Triple integrals in Cartesian, cylindrical, spherical coordinates: Worked examples

4.4.20 · D3 · Maths › Multivariable Calculus › Triple integrals in Cartesian, cylindrical, spherical coordi

Scenario matrix

Kuch bhi compute karne se pehle, chalte hain har distinct situation ki list banate hain jo ek triple integral tumhare saamne rakh sakta hai. Neeche ke har worked example mein us cell ka tag lagega jise woh fill karta hai.

Cell	Kya cheez use distinct banati hai	Example
C1 Cartesian, straight walls	limits constants ya simple lines/planes hain	Ex 1
C2 Cartesian, order matters	$d z d y d x$ ya koi aur order choose karna difficulty badal deta hai	Ex 2
CYL Cylindrical, axis symmetry	region $z$ -axis ke around rotate karta hai	Ex 3
SPH Spherical, full round	boundary $ρ =$ const hai, integrand mein $x^{2} + y^{2} + z^{2}$ hai	Ex 4
DEG Degenerate / zero region	limits collapse ho jaate hain — volume $0$ hai; ek sanity trap	Ex 5
LIM Limiting behaviour	koi parameter $\to 0$ ya $\to \infty$ le jaao aur check karo	Ex 6
SIGN Negative coordinates / sign-changing integrand	region ek axis cross karta hai; $f$ pehle $+$ phir $-$ hoti hai	Ex 7
WORD Real-world word problem	density $\to$ mass ya centre of mass	Ex 8
TWIST Exam-style twist	galat-lagta coordinate, ya order reverse karna taaki solve ho sake	Ex 9

Is page se jaane se pehle har cell cover honi chahiye. Chalein shuru karte hain.

Ex 1 — Cartesian, straight walls (cell C1)

Worked example Ek slab jiska oopar ka dhakna thoda tircha hai

Us solid $E$ ka volume nikalo jo $0 \leq x \leq 2$ , $0 \leq y \leq 3$ , aur $0 \leq z \leq x + 1$ se bounded hai.

Forecast: Aage padhne se pehle guess karo — kya answer flat box $2 \times 3 \times 1 = 6$ se bada hai ya chhota? (Dhakna $z = x + 1$ upar $1$ se $3$ tak jaata hai, toh average height $2$ honi chahiye. Apna guess likh lo.)

Step 1 — inner $z$ . $x, y$ freeze karo; ek arrow upar shoot karo. Woh $z = 0$ pe enter karta hai aur dhakne $z = x + 1$ pe exit karta hai. $\int_{0}^{x + 1} d z = x + 1$ Yeh step kyun? Innermost variable wahi hota hai jiska top/bottom dusron pe depend karta hai; yahan dhakna $x$ pe depend karta hai, isliye $z$ andar jaata hai.

Step 2 — middle $y$ . $x y$ -plane pe shadow rectangle $0 \leq y \leq 3$ hai; $y$ ke limits constant hain. $\int_{0}^{3} (x + 1) d y = 3 (x + 1)$ Yeh step kyun? $x + 1$ mein koi $y$ nahi, toh $y$ pe integrate karna sirf width $3$ se multiply karna hai.

Step 3 — outer $x$ . Constant range $0 \to 2$ . $\int_{0}^{2} 3 (x + 1) d x = 3 [\frac{x ^{2}}{2} + x]_{0}^{2} = 3 (2 + 2) = 12$ Yeh step kyun? Outer variable woh aakhri wala hota hai jo bacha hai, isliye use solid ke poore $x$ -extent pe run karna hoga, yahan constant interval $0 \leq x \leq 2$ . (Is box mein extent constant hai; ek tirche region mein outer range alag ho sakti hai, lekin yeh hamesha outer variable ka total spread hota hai $E$ ke across, jab $y, z$ already integrate ho chuke hon.)

Verify: $x \in [0, 2]$ pe average lid height $\frac{1}{2} \int_{0}^{2} (x + 1) d x = \frac{1}{2} \cdot 4 = 2$ hai (hum interval length $2$ se divide karte hain). Base area $= 2 \times 3 = 6$ . Volume $= 6 \times 2 = 12$ . ✓ Match karta hai — aur yeh flat box ke $6$ se bada nikla, jaise forecast tha.

