Applications — mass, centre of mass, moments of inertia: Worked examples

4.4.22 · D3 · Maths › Multivariable Calculus › Applications — mass, centre of mass, moments of inertia

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Scenario matrix

Is topic ke har problem ko inhi cells mein se ek mein daala ja sakta hai. Neeche har example ko uski cell ke saath tag kiya gaya hai.

Cell	Kya cheez isko alag banati hai	Example
C1 Constant density, simple shape	$ρ$ cancel ho jaata hai → pure geometry (centroid)	Ex 1
C2 Variable density	$ρ (x, y)$ integral ke andar rehna chahiye	Ex 2
C3 Region touching all four quadrants / sign of $x, y$	lever arms dono $x$ aur $y$ mein negative ho jaate hain — kya woh cancel hote hain?	Ex 3
C4 Symmetry shortcut	ek moment symmetry ki wajah se zero hai — pahchano, kaam bachao	Ex 4
C5 Polar coordinates (curved region)	zaroori Jacobian $r$	Ex 5
C6 Degenerate / limiting shape	thin rod, ya $a \to 0$ — check karo ki formula survive karti hai	Ex 6
C7 Real-world word problem	words ko region + density mein translate karo	Ex 7
C8 Perpendicular-axis / radius of gyration	$I_{x}, I_{y}$ combine karo; $R_{g}$ nikalo	Ex 8
C9 Triple integral (3-D solid)	$d m = ρ d V$ , ek dimension aur	Ex 9
C10 Exam twist — density blows up / non-uniform axis	ek trap jo autopilot ko sazaa deta hai	Ex 10

Ab hum har cell fill karte hain.

Example 1 — C1: constant density, ek triangle ka centroid

Worked example Statement

Ek uniform plate ( $ρ =$ constant) woh triangle hai jiske corners $(0, 0)$ , $(2, 0)$ , $(0, 3)$ hain. Uska centre of mass nikalo.

Forecast: tumhara kya lagta hai ki balance point kahan baithega? Padhne se pehle guess karo. (Hint: triangle ke medians ek point par milte hain jo har side se ek-tihaai door hota hai.)

Step 1 — Region describe karo. Tilted edge $(2, 0)$ se $(0, 3)$ tak jaati hai. Uski line hai $\frac{x}{2} + \frac{y}{3} = 1$ , yaani $y = 3 - \frac{3}{2} x$ . Toh har $x$ ke liye $0$ se $2$ ke beech, $y$ $0$ se $3 - \frac{3}{2} x$ tak jaata hai. Yeh step kyun? Integrate karne se pehle exact limits jaanni chahiye — picture (vertical dashed strip dekho) batati hai ki $y$ base se shuru hoke tilted red line par band hota hai.

Step 2 — Mass. Kyunki $ρ$ constant hai, $m = ρ \cdot (area) = ρ \cdot \frac{1}{2} (2) (3) = 3 ρ$ . Yeh step kyun? Uniform plate ke liye mass sirf density × area hai — koi integral nahi chahiye, lekin $\iint ρ d A$ bhi same deta hai.

Step 3 — Moment $M_{y} = \iint x ρ d A$ (upar box mein define hua — lever arm $x$ ). $M_{y} = ρ \int_{0}^{2} \int_{0}^{3 - \frac{3}{2} x} x d y d x = ρ \int_{0}^{2} x (3 - \frac{3}{2} x) d x .$ Inner integral: $\int_{0}^{3 - \frac{3}{2} x} d y = 3 - \frac{3}{2} x$ (strip ki height). Outer: $ρ \int_{0}^{2} (3 x - \frac{3}{2} x^{2}) d x = ρ [\frac{3}{2} x^{2} - \frac{1}{2} x^{3}]_{0}^{2} = ρ (6 - 4) = 2 ρ$ . Yeh step kyun? $M_{y}$ har strip ko uski $x$ -position se weight karta hai — $y$ -axis tak ka lever arm.

Step 4 — $\overset{x}{ˉ} = M_{y} / m = 2 ρ /3 ρ = \frac{2}{3}$ . Dusre moment $M_{x} = \iint y ρ d A$ ke same computation se $M_{x} = 3 ρ$ aur $\overset{y}{ˉ} = M_{x} / m = 1$ milta hai. Yeh step kyun? $M_{x}$ (lever arm $y$ , upar box mein define hua) vertical balance deta hai; har moment ko total mass se divide karne par "total leverage" se "woh single position" milti hai jo isse reproduce kare.

