Changing order of integration: Worked examples — Hinglish deep dive

4.4.18 · D3 · Maths › Multivariable Calculus › Changing order of integration

Yeh page Changing order of integration ki practice ground hai. Parent note ne tumhe 5-step recipe sikhaya; yahan hum isse har tarah ke region ke against run karte hain jo yeh topic de sakta hai — aur ek bhi case skip nahi karte. Kisi bhi example se pehle, hum scenarios ka ek matrix banate hain taaki tum poora battlefield dekh sako.

Scenario matrix

Cell	Region ka shape	Woh twist jise tumhe survive karna hai	Example
C1	Triangle (line + axes)	inner ka koi elementary antiderivative nahi — bachane ke liye reorder karo	Ex 1
C2	Lens (do curves cross karti hain: line vs parabola)	naye variable ke liye dono boundaries ko re-solve karna hoga, correct branch choose karo	Ex 2
C3	Region jise ek taraf split chahiye	ek orientation = 2 integrals, doosra = 1 clean integral	Ex 3
C4	Strip jo alag-alag heights par different curves par exit karta hai	exit boundary switch karti hai — tumhe ZAROOR split karna hoga	Ex 4
C5	Degenerate corner / zero-width tip	boundaries ek point par milti hain; check karo limit theek hai, undefined nahi	Ex 5
C6	Circle / disk piece	reorder karo AUR recognize karo ki polar aur bhi better hai	Ex 6
C7	Word problem (real-world total)	physical region ko limits mein translate karo, phir compute karne ke liye reorder karo	Ex 7
C8	Exam twist: unknown/general $f$ , symbolic answer	$f$ ko kabhi evaluate kiye bina reorder karo	Ex 8

Har cell C1–C8 neeche cover kiya gaya hai. Numeric answers sab verify block mein machine-checked hain.

Ex 1 — Cell C1: triangle, impossible inner integral

Worked example Statement

$I = \int_{0}^{1} \int_{x}^{1} e^{y^{2}} d y d x$ evaluate karo.

Forecast: Padhne se pehle guess karo — kya answer mein $e$ hoga? Kya reordering se $y$ -integral doable ho jaayega? (Yeh socho.)

Step 1 — Limits padho. Outer $0 \leq x \leq 1$ ; inner $x \leq y \leq 1$ . Yeh step kyun? Limits HI region hain. Unhe zor se padhne se hume boundary curves milti hain: har vertical strip ka bottom $y = x$ line par chalta hai, top $y = 1$ line par.

Step 2 — Sketch (Figure s01 dekho). $y = x$ ke upar, $y = 1$ ke neeche, $x = 0$ ke daayein: corners $(0, 0), (0, 1), (1, 1)$ wala ek triangle.

Yeh step kyun?

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

ka no elementary antiderivative hai, isliye given order ek dead end hai. Sketch humein alternative order batata hai.

Step 3 — Naya outer variable $y$ . Picture se poora region $0 \leq y \leq 1$ span karta hai. Yeh step kyun? Sabse neecha tile $y = 0$ par hai (corner), sabse upar $y = 1$ par (top edge). Outer limits pure numbers hone chahiye.

Step 4 — Naya inner variable $x$ . Ek fixed height $y$ par ek horizontal strip (figure mein red) slide karo. Yeh left wall $x = 0$ par enter karta hai aur slanted boundary $y = x$ , yaani $x = y$ par exit karta hai. Toh $0 \leq x \leq y$ . Yeh step kyun? Hume boundary $y = x$ ko naye inner variable ke liye ek formula ke roop mein re-express karna hai: $y = x$ ko $x$ ke liye solve karo $\Rightarrow x = y$ .