Ex 2 — Cartesian, order matters (cell C2)

Worked example Same solid, mushkil order

Tetrahedron $x, y, z \geq 0, x + y + z \leq 1$ ko dobara compute karo lekin order $d x d z d y$ mein integrate karo. (Parent ne $d z d y d x$ kiya tha aur $\frac{1}{6}$ mila tha.)

Forecast: Answer zaroor $\frac{1}{6}$ hi rahega — solid nahi badla. Baat yeh hai ki limits cleanly kaise re-derive hoti hain, yeh dekhna hai.

Step 1 — inner $x$ . $y, z$ freeze karo; constraint $x + y + z \leq 1$ se $x$ milta hai $0$ se $1 - y - z$ tak. $\int_{0}^{1 - y - z} d x = 1 - y - z$ Yeh step kyun? Hum jis variable ko bhi innermost banayein, uska exit face wahi plane $x + y + z = 1$ hai us variable ke liye solve karke.

Step 2 — middle $z$ . $x$ gone hai, $y z$ -plane pe shadow $y, z \geq 0, y + z \leq 1$ hai. $y$ freeze karo: $z$ runs $0 \to 1 - y$ . $\int_{0}^{1 - y} (1 - y - z) d z = [(1 - y) z - \frac{z ^{2}}{2}]_{0}^{1 - y} = \frac{( 1 - y ) ^{2}}{2}$ Yeh step kyun? Shadow ki boundary paane ke liye constraint mein $x = 0$ set karo: $y + z = 1$ milta hai.

Step 3 — outer $y$ . $y : 0 \to 1$ . $\int_{0}^{1} \frac{( 1 - y ) ^{2}}{2} d y = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$ Yeh step kyun? Outer variable solid ke poore $y$ -spread pe run karta hai, jo is tetrahedron ke liye constant interval $0 \leq y \leq 1$ hai; ab tak $x, z$ already integrate ho chuke hain toh sirf ek number bachta hai.

Verify: $\frac{1}{6}$ — parent ke $d z d y d x$ answer se identical. ✓ Lesson: integration ka order ek fixed solid ke liye value kabhi nahi badalta; sirf limits ki ugliness badal sakti hai.

Ex 3 — Cylindrical, axis symmetry (cell CYL)

Figure — Triple integrals in Cartesian, cylindrical, spherical coordinates

Worked example Ek paraboloid ke neeche, plane ke upar volume

Region $E$ ka volume nikalo jo paraboloid $z = 4 - x^{2} - y^{2}$ ke andar aur $z = 0$ ke upar hai.

Forecast: Paraboloid radius $2$ ka ek bowl cap karta hai ( $z = 0$ set karo: $x^{2} + y^{2} = 4$ ) aur height $4$ hai. Guess karo: kya yeh us cylinder $π \cdot 2^{2} \cdot 4 = 16 π$ se bada ya chhota hai jo ise contain karta hai? (Ek dome apne bounding cylinder se neeche hota hai.)

Figure mein: cyan curve paraboloid $z = 4 - r^{2}$ hai edge-on dekha gaya; neeche safed line floor $z = 0$ hai; amber vertical arrow $r = 1$ pe inner- $z$ integration path dikhata hai — yeh floor pe start hota hai aur cyan lid pe rukta hai, yahi matlab hai "shoot an arrow up" ka.

Step 1 — cylindrical mein switch karo. $x = r cos θ, y = r sin θ$ rakho. Tab $x^{2} + y^{2} = r^{2}$ , isliye lid hai $z = 4 - r^{2}$ aur $d V = r d r d θ d z$ . Hum order $d z d r d θ$ mein integrate karenge: innermost $z$ ( $r, θ$ hold karo), phir $r$ ( $θ$ hold karo), phir outer $θ$ . $r$ factor kyun? Polar "pixel" ek chhota wedge hota hai width $d r$ aur arc length $r d θ$ ke saath, isliye uska base area $r d r d θ$ hai (na ki $d r d θ$ ); height $d z$ se multiply karne par $d V = r d r d θ d z$ milta hai. Bada $r$ ⇒ chauda wedge ⇒ zyada volume. Woh $r$ cylindrical Jacobian hai. Note karo ki $d V$ differentials group karta hai; hum actually kis order mein integrate karein yeh hamari free choice hai, upar bataya gaya hai. Cylindrical kyun use karein? Region $z$ -axis ke around round hai; har horizontal slice ek disk hai. Cylindrical ugly $x^{2} + y^{2}$ ko ek single $r^{2}$ mein badal deta hai.

