Double integrals over general regions — Type I and II

4.4.17 · Maths › Multivariable Calculus

WHY hume "general regions" chahiye

WHY ye matter karta hai: almost koi bhi real region rectangle nahi hoti. Areas, volumes, masses, centres of mass, aur probabilities sab triangles, circles, parabolic caps, etc. ke upar hote hain. Type I/II wo bread-and-butter method hai jo in sab ko computable banata hai.

Type I aur Type II: slice karne ke do tarike

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

DERIVATION — iterated integral kahan se aata hai

Hum Type I ko first principles se derive karte hain (Fubini + extend-by-zero).

Step 1 — Enclose. $D$ ko rectangle $R = [a, b] \times [c, d]$ ke andar rakh do jahan $c \leq g_{1} (x)$ aur $g_{2} (x) \leq d$ sab $x \in [a, b]$ ke liye ho. Ye step kyun? Hum sirf rectangles par iterate karna jaante hain, isliye hum $R$ par kaam karte hain.

Step 2 — Extend by zero. $F = f$ define karo $D$ par, $F = 0$ $D$ ke bahar. Tab $\iint_{D} f d A = \iint_{R} F d A$ . Kyun? Zero region koi volume add nahi karta, isliye values agree karti hain.

Step 3 — Rectangle par Fubini apply karo. $\iint_{R} F d A = \int_{a}^{b} (\int_{c}^{d} F (x, y) d y) d x .$ Kyun? Fubini hume constant limits wale rectangle par iterate karne deta hai.

Step 4 — Inner integral collapse karo. Kisi fixed $x$ ke liye, $F (x, y) = 0$ jab tak $g_{1} (x) \leq y \leq g_{2} (x)$ na ho. Isliye $\int_{c}^{d} F (x, y) d y = \int_{g_{1} (x)}^{g_{2} (x)} f (x, y) d y .$ Kyun? $[g_{1}, g_{2}]$ ke bahar integrand literally $0$ hai, jo kuch contribute nahi karta.

Step 5 — Combine karo: $\iint_{D} f d A = \int_{a}^{b} \int_{g_{1} (x)}^{g_{2} (x)} f (x, y) d y d x . ■$

Type II bilkul same hai, bas $x$ aur $y$ ki roles swap hain (horizontal strips).

WORKED EXAMPLES

Worked example 1 — Triangle, dono tarike se

$f (x, y) = x + 2 y$ ko triangle $D$ ke upar integrate karo jiske vertices $(0, 0), (1, 0), (1, 1)$ hain (region line $y = x$ ke neeche, $x \leq 1$ ke saath).

Type I ke roop mein (vertical strips). Fixed $x \in [0, 1]$ ke liye, $y$ $0$ se line $y = x$ tak jaata hai. Kyun? Bottom $x$ -axis hai ( $y = 0$ ); top $y = x$ hai. $\int_{0}^{1} \int_{0}^{x} (x + 2 y) d y d x .$ Inner: $\int_{0}^{x} (x + 2 y) d y = [x y + y^{2}]_{0}^{x} = x^{2} + x^{2} = 2 x^{2} .$ Outer: $\int_{0}^{1} 2 x^{2} d x = \frac{2}{3} .$

Type II ke roop mein (horizontal strips). Fixed $y \in [0, 1]$ ke liye, $x$ line $x = y$ se $x = 1$ tak jaata hai. Kyun? Horizontal strip par left boundary $y = x \Rightarrow x = y$ hai; right boundary $x = 1$ hai. $\int_{0}^{1} \int_{y}^{1} (x + 2 y) d x d y .$ Inner: $\int_{y}^{1} (x + 2 y) d x = [\frac{x ^{2}}{2} + 2 y x]_{y}^{1} = (\frac{1}{2} + 2 y) - (\frac{y ^{2}}{2} + 2 y^{2}) = \frac{1}{2} + 2 y - \frac{5 y ^{2}}{2} .$ Outer: $\int_{0}^{1} (\frac{1}{2} + 2 y - \frac{5 y ^{2}}{2}) d y = \frac{1}{2} + 1 - \frac{5}{6} = \frac{2}{3} .$ ✅ Same answer — order value nahi badalta, sirf difficulty badal sakti hai.

COMMON MISTAKES (steel-manned)

Recall Feynman: 12-saal ke bachche ko explain karo

Socho ek pahadi tent hai aur tum uske neeche ki jagah jaanna chahte ho. Tent ke neeche ke floor ko thin strips mein kaat lo, jaise ek bread loaf. Har strip ke liye tum measure karte ho ki tent kitna tall hai uske saath aur jod lete ho — wahi inner integral hai. Phir tum sab strips ko pure floor mein jod lete ho — wahi outer integral hai. Agar tum bread ko lambe taraf (vertical strips) kaat lo toh woh Type I hai; agar chote taraf (horizontal strips) kaat lo toh woh Type II hai. Dono taraf tumhe same jagah milti hai; tum sirf wohi slicing chunte ho jo measure karna aasaan ho.

Type I region kya hota hai?

D = {a \leq x \leq b, g_{1} (x) \leq y \leq g_{2} (x)}

, integrate hota hai

\int_{a}^{b} \int_{g_{1} (x)}^{g_{2} (x)} f d y d x

ke roop mein (vertical strips).

Type II region kya hota hai?

D = {c \leq y \leq d, h_{1} (y) \leq x \leq h_{2} (y)}

, integrate hota hai

\int_{c}^{d} \int_{h_{1} (y)}^{h_{2} (y)} f d x d y

ke roop mein (horizontal strips).

Outer integral ki limits constants kyun honi chahiye?

Kyunki outer integral ek number par evaluate hona chahiye; uski limits mein variable hone se ek undefined expression reh jaayega.

Double integral se region ka area kaise compute karte hain?

Area = \iint_{D} 1 d A

Non-rectangular region par integral define karne ka kaunsa "trick" hai?

D

ko ek rectangle mein band karo aur

f

D

ke bahar

0

tak extend karo; bahar ka area kuch contribute nahi karta.

Integration ka order switch karna essential (sirf convenient nahi) kab hota hai?

Jab diye hue order mein inner antiderivative exist nahi karta, jaise

\int\int e^{y^{2}} d y d x

— pehle

x

mein integrate karne ke liye swap karo.

Triangle (0,0),(1,0),(1,1) ke liye,

y

ke Type I inner limits kya hain?

0

x

tak.

Ye decide kaise karte hain ki kaun si curve

y

ke liye upper bound hai?

Ek test

x

-value plug karo aur dono curves ki

y

-values compare karo; jo bada ho woh top (

g_{2}

) hai.

Connections

Fubini's Theorem — wo engine jo double integral ko iterated integral mein turn karne ko justify karta hai.
Double integrals over rectangles — simpler base case jis par Type I/II reduce hota hai.
Changing the order of integration — direct application; sketch karo phir limits dobara padhو.
Area between two curves — $f \equiv 1$ set karne se milta hai.
Polar coordinates double integrals — alternative jab region circular/annular ho.
Centre of mass and moments — in integrals ko weight $f = ρ (x, y)$ ke saath use karta hai.
Jacobian and change of variables — slicing ko skewed coordinate grids tak generalise karta hai.