Step 5 — Rewrite & evaluate. $I = \int_{0}^{1} \int_{0}^{y} e^{y^{2}} d x d y = \int_{0}^{1} e^{y^{2}} [x]_{0}^{y} d y = \int_{0}^{1} y e^{y^{2}} d y .$ $u = y^{2}, d u = 2 y d y$ substitute karo: $I = \frac{1}{2} \int_{0}^{1} e^{u} d u = \frac{1}{2} (e - 1) .$

Verify: Numerically $\frac{1}{2} (e - 1) \approx 0.8591$ . Sanity check: integrand $e^{y^{2}} \geq 1$ area $\frac{1}{2}$ ke triangle par, toh $I \geq \frac{1}{2}$ — consistent. ✓

Ex 2 — Cell C2: ek line aur parabola ke beech lens

Worked example Statement

$J = \int_{0}^{1} \int_{x^{2}}^{x} (x + y) d y d x$ evaluate karo.

Forecast: Region $y = x$ aur $y = x^{2}$ ke beech ka sliver hai. Guess karo: reorder karne ke baad, kya left edge line hogi ya parabola?

Step 1 — Padho. $0 \leq x \leq 1$ , $x^{2} \leq y \leq x$ . Yeh step kyun? Har vertical strip ka bottom parabola $y = x^{2}$ par chalta hai; top line $y = x$ par. Yeh $x = 0$ aur $x = 1$ par coincide karte hain.

Step 2 — Sketch (Figure s02). Line (upper) aur parabola (lower) ke beech lens-shaped region, corners $(0, 0), (1, 1)$ .

Yeh step kyun? Hume jaanna hai ki fixed height

y

ke liye, kaun si curve left par hai aur kaun si right par.

Step 3 — Naya outer $y$ : $0 \leq y \leq 1$ . Yeh step kyun? Lens ke lowest aur highest points do corners hain.

Step 4 — Naya inner $x$ . Fixed $y \in (0, 1)$ ke liye, har boundary ko $x$ ke liye solve karo:

line $y = x \Rightarrow x = y$ ,
parabola $y = x^{2} \Rightarrow x = y$ (positive branch — region first quadrant mein hai).

Kyunki $0 < y < 1$ se $y < y$ milta hai, strip $y \leq x \leq y$ run karti hai. Yeh step kyun? Hume dono boundaries re-solve karni HAIN aur woh branch choose karni hai jo $R$ mein ho. Test $y = \frac{1}{4}$ : strip $[0.25, 0.5]$ — dono lens ke andar. ✓

Step 5 — Dono taraf evaluate karo (unhe match karna chahiye). Original order: $J = \int_{0}^{1} \int_{x^{2}}^{x} (x + y) d y d x = \int_{0}^{1} [x y + \frac{y ^{2}}{2}]_{x^{2}}^{x} d x = \int_{0}^{1} (\frac{3}{2} x^{2} - x^{3} - \frac{x ^{4}}{2}) d x .$ $= \frac{1}{2} - \frac{1}{4} - \frac{1}{10} = \frac{3}{20} .$ Reordered check: $J = \int_{0}^{1} \int_{y}^{y} (x + y) d x d y = \frac{3}{20}$ bhi.

Verify: $J = \frac{3}{20} = 0.15$ , dono orders agree karte hain. ✓

Ex 3 — Cell C3: ek order = do integrals, doosra = ek

Worked example Statement

Order reverse karo aur $S = \int_{0}^{2} \int_{0}^{x} y d y d x + \int_{2}^{4} \int_{0}^{4 - x} y d y d x$ evaluate karo. Yeh vertices $(0, 0), (4, 0), (2, 2)$ wale triangle par total hai, do vertical-strip integrals ke roop mein likha gaya hai (ise $x = 2$ par split karna pada jahan top edge $y = x$ se $y = 4 - x$ par switch hoti hai).

Forecast: Guess karo: kya horizontal strips is triangle ko EK integral mein describe kar sakti hain (bina split ke)?

Step 1 — Padho. Left piece: $0 \leq x \leq 2, 0 \leq y \leq x$ . Right piece: $2 \leq x \leq 4, 0 \leq y \leq 4 - x$ . Yeh step kyun? Do pieces isliye kyunki vertical strip ki top boundary $x < 2$ ke liye $y = x$ hai lekin $x > 2$ ke liye $y = 4 - x$ — exit curve apex par switch karti hai.