Step 2 — inner $z$ . $r, θ$ hold karo aur amber arrow follow karo: $z = 0$ pe enter, $z = 4 - r^{2}$ pe exit. $\int_{0}^{4 - r^{2}} r d z = r (4 - r^{2})$ Yeh step kyun? $r, θ$ frozen hain toh us jagah ki height exactly $4 - r^{2}$ hai. $d V$ ka $r$ saath mein le chalte hain.

Step 3 — middle $r$ . Ab $θ$ hold karo; shadow disk ki radius $2$ hai, toh $r : 0 \to 2$ . $\int_{0}^{2} r (4 - r^{2}) d r = \int_{0}^{2} (4 r - r^{3}) d r = [2 r^{2} - \frac{r ^{4}}{4}]_{0}^{2} = 8 - 4 = 4$ Yeh step kyun? $r$ projected shadow ke upar range karta hai, jo radius $2$ ki disk hai.

Step 4 — outer $θ$ . Full turn: $θ : 0 \to 2 π$ . $\int_{0}^{2 π} 4 d θ = 8 π$ Yeh step kyun? Kuch bhi $θ$ pe depend nahi karta, aur region axis ke around poora wrap karta hai, toh $θ$ pura $2 π$ sweep karta hai.

Verify: $8 π \approx 25.1$ . Yeh bounding cylinder $16 π$ ka exactly aadha hai — ek paraboloid dome apne box ka aadha fill karta hai, yeh ek jaana-maana fact hai. ✓ Cylinder se chhota, jaise forecast tha.

Ex 4 — Spherical, full round (cell SPH)

Worked example Radial density ke saath mass

Radius $a$ ki ek ball ki density $δ (x, y, z) = x^{2} + y^{2} + z^{2}$ hai (surface ke paas zyada ghani). Uski total mass nikalo. (Density ko $δ$ likhte hain, $ρ$ nahi, taaki spherical radial coordinate $ρ$ se clearly alag rahe.)

Forecast: Density baahir ki taraf badhti hai, isliye mass "uniform density $\times$ volume" se zyada hogi agar hum average density use karein. Bas form guess karo: yeh (constant) $\times a^{5}$ jaisa dikhega. $a^{5}$ kyun? Density ke units length $^{2}$ hain, volume length $^{3}$ — product length $^{5}$ . Yeh guess note karo.

Figure mein: cyan wireframe sphere $ρ = a$ hai; amber radial spoke $ρ$ ki ek value hai jo surface tak point karti hai; $ϕ$ $+ z$ axis se neeche measure hota hai aur $θ$ uske around ghoomta hai — yeh teen coordinates ball ke har point ko banate hain.

Step 1 — spherical mein sab kuch. Origin se doori radial coordinate $ρ$ hai, isliye $x^{2} + y^{2} + z^{2} = ρ^{2}$ aur density $δ = ρ^{2}$ hai. Boundary single equation $ρ = a$ hai, aur $d V = ρ^{2} sin ϕ d ρ d ϕ d θ$ . $ρ^{2} sin ϕ$ factor kyun? Chhote spherical box ke teen perpendicular edges hain: radial edge $d ρ$ , meridian arc $ρ d ϕ$ , aur latitude-circle arc jiska radius $ρ sin ϕ$ hai aur length $ρ sin ϕ d θ$ . Teenon lengths multiply karo: $d ρ \cdot (ρ d ϕ) \cdot (ρ sin ϕ d θ) = ρ^{2} sin ϕ d ρ d ϕ d θ$ . $sin ϕ$ poles pe $0$ ho jaata hai kyunki latitude circles wahan points mein collapse ho jaate hain. Woh product spherical Jacobian hai. $Mass = \int_{0}^{2 π} \int_{0}^{π} \int_{0}^{a} density δ ρ^{2} \cdot d V ρ^{2} sin ϕ d ρ d ϕ d θ$ Spherical kyun use karein? Integrand ( $x^{2} + y^{2} + z^{2}$ ) aur region (ek ball) dono har direction mein round hain — yeh spherical ka textbook flag hai.