Verify: triangle ka centroid uske vertices ka average hota hai: $\overset{x}{ˉ} = \frac{0 + 2 + 0}{3} = \frac{2}{3}$ , $\overset{y}{ˉ} = \frac{0 + 0 + 3}{3} = 1$ . ✓ Bilkul match karta hai. Notice karo ki $ρ$ cancel ho gaya — yahi cheez isko centroid banati hai (dekho Centroids and the Pappus Theorems).

Example 2 — C2: variable density strip

Worked example Statement

Ek plate $0 \leq x \leq 1$ , $0 \leq y \leq 1$ (unit square) par hai jisme density $ρ (x, y) = x$ hai (right side par zyada heavy, left edge par zero). $m$ aur $\overset{x}{ˉ}$ nikalo.

Forecast: zyada mass right side par hai, toh $\overset{x}{ˉ}$ $0.5$ se bada hoga ya chhota?

Step 1 — Mass. $m = \int_{0}^{1} \int_{0}^{1} x d y d x = \int_{0}^{1} x d x = \frac{1}{2} .$ Yeh step kyun? $x$ (jo $y$ mein constant hai) ka inner $y$ -integral sirf height $1$ se multiply kar deta hai.

Step 2 — $M_{y} = \iint x \cdot ρ d A = \iint x \cdot x d A$ . $M_{y} = \int_{0}^{1} \int_{0}^{1} x^{2} d y d x = \int_{0}^{1} x^{2} d x = \frac{1}{3} .$ Yeh step kyun? Yahan "distance" $x$ aur density $x$ dono appear karte hain, jisse $x^{2}$ milta hai — density factor mat bhulo (parent note mein yahi classic mistake hai).

Step 3 — $\overset{x}{ˉ} = M_{y} / m = \frac{1/3}{1/2} = \frac{2}{3}$ . Yeh step kyun? Moment ko total mass se divide karne par woh single $x$ milta hai jahan saari mass baith sakti thi aur same balance deti.

Verify: $\overset{x}{ˉ} = \frac{2}{3} > 0.5$ ✓ — balance point heavy side ki taraf right mein jhukata hai, bilkul forecast jaisa. Agar hum (galti se) density ignore kar dete, geometric centre $\overset{x}{ˉ} = 0.5$ milta; density ne isse $\frac{2}{3}$ tak khiinch liya.

Example 3 — C3: CHARON quadrants mein phayeli region (dono lever arms ke signs)

Worked example Statement

Plate: rectangle $- 1 \leq x \leq 3$ , $- 2 \leq y \leq 2$ , uniform density $ρ$ . Yeh dono axes ko straddle karta hai, toh tiles charon quadrants mein hain. $\overset{x}{ˉ}$ aur $\overset{y}{ˉ}$ nikalo. Kya negative $x$ aur negative $y$ values matter karte hain?

Forecast: plate right mein zyada door tak jaati hai ( $x$ tak $3$ ) left se ( $x$ tak $- 1$ ), toh $\overset{x}{ˉ} > 0$ . Lekin yeh top–bottom symmetric hai ( $y$ $- 2$ se $2$ tak), toh $\overset{y}{ˉ}$ kya hona chahiye?

Step 1 — Mass. Area $= (3 - (- 1)) \times (2 - (- 2)) = 4 \times 4 = 16$ , toh $m = 16 ρ$ . Yeh step kyun? Dono moments ko is total mass se divide karenge taaki balance point mile; mass woh "weight" hai jo leverage ko position mein convert karta hai. Uniform plate ke liye yeh density × area hai.

Step 2 — $M_{y} = ρ \iint x d A$ (horizontal balance). Lever arm $x$ do left quadrants mein negative hai (figure mein violet) aur do right quadrants mein positive (orange). Yeh partly cancel karte hain. $M_{y} = ρ \int_{- 2}^{2} \int_{- 1}^{3} x d x d y = ρ \int_{- 2}^{2} [\frac{x ^{2}}{2}]_{- 1}^{3} d y = ρ \int_{- 2}^{2} \frac{9 - 1}{2} d y = ρ \cdot 4 \cdot 4 = 16 ρ .$ Yeh step kyun? $\int_{- 1}^{3} x d x = \frac{9}{2} - \frac{1}{2} = 4$ — $x \in [- 1, 0]$ ka negative contribution real hai aur subtract karta hai, lekin bada positive side jeet jaata hai.