Step 2 — Sketch (Figure s03). Ek triangle: $x$ -axis par $0$ se $4$ tak base, $(2, 2)$ par apex.

Yeh step kyun? Picture dikhata hai ki height

y

par horizontal strip left edge

y = x \Rightarrow x = y

par enter karta hai aur right edge

y = 4 - x \Rightarrow x = 4 - y

par exit karta hai, SAARE

y

ke liye — koi split nahi chahiye.

Step 3 — Naya outer $y$ : $0 \leq y \leq 2$ . Yeh step kyun? Apex height $2$ hai, base $0$ par.

Step 4 — Naya inner $x$ : $y \leq x \leq 4 - y$ . Yeh step kyun? Dono edges $x$ ke liye solve ki gayi hain; ek single description kaam karti hai.

Step 5 — Single integral evaluate karo. $S = \int_{0}^{2} \int_{y}^{4 - y} y d x d y = \int_{0}^{2} y [(4 - y) - y] d y = \int_{0}^{2} y (4 - 2 y) d y = \int_{0}^{2} (4 y - 2 y^{2}) d y .$ $= [2 y^{2} - \frac{2}{3} y^{3}]_{0}^{2} = 8 - \frac{16}{3} = \frac{8}{3} .$

Verify: $S = \frac{8}{3} \approx 2.667$ . Do original pieces ko sum karke cross-check: $\int_{0}^{2} \frac{x ^{2}}{2} d x + \int_{2}^{4} \frac{( 4 - x ) ^{2}}{2} d x = \frac{4}{3} + \frac{4}{3} = \frac{8}{3}$ . ✓ (Reordering ne 2 integrals ko 1 mein badal diya.)

Ex 4 — Cell C4: exit boundary switch karti hai (doosre taraf split karna ZAROORI hai)

Worked example Statement

$T = \int_{0}^{2} \int_{x^{2}}^{2 x} f (x, y) d y d x$ mein order reverse karo (region $y = x^{2}$ aur $y = 2 x$ ke beech).

Forecast: Curves $(0, 0)$ aur $(2, 4)$ par cross karti hain. Guess karo: jab hum horizontal strips use karte hain, kya ek exit curve poori height ke liye kaam karti hai, ya switch hoti hai?

Step 1 — Padho. $0 \leq x \leq 2$ , $x^{2} \leq y \leq 2 x$ . $[0, 2]$ par $x^{2} \leq 2 x$ , toh parabola line ke neeche hai. Yeh step kyun? Region establish karta hai: line $y = 2 x$ ke neeche, parabola $y = x^{2}$ ke upar.

Step 2 — Sketch (Figure s04). $(0, 0)$ se $(2, 4)$ tak ek curved sliver.

Yeh step kyun? Hum check karte hain ki horizontal strip kahan enter/exit karti hai.

x

ke liye solve karo: line

y = 2 x \Rightarrow x = y /2

; parabola

y = x^{2} \Rightarrow x = y

(positive branch).

0 < y < 4

ke liye,

y /2 < y

— strip

y /2 \leq x \leq y

run karti hai, aur yeh POORE range

0 \leq y \leq 4

ke liye hold karta hai.

Step 3 — Naya outer $y$ : $0 \leq y \leq 4$ (apex $(2, 4)$ par). Step 4 — Naya inner $x$ : $\frac{y}{2} \leq x \leq y$ . Yeh step kyun? Dono boundaries $x$ ke liye re-solve ki gayi hain; $\cdot$ ka correct branch choose kiya gaya (region first quadrant mein). Test $y = 1$ : strip $[0.5, 1]$ ; midpoint $(0.7, 1)$ check karo satisfies $0.49 \leq 1 \leq 1.4$ ✓.