Step 2 — integral factor karo. Limits sab constant hain aur integrand ek $ρ$ -part aur ek $ϕ$ -part mein split hota hai: $= 2 π \int_{0}^{2 π} d θ \cdot 2 \int_{0}^{π} sin ϕ d ϕ \cdot a^{5} /5 \int_{0}^{a} ρ^{4} d ρ$ Yeh step kyun? Fubini teen 1D integrals multiply karne deta hai jab sab kuch separate ho — sabse tez raasta.

Step 3 — multiply karo. $Mass = 2 π \cdot 2 \cdot \frac{a ^{5}}{5} = \frac{4 π a ^{5}}{5}$

Verify: Units check — density (length $^{2}$ ) $\times$ volume (length $^{3}$ ) = length $^{5}$ : $a^{5}$ se match karta hai. ✓ Ek uniform ball se compare karo jisme boundary density $a^{2}$ ho: woh deta $a^{2} \cdot \frac{4}{3} π a^{3} = \frac{4}{3} π a^{5} = \frac{20}{15} π a^{5}$ , aur hamara answer $\frac{4}{5} π a^{5} = \frac{12}{15} π a^{5}$ chhota hai — sahi hai, kyunki density andar har jagah $a^{2}$ se chhoti hai. ✓

Related to Center of mass and moments of inertia aur Spherical coordinates geometry.

Ex 5 — Degenerate / zero region (cell DEG)

Worked example Woh trap jahan answer exactly zero hota hai

"Solid $E$ ka volume nikalo jahan $0 \leq z \leq 0$ , $0 \leq x \leq 1$ , $0 \leq y \leq 1$ ho." Kya hai yeh?

Forecast: $z$ -limits dono $0$ hain. Guess karo: kya yeh ek real solid hai, ya bina thickness ka ek flat square hai?

Step 1 — inner $z$ . Arrow upar $z = 0$ se $z = 0$ tak — woh kabhi floor nahi chhodta. $\int_{0}^{0} d z = 0$ Yeh step kyun? Jab kisi integral ki lower aur upper limit ek hi ho, woh integral $0$ hoti hai chahe bahar kuch bhi ho.

Step 2 — baaki sab matter nahi karta. Ek baar ek factor $0$ ho gaya, toh poora product $0$ hai: $\int_{0}^{1} \int_{0}^{1} 0 d y d x = 0$ Yeh step kyun? $0$ ko finite base area $1 \times 1$ se multiply karna phir bhi $0$ deta hai.

Verify: Ek flat square ki koi thickness nahi, isliye woh zero volume enclose karta hai. ✓ Yeh woh sanity check hai jo tum tab run karte ho jab tumhare limits collapse ho jaayein: agar $z_{1} (x, y) = z_{2} (x, y)$ poore region pe ho, ruk jaao — answer $0$ hai, aur koi nonzero result algebra slip ka matlab hai.

Ex 6 — Limiting behaviour (cell LIM)

Worked example Ek cone jiska tip angle shrink karta hai

Ice-cream cone $0 \leq ρ \leq a$ , $0 \leq ϕ \leq α$ (sab $θ$ ) ka volume nikalo, jahan $α$ north pole se half-angle hai. Phir limits $α \to 0$ aur $α \to π$ lo.

Forecast: Jaise $α \to 0$ , cone ek needle ban jaata hai — volume $\to 0$ hona chahiye. Jaise $α \to π$ , woh poori ball nigal leta hai — volume $\to \frac{4}{3} π a^{3}$ hona chahiye. Woh formula predict karo jo dono karta ho.

Figure mein: cyan arc spherical cap $ρ = a$ hai; origin se do safed lines cone ki slanted sides hain; amber vertical line $+ z$ axis hai, aur half-angle $α$ woh opening hai jo us axis se neeche cone wall tak measure hoti hai.

Step 1 — spherical mein set up karo. Figure mein wedge mein $ϕ$ sirf $0$ se $α$ tak swept hai. Ex 4 se yaad karo ki $d V = ρ^{2} sin ϕ d ρ d ϕ d θ$ (radial edge $d ρ$ , meridian arc $ρ d ϕ$ , latitude arc $ρ sin ϕ d θ$ ): $V (α) = \int_{0}^{2 π} \int_{0}^{α} \int_{0}^{a} ρ^{2} sin ϕ d ρ d ϕ d θ$ Yeh step kyun? $z$ -axis ke around ek cone ek fixed polar angle $ϕ = α$ se bounded hai — spherical ise ek constant se describe karta hai, aur kuch nahi kar sakta.

