{1}{3})=\frac{1}{3}.$ Doosra order $x\leftrightarrow y$ symmetry se identical hai, region aur integrand dono ki, aur phir $\frac{1}{3}$ deta hai. Common value $=\boxed{\frac{1}{3}}$. Yeh Fubini's Theorem action mein hai.

Kaun sa order ek non-elementary inner integrand ko khatam karta hai — aur kyun? ::: Switch karo taaki messy factor (jaise $e^{y^2},\sin (y^3),e^x/x$) inner integration ke dauran ek constant ban jaaye, aur strip length woh derivative factor supply kare jo outer integral ko chahiye.