Step 5 — Result. $T = \int_{0}^{4} \int_{y /2}^{y} f (x, y) d x d y .$ Concrete banane ke liye $f = 1$ lo (area). Original: $\int_{0}^{2} (2 x - x^{2}) d x = 4 - \frac{8}{3} = \frac{4}{3}$ . Reordered: $\int_{0}^{4} (y - \frac{y}{2}) d y = \frac{2}{3} \cdot 8 - \frac{1}{4} \cdot 16 = \frac{16}{3} - 4 = \frac{4}{3}$ . ✓

Verify: area $= \frac{4}{3}$ dono orders mein. ✓ (Yahan exit curve ne switch NAHI kiya — Ex 3 se contrast karo jahan kiya. Picture padhna hi batata hai kaun sa.)

Ex 5 — Cell C5: degenerate tip, check karo limit theek hai

Worked example Statement

$U = \int_{0}^{1} \int_{0}^{1 - x^{2}} x d y d x$ mein order reverse karo — ek quarter disk. Degenerate corners par dhyaan do jahan strip width zero ho jaata hai.

Forecast: $x = 1$ par top boundary $1 - x^{2} = 0$ : vertical strip ki height zero hai. Guess karo: kya isse kisi bhi order mein koi trouble hoti hai?

Step 1 — Padho. $0 \leq x \leq 1$ , $0 \leq y \leq 1 - x^{2}$ : first quadrant mein unit circle ka quarter. Yeh step kyun? $y = 1 - x^{2}$ upper arc $x^{2} + y^{2} = 1$ hai.

Step 2 — Sketch (Figure s05). Quarter disk, corners $(0, 0), (1, 0), (0, 1)$ .

Yeh step kyun?

x \to 1

par strip height

\to 0

(

(1, 0)

par ek degenerate tip). Integrands finite rehte hain, toh integral ek proper (non-improper) integral hai — koi blow-up nahi.

Step 3 — Naya outer $y$ : $0 \leq y \leq 1$ . Step 4 — Naya inner $x$ : arc $y = 1 - x^{2} \Rightarrow x = 1 - y^{2}$ solve karo (positive branch). Toh $0 \leq x \leq 1 - y^{2}$ . Yeh step kyun? Horizontal strip $x = 0$ wall par enter karta hai, arc $x = 1 - y^{2}$ par exit karta hai. $y \to 1$ par yeh width $\to 0$ — WAHI degenerate tip, ab top par. Formula wahan bhi finite hai, confirming koi problem nahi.

Step 5 — Evaluate. $U = \int_{0}^{1} \int_{0}^{1 - y^{2}} x d x d y = \int_{0}^{1} \frac{1}{2} (1 - y^{2}) d y = \frac{1}{2} [y - \frac{y ^{3}}{3}]_{0}^{1} = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3} .$

Verify: Original order: $\int_{0}^{1} x 1 - x^{2} d x = [- \frac{1}{3} (1 - x^{2})^{3/2}]_{0}^{1} = \frac{1}{3}$ . Dono $\frac{1}{3}$ dete hain; degenerate tip ne koi defect contribute nahi kiya. ✓

Ex 6 — Cell C6: disk piece jahan polar jeet jaata hai

Worked example Statement

Same quarter disk par $V = \int_{0}^{1} \int_{0}^{1 - x^{2}} (x^{2} + y^{2}) d y d x$ consider karo. Reorder karo — phir note karo ki Polar Coordinates in Double Integrals aur bhi cleaner hai.

Forecast: Guess karo ki Cartesian reorder pleasant hai, ya tum $r, θ$ par switch karna prefer karoge.

Step 1 — Padho & sketch karo. Same quarter disk as Ex 5 (Figure s05 reuse karo). Yeh step kyun? Region identical hai; sirf integrand $x^{2} + y^{2}$ mein badal gaya.

Step 2 — Reorder (Cartesian). Ex 5 wali solving se: $V = \int_{0}^{1} \int_{0}^{1 - y^{2}} (x^{2} + y^{2}) d x d y .$ Yeh step kyun? Dikhata hai ki reorder valid hai, lekin inner integral $\int (x^{2} + y^{2}) d x = \frac{x ^{3}}{3} + y^{2} x$ ko $x = 1 - y^{2}$ par evaluate karna messy hai.