Step 2 — factor karo aur integrate karo. $= 2 π \int_{0}^{2 π} d θ \cdot 1 - c o s α \int_{0}^{α} sin ϕ d ϕ \cdot a^{3} /3 \int_{0}^{a} ρ^{2} d ρ$ Yeh step kyun? $\int_{0}^{α} sin ϕ d ϕ = [- cos ϕ]_{0}^{α} = 1 - cos α$ — yahi ek piece hai jo $α$ pe depend karta hai.

Step 3 — assemble karo. $V (α) = \frac{2 π a ^{3}}{3} (1 - cos α)$ Yeh step kyun? Step 2 ke teen factors ko kisi bhi order mein multiply karo (ab woh sirf numbers hain): $2 π \cdot (1 - cos α) \cdot \frac{a ^{3}}{3}$ . Unhe collect karne par ek tidy formula milta hai single parameter $α$ mein, jo exactly wahi hai jo humein limits lene ke liye chahiye.

Verify (poora point yahi hai):

$α \to 0$ : $cos α \to 1$ , toh $V \to 0$ . Needle ka koi volume nahi. ✓
$α = \frac{π}{2}$ (hemisphere): $cos \frac{π}{2} = 0$ , $V = \frac{2 π a ^{3}}{3} = \frac{1}{2} \cdot \frac{4}{3} π a^{3}$ — exactly aadhi ball. ✓
$α \to π$ : $cos π = - 1$ , $V \to \frac{2 π a ^{3}}{3} \cdot 2 = \frac{4}{3} π a^{3}$ — poori ball. ✓

Ek clean formula teeno limits pass karta hai — issi se pata chalta hai yeh sahi hai.

Ex 7 — Negative coordinates & ek sign-changing integrand (cell SIGN)

Worked example Jab integrand pehle positive phir negative ho jaata hai

$I = ∭_{E} x d V$ evaluate karo box $E : - 1 \leq x \leq 2, 0 \leq y \leq 1, 0 \leq z \leq 1$ pe. Region $x = 0$ cross karta hai, aur integrand $x$ negative hai $x < 0$ ke liye, positive $x > 0$ ke liye.

Forecast: $x < 0$ ke liye term $x d V$ negative hai, $x > 0$ ke liye positive. Box positive $x$ mein zyada door jaata hai (out to $2$ ) negative $x$ se (sirf $- 1$ tak), isliye guess karo: net result positive.

Step 1 — integral separate karo. Limits sab constant hain aur $f = x$ mein koi $y, z$ nahi, isliye yeh factor ho jaata hai: $I = (\int_{- 1}^{2} x d x) (\int_{0}^{1} d y) (\int_{0}^{1} d z)$ Yeh step kyun? Jab box ke constant limits hon aur $f$ sirf ek variable pe depend kare, Fubini triple integral ko 1D integrals ke product mein badal deta hai.

Step 2 — $x$ -integral, sign dekh ke. $x = 0$ pe split karo taaki cancellation dikhaye: $\int_{- 1}^{2} x d x = - 1/2 (negative part) \int_{- 1}^{0} x d x + + 2 (positive part) \int_{0}^{2} x d x = - \frac{1}{2} + 2 = \frac{3}{2}$ Yeh step kyun? Sign change pe split karna physics ko explicit banata hai: chhota negative slab ( $- \frac{1}{2}$ ) bade positive slab ( $+ 2$ ) se haar jaata hai. (Directly, $[\frac{x ^{2}}{2}]_{- 1}^{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2}$ .)

Step 3 — trivial $y, z$ factors se multiply karo. $I = \frac{3}{2} \cdot 1 \cdot 1 = \frac{3}{2}$ Yeh step kyun? $\int_{0}^{1} d y$ aur $\int_{0}^{1} d z$ dono $1$ ke barabar hain, isliye woh value nahi badlate — woh sirf confirm karte hain ki cross-section unit square hai.

Verify: Positive, jaise forecast tha. Mean se sanity check: $x$ ka average $[- 1, 2]$ pe midpoint $\frac{- 1 + 2}{2} = \frac{1}{2}$ hai; interval length $3$ se multiply karne par $\frac{1}{2} \cdot 3 = \frac{3}{2}$ milta hai. ✓ Lesson: sign-changing integrand ka matlab nahi hai ki pieces mein split karna zaroori hai — lekin splitting dikhata hai kyun answer zero nahi hai. Agar box symmetric hota ( $- 1 \leq x \leq 1$ ), toh dono halves cancel ho jaate aur $I = 0$ hota.

