Question bank — Double integrals over general regions — Type I and II

False — agar $f$ $D$ par integrable hai (jaise continuous, ya absolutely convergent), to Fubini's Theorem guarantee karta hai ki value identical hai; sirf do iterated integrals ki difficulty alag hoti hai, woh number nahi jo woh produce karte hain.

False — ek annulus (do circles ke beech ek washer) ek vertical strip ko region chhodne aur re-enter karne par majboor karta hai (Figure 2 dekho), isliye koi bhi single top/bottom curves $g_1,g_2$ ka pair kaam nahi karta aur use kai Type I pieces mein split karna padta hai.

True — aisa region (jaise triangle ya disc) dono taraf sweep kiya ja sakta hai, isliye tum woh direction chunte ho jiska inner antiderivative elementary ho, bilkul waise jaise neeche "Spot the error" section mein $e^{y^2}$ rescue hai.

True — ${\iint }_{D}1dA={\int }_a^b(g_2(x)-g_1(x))dx$, jo precisely single-variable calculus ka Area between two curves formula hai.

False — outer limits pure constants $a,b$ hain; outer integral ek number ban ke khatam hona chahiye, isliye usme koi bhi variable bilkul nahi ho sakta.

False — do boundary curves ka endpoints par milna sirf yeh matlab hai ki strip wahan zero height hai (ek pointed tip), jo triangles aur lens-shaped regions ke liye bilkul normal hai.

False — ek clean top curve $g_2$ hona Type I ke liye convenient hai, lekin wohi region Type II bhi ho sakta hai agar uski left aur right boundaries $h_1,h_2$ har ek $y$ ki single functions hain; "Type I ke liye nice hona" Type II ko forbid nahi karta.

Outer limit $x$ hai, jo illegal hai — outer integral ek number evaluate hona chahiye, isliye uski limits ($a,b$) constants honi chahiye; sirf inner limit ($g_2(x)$) outer variable carry kar sakti hai.

$0<x<1$ par hame $x^2<x$ milta hai, isliye $x$ top curve $g_2$ hai — higher curve se lower curve par integrate karna orientation reverse kar deta hai aur answer negate ho jaata hai; hamesha ek sample point test karo jaise $x=\frac{1}{2}$.

Value order-independent hai, lekin limits nahi hain — horizontal aur vertical strips alag boundaries ko hit karti hain, isliye tumhe $D$ redraw karni hai aur limits dobara padhni hain, jaisa Changing the order of integration mein stressed hai.

Yeh us order mein impossible hai kyunki $e^{y^2}$ ka $y$ mein koi elementary antiderivative nahi hai; Type II mein switch karna ($x$ from $0$ to $y$) ise ${\int }_0^{1}y{e}^{y^2}dy=\frac{e-1}{2}$ bana deta hai, isliye yeh bilkul doable hai.

$y$ mein inner integral ko $y$ bilkul eliminate kar dena chahiye dono limits substitute karke — leftover $y$ matlab hai bounds galat plug in kiye gaye (ek bhula hua evaluation) aur outer $x$-integral aage proceed nahi kar sakta.

Constant $y$-limits $[0,1]$ enclosing square describe karte hain, jo galat tarike se $y=x$ line ke upar ke wedge ko include karta hai; triangle ko sloped top limit $g_2(x)=x$ chahiye taaki woh extra wedge exclude ho aur kabhi contribute na kare.

Intersection points ($y=1$ aur $y=-2$) ke bina outer $y$-limits $c,d$ unknown hain, isliye region undefined hai — pehle $y^2=2-y$ solve karna zaroori hai un bottom aur top constants $c=-2,d=1$ ko fix karne ke liye.

Inner integral pehle kiya jaata hai jabki outer variable temporarily fixed rakha jaata hai, isliye uski limits woh frozen value reference kar sakti hain; ek baar finish hone ke baad, sirf outer variable bachta hai aur use fixed numbers $a,b$ (ya $c,d$) ke beech integrate karna hota hai.

Integrand ko $D$ ke bahar $0$ define karna matlab hai ki curved boundary se pare ka area koi volume add nahi karta, isliye rectangle integral region integral ke barabar hota hai — yahi woh cheez hai jo Fubini's Theorem ko rectangle par iterated integral mein collapse hone deti hai curve-dependent limits ke saath.

Hum sirf rectangle par constant limits ke saath iterate karna jaante hain; $D$ ko $R$ mein enclose karna aur zero-extend karna woh bridge hai jo hard curved-region case ko solved rectangular wale mein convert karta hai.

Inner integrand kisi aisi cheez se badal sakta hai jiska koi elementary antiderivative nahi hai aise mein jiska hai, ya ek helpful factor gain kar sakta hai (jaise $y{e}^{y^2}$ mein $y$) jo ek substitution unlock kar deta hai.

Agar ek direction mein sweep karna ek strip ko $D$ mein ek single curves ke pair ke across enter aur leave karta hai, lekin doosri direction ek corner cross karti hai jo swap karta hai ki kaun si curve use bound karti hai, to smart direction $D$ ko pieces mein split karne se bachata hai.

Kyunki $f\equiv 1$ inner integral ko sirf strip height $g_2(x)-g_1(x)$ banata hai; $x$ par heights sum karna literally single-variable Area between two curves integral hai, isliye dono frameworks ko agree karna hi hai.

Har strip ki zero height hai, isliye region zero area wala ek curve hai aur integral $0$ hai regardless of $f$ — ek degenerate lekin valid limiting case.

Hole se guzarne wali ek vertical strip outer circle mein enter karti hai, khaali centre mein exit karti hai, phir re-enter karti hai (Figure 2) — do disjoint $y$-ranges — isliye koi single top/bottom pair kaam nahi karta; pieces mein split karna (ya Polar coordinates double integrals use karna) clean fix hai.

Tum $D$ ko har crossing par sub-regions mein split karte ho jahan top/bottom (ya left/right) ordering consistent ho, phir alag integrals sum karte ho — ek single limit pair ek swap span nahi kar sakta jismein kaun si curve upar hai.

Zero — zero area wala set koi bhi volume contribute nahi karta chahe wahan $f$ kitna bhi bada ho, kyunki strips ki "thickness" kuch nahi tak shrink ho jaati hai.

Haan, lekin outer integral improper ho jaata hai (${\int }_0^{\infty }$) aur answer tabhi trustworthy hai jab integral absolutely converge kare (${\iint }_D|f|dA<\infty$); us condition ke saath tum same strip logic rakhte ho aur limit lete ho, aur boundary curve $g_2(x)=e^{-x}$ abhi bhi inner limit deti hai.

Yeh signed volume represent karta hai — jahan $f<0$ woh parts subtract karte hain — isliye number zero ya negative ho sakta hai chahe region nonempty ho; saccha geometric volume ${\iint }_D|f|dA$ chahiye hoga.