Step 3 — Polar par switch karo. Quarter disk par: $x^{2} + y^{2} = r^{2}$ , $d A = r d r d θ$ , with $0 \leq r \leq 1$ , $0 \leq θ \leq \frac{π}{2}$ . Yeh step kyun? Polar kyun, aur zyada Cartesian gymnastics kyun nahi? Kyunki region ek circular arc hai aur integrand sirf $r$ par depend karta hai — polar limits ko constants banata hai aur integral ko separate karta hai. Yahi woh hai jo Change of Variables (Jacobian) formalise karta hai ( $r$ factor Jacobian deta hai).

Step 4 — Evaluate. $V = \int_{0}^{π /2} \int_{0}^{1} r^{2} \cdot r d r d θ = \int_{0}^{π /2} [\frac{r ^{4}}{4}]_{0}^{1} d θ = \frac{1}{4} \cdot \frac{π}{2} = \frac{π}{8} .$

Verify: $V = \frac{π}{8} \approx 0.3927$ . (Cartesian reorder grinding ke baad same deta hai.) ✓

Ex 7 — Cell C7: real-world word problem

Worked example Statement

Ek triangular metal plate ke vertices $(0, 0), (2, 0), (0, 2)$ hain (units: cm). Iska surface density $ρ (x, y) = e^{y^{2}}$ g/cm² hai. Mass $M = \iint_{R} e^{y^{2}} d A$ hai. Ise convenient order mein set up karo aur compute karo.

Forecast: Guess karo: inner integral doable ho sake isliye kaun sa variable outer hona chahiye?

Step 1 — Plate ko ek region mein translate karo. Triangle $x = 0$ , $y = 0$ , aur $(2, 0), (0, 2)$ se hoke jaane wali line se bounded hai: $x + y = 2$ , yaani $0 \leq x \leq 2 - y$ . Toh $R : 0 \leq y \leq 2, 0 \leq x \leq 2 - y$ . Yeh step kyun? Word problem sirf ek region + integrand disguise mein hota hai; hume dono extract karne hain.

Step 2 — Sketch (Figure s06).

Yeh step kyun? Hypotenuse

x + y = 2

confirm karta hai aur kaun sa order

e^{y^{2}}

ko integrable banata hai yeh bhi.

Step 3 — $y$ ko OUTER choose karo. Agar $y$ inner hota toh hume $\int e^{y^{2}} d y$ face karna padta — impossible. $y$ outer hone se, inner $x$ -integral $e^{y^{2}}$ ko ek constant maanta hai. Yeh step kyun? Ex 1 wala hi lesson: "impossible" variable ko BAHAR rakho.

Step 4 — Inner limits. Fixed $y$ ke liye, $x$ $0$ se $2 - y$ tak run karta hai. Yeh step kyun? Left wall $x = 0$ , right boundary hypotenuse $x = 2 - y$ .

Step 5 — Evaluate. $M = \int_{0}^{2} \int_{0}^{2 - y} e^{y^{2}} d x d y = \int_{0}^{2} (2 - y) e^{y^{2}} d y .$ Split karo: $\int_{0}^{2} 2 e^{y^{2}} d y - \int_{0}^{2} y e^{y^{2}} d y$ . Doosra elementary hai: $\frac{1}{2} (e^{4} - 1)$ . Pehla genuinely non-elementary integral hai (dekho Improper and Non-elementary Integrals); ise $2 \int_{0}^{2} e^{y^{2}} d y$ likho, ek value $\approx 2 (16.4526) = 32.905$ . $M \approx 32.905 - \frac{1}{2} (e^{4} - 1) = 32.905 - 26.799 = 6.106 g .$ Yeh step kyun? Yahan bhi reordering ne elementary chunk ko isolate kiya; baaki term honestly non-elementary hai aur numerically evaluate karni padegi.

Verify: $M \approx 6.11$ g. Units se sanity: g/cm² × cm² = g ✓. Bound se sanity: area $= 2$ cm², density $1$ aur $e^{4} \approx 54.6$ ke beech, aur mass $6.1$ $[2, 109]$ mein hai ✓.