Ex 8 — Real-world word problem (cell WORD)

Worked example Ek cone ka balance point kahan hai?

Uniform density ka ek solid cone apni tip origin pe rakh ke upar khulta hai: yeh region $E$ hai jahan $0 \leq z \leq h$ , aur height $z$ pe uski radius $\frac{R}{h} z$ hai (radius linearly $R$ tak badhti hai top pe). Uske centre of mass ka $z$ -coordinate nikalo, $\overset{z}{ˉ} = \frac{∭ _{E} z d V}{∭ _{E} d V}$ .

Forecast: Chaudi top ke paas zyada material hai narrow tip ke mukable, isliye balance point midpoint $h /2$ se upar hona chahiye. $h$ ka koi fraction guess karo.

Step 1 — cylindrical mein describe karo. Height $z$ pe cross-section radius $\frac{R}{h} z$ ki disk hai, toh $r$ runs $0 \to \frac{R}{h} z$ , $θ$ full turn, $z : 0 \to h$ . Yaad karo $d V = r d r d θ d z$ — extra $r$ polar-wedge area factor hai. Hum order $d r d z d θ$ mein integrate karenge: innermost $r$ ( $z, θ$ hold karo), phir $z$ , phir $θ$ . Yeh step kyun? Cone $z$ -axis ke around round hai; height se slice karne par har slice ek disk ban jaati hai jiska radius $z$ ka jaana-maana function hai.

Step 2 — denominator (volume). $V = \int_{0}^{2 π} \int_{0}^{h} \int_{0}^{R z / h} r d r d z d θ = 2 π \int_{0}^{h} \frac{1}{2} (\frac{R z}{h})^{2} d z = π \frac{R ^{2}}{h ^{2}} \cdot \frac{h ^{3}}{3} = \frac{π R ^{2} h}{3}$ Yeh step kyun? Famous cone volume $\frac{1}{3} π R^{2} h$ appear hota hai — ek built-in check ki humari limits sahi hain.

Step 3 — numerator ( $∭ z d V$ ). Extra $z$ daalo: $\int_{0}^{2 π} \int_{0}^{h} \int_{0}^{R z / h} z r d r d z d θ = 2 π \int_{0}^{h} z \cdot \frac{1}{2} (\frac{R z}{h})^{2} d z = π \frac{R ^{2}}{h ^{2}} \int_{0}^{h} z^{3} d z = π \frac{R ^{2}}{h ^{2}} \cdot \frac{h ^{4}}{4} = \frac{π R ^{2} h ^{2}}{4}$ Yeh step kyun? $z = 0$ ke around moment har slice ko uski height $z$ se weight karta hai; $z^{3}$ (moment se ek $z$ , badhti radius se $z^{2}$ ) wahi cheez hai jo balance ko upar tilt karti hai.

Step 4 — divide karo. $\overset{z}{ˉ} = \frac{π R ^{2} h ^{2} /4}{π R ^{2} h /3} = \frac{3 h}{4}$ Yeh step kyun? Centre of mass moment divided by total mass hai (yahan uniform density top aur bottom mein cancel ho jaati hai), ek pure geometric ratio bachta hai.

Verify: $\overset{z}{ˉ} = \frac{3}{4} h$ , jo midpoint $\frac{1}{2} h$ se upar hai — exactly jaise forecast tha, kyunki upar ka chauda end zyada mass rakhta hai. Solid cone ke liye standard textbook result: centre of mass tip se base tak $\frac{3}{4}$ raasta hai. ✓ Units: $\overset{z}{ˉ}$ ke units length hain. ✓

Related to Center of mass and moments of inertia.

Ex 9 — Exam-style twist (cell TWIST)

Worked example Woh integral jo sirf reverse karne ke baad kaam karta hai

Evaluate karo $I = \int_{0}^{1} \int_{x}^{1} \int_{0}^{1 - y} 1 d z d y d x .$ Jaisa likha hai theek hai, lekin twist yeh hai: yeh kaunsa solid hai, aur kya hum order reverse karke answer geometrically confirm kar sakte hain?

Forecast: Yeh pure volume hai ( $f = 1$ ). $y$ -limits $x \leq y \leq 1$ aur $z$ -limit $z \leq 1 - y$ koi curved-based prism describe karte hain. Guess karo kya answer ek "nice" fraction hai.

Step 1 — inner $z$ . $x, y$ hold karo. $\int_{0}^{1 - y} d z = 1 - y$ Yeh step kyun? $x, y$ freeze karo; roof $z = 1 - y$ hai, floor $z = 0$ .

Step 2 — middle $y$ , $x$ se $1$ tak. $\int_{x}^{1} (1 - y) d y = [y - \frac{y ^{2}}{2}]_{x}^{1} = (1 - \frac{1}{2}) - (x - \frac{x}{2}) = \frac{1}{2} - x + \frac{x}{2}$ Yeh step kyun? Lower curve $y = x$ wahi cheez hai jo base region ko curved banati hai — yeh parabola $x = y^{2}$ hai sideways dekha gaya.

Step 3 — outer $x : 0 \to 1$ . Ab term by term integrate karo, $\int_{0}^{1} x d x = \frac{2}{3}$ aur $\int_{0}^{1} x d x = \frac{1}{2}$ use karke: $I = \int_{0}^{1} (\frac{1}{2} - x + \frac{x}{2}) d x = \frac{1}{2} - \frac{2}{3} + \frac{1}{4} = \frac{6 - 8 + 3}{12} = \frac{1}{12}$ Yeh step kyun? Outer variable base region ke poore $x$ -spread pe run karta hai, yahan $0 \leq x \leq 1$ ; $y, z$ already gone hain toh sirf ek number bachta hai, isliye hum teen terms simply integrate karte hain.

Verify — order reverse karo. $x y$ -plane mein base region ${0 \leq x \leq y^{2}, 0 \leq y \leq 1}$ hai (from $x \leq y^{2}$ , i.e. $y \geq x$ ). $x$ andar rakh ke dobara karo: $I = \int_{0}^{1} \int_{0}^{y^{2}} \int_{0}^{1 - y} d z d x d y = \int_{0}^{1} y^{2} (1 - y) d y = \frac{1}{3} - \frac{1}{4} = \frac{1}{12} .$ Dono orders $\frac{1}{12}$ dete hain. ✓ Lesson: jab limits awkward lagtein, base region redraw karo aur order swap karna ek full, independent check hai — aur yahan yeh twist bhi reveal karta hai: solid ka parabolic base $x = y^{2}$ hai, jo order flip karne se pehle invisible tha.

Coverage check

Recall Kya humne matrix ki har cell hit ki?

C1 → Ex 1 · C2 → Ex 2 · CYL → Ex 3 · SPH → Ex 4 · DEG → Ex 5 · LIM → Ex 6 · SIGN → Ex 7 · WORD → Ex 8 · TWIST → Ex 9. Har scenario class cover hai.

Recall Quick self-test

$z = 4 - r^{2}$ ke neeche uske shadow disk pe volume ::: $8 π$ Radius $a$ ki ball jisme density $x^{2} + y^{2} + z^{2}$ ho uski mass ::: $\frac{4 π a ^{5}}{5}$ Half-angle $α$ ke liye cone wedge volume $V (α)$ ::: $\frac{2 π a ^{3}}{3} (1 - cos α)$ Box $- 1 \leq x \leq 2, 0 \leq y, z \leq 1$ pe $∭ x d V$ ::: $\frac{3}{2}$ Solid cone ki centre of mass height ::: tip se $\frac{3}{4} h$ upar Reversed-order integral $I$ ki value ::: $\frac{1}{12}$

Connections

Triple integrals in Cartesian, cylindrical, spherical coordinates — parent note; yeh page uska full worked-example companion hai.
Double integrals & polar coordinates — Ex 3 aur Ex 8 2D case se $r d r d θ$ reuse karte hain.
Jacobian and change of variables — har $d V$ factor ( $r$ , $ρ^{2} sin ϕ$ ) disguise mein ek Jacobian hai.
Spherical coordinates geometry — Ex 4 aur Ex 6 $ρ, ϕ, θ$ picture pe lean karte hain.
Center of mass and moments of inertia — Ex 8 ek direct centre-of-mass computation hai.
Divergence theorem — baad mein, yeh volume integrals flux calculations ke "andar" ban jaate hain.