Ex 8 — Cell C8: exam twist, general $f$ , symbolic reorder

Worked example Statement

Integration ka order reverse karo, evaluate kiye bina, $W = \int_{0}^{2} \int_{0}^{y /2} f (x, y) d x d y + \int_{2}^{4} \int_{0}^{2 - y /2 + ?} \dots$ — actually yeh clean twist hai: $W = \int_{0}^{1} \int_{- y}^{y} f (x, y) d x d y$ reverse karo. Yeh region $x = 0$ ke across symmetric hai, toh yeh $y$ -axis ke left aur right dono quadrants span karta hai.

Forecast: $x$ ke dono signs appear hote hain. Guess karo: reorder karne ke baad, kya region $x = 0$ par split hoga, aur do branches kya hain?

Step 1 — Padho. $0 \leq y \leq 1$ , $- y \leq x \leq y$ : parabola $y = x^{2}$ ke neeche ka region upar ki taraf opening, $y = 1$ par capped, $y$ -axis ke baare mein symmetric. Yeh step kyun? $x = \pm y$ , $y = x^{2}$ ki do branches hain; $x$ ke dono signs hote hain, toh hume negative- $x$ aur positive- $x$ tiles dono handle karne hain.

Step 2 — Sketch (Figure s07).

Yeh step kyun? Dikhata hai ki region

- 1 \leq x \leq 1

par ek parabolic cap hai, neeche

y = x^{2}

se aur upar

y = 1

se bounded.

Step 3 — Naya outer $x$ : overall $- 1 \leq x \leq 1$ . Yeh step kyun? Sabse wide tiles $y = 1$ par hain jahan $x = \pm 1$ . Note karo ki outer variable ab negative values leta hai — ek aisa case jo naive "min=0" ki aadat miss kar deti.

Step 4 — Naya inner $y$ : fixed $x$ ke liye, vertical strip parabola $y = x^{2}$ par enter karti hai aur $y = 1$ par exit karti hai. Toh $x^{2} \leq y \leq 1$ . Yeh single description $x$ ke DONO signs cover karti hai kyunki $x^{2}$ even hai — koi split nahi chahiye. Yeh step kyun? Boundary $x = \pm y$ ko $y$ ke liye re-solve karne par single curve $y = x^{2}$ milti hai, jo do branches ko merge kar deti hai.

Step 5 — Result. $W = \int_{- 1}^{1} \int_{x^{2}}^{1} f (x, y) d y d x .$ Yeh step kyun? Poori tarah symbolic; Fubini's Theorem se is general region par kisi bhi integrable $f$ ke liye valid.

Verify ( $f = 1$ lo, yaani area): original $\int_{0}^{1} 2 y d y = 2 \cdot \frac{2}{3} = \frac{4}{3}$ ; reordered $\int_{- 1}^{1} (1 - x^{2}) d x = 2 - \frac{2}{3} = \frac{4}{3}$ . Dono $= \frac{4}{3}$ , reorder confirm karta hai (aur yeh bhi ki negative $x$ correctly include kiya gaya). ✓

Recall Which cell does each trap live in?

"Left the outer limit as a curve" sabse zyada kaun se cell type mein hota hai? ::: Word/general ones mein (C7, C8) — OUTER limits ko hamesha constants tak reduce karo. Reordering 2 integrals ko 1 mein kab badalta hai? ::: Jab exit curve ek orientation mein switch karti hai lekin doosre mein nahi (C3). Square root ka negative branch correct kab hota hai? ::: Jab region $y$ -axis ke baayein ho (C8) — woh branch choose karo jo $R$ mein land kare.

Aage kahan jaayein

Reorder mechanism Fubini's Theorem se licensed hai.
Circular/annular regions: Polar Coordinates in Double Integrals aur general Change of Variables (Jacobian) prefer karo.
Teen-variable analogue: Triple Integrals — Order of Integration.