Step 3 — $M_{x} = ρ \iint y d A$ (vertical balance). Ab lever arm $y$ $x$ -axis ke neeche negative aur upar positive hai, symmetric range $[- 2, 2]$ par: $M_{x} = ρ \int_{- 1}^{3} \int_{- 2}^{2} y d y d x = ρ \int_{- 1}^{3} [\frac{y ^{2}}{2}]_{- 2}^{2} d x = ρ \int_{- 1}^{3} \frac{4 - 4}{2} d x = 0.$ Yeh step kyun? Kyunki $y$ $0$ ke baare mein symmetrically range karta hai, height $+ y$ par har tile ka mirror $- y$ par cancel ho jaata hai — ek zero moment, toh ek baar spot kar lo toh koi integral ka toil sach mein zaroori nahi.

Step 4 — Centre of mass. $\overset{x}{ˉ} = \frac{M _{y}}{m} = \frac{16 ρ}{16 ρ} = 1$ , aur $\overset{y}{ˉ} = \frac{M _{x}}{m} = \frac{0}{16 ρ} = 0$ . Yeh step kyun? Har total moment (leverage) ko total mass se divide karne par woh ek point milta hai jo har axis ke baare mein same leverage reproduce karta hai — balance point.

Verify: rectangle ka geometric centre $(\frac{- 1 + 3}{2}, \frac{- 2 + 2}{2}) = (1, 0)$ hai ✓. Toh $\overset{x}{ˉ} = 1 > 0$ (right mein jhukata hai, forecast jaisa) aur $\overset{y}{ˉ} = 0$ (top–bottom symmetry, forecast jaisa). Lesson: dono directions mein signed lever arms automatically cancel ho jaate hain — tum kabhi integral ko quadrant ke hisaab se split mat karo; algebra khud har sign handle kar leta hai.

Example 4 — C4: symmetry ek moment ko vanish kara deti hai

Worked example Statement

Ek uniform semicircular plate of radius $a$ apni flat edge $x$ -axis par rakhke baitha hai: $x^{2} + y^{2} \leq a^{2}$ , $y \geq 0$ . $\overset{x}{ˉ}$ bina integrate kiye nikalo, phir $\overset{y}{ˉ}$ nikalo.

Forecast: shape $y$ -axis ke across mirror image hai. Woh $\overset{x}{ˉ}$ ko kya force karta hai?

Step 1 — Symmetry dekho. Har tile $(x, y)$ par ek mirror tile $(- x, y)$ hai jisme same density hai (uniform). Unke $x$ -moments $x ρ d A$ aur $(- x) ρ d A$ cancel ho jaate hain (figure mein mirrored violet/orange tiles dekho). Yeh step kyun? Symmetry pehchanna poora integral bachata hai — $M_{y} = 0$ , isliye $\overset{x}{ˉ} = 0$ .

Step 2 — $\overset{y}{ˉ}$ ke liye kaam karna padega; polar use karo. $x^{2} + y^{2} \leq a^{2}$ , $y \geq 0$ matlab $0 \leq r \leq a$ , $0 \leq θ \leq π$ , jisme $y = r sin θ$ aur $d A = r d r d θ$ . $m = ρ \cdot \frac{1}{2} π a^{2}, M_{x} = ρ \int_{0}^{π} \int_{0}^{a} (r sin θ) r d r d θ .$ Polar kyun? Circular boundary polar mein trivial hai ( $r = a$ ) lekin rectangular mein ugly. Dekho Polar Coordinates and the Jacobian.

Step 3 — $M_{x}$ evaluate karo. $M_{x} = ρ \int_{0}^{π} sin θ d θ \int_{0}^{a} r^{2} d r = ρ \cdot [- cos θ]_{0}^{π} \cdot \frac{a ^{3}}{3} = ρ \cdot 2 \cdot \frac{a ^{3}}{3} = \frac{2 ρ a ^{3}}{3} .$

Step 4 — $\overset{y}{ˉ} = \frac{M _{x}}{m} = \frac{2 ρ a ^{3} /3}{ρ π a ^{2} /2} = \frac{4 a}{3 π} .$

Verify: semicircle ke centroid ki standard textbook value $\overset{y}{ˉ} = \frac{4 a}{3 π} \approx 0.4244 a$ hai, jo halfway ( $0.5 a$ ) se neeche hai — sahi hai, kyunki zyada area flat base ke paas ikatta hota hai — aur $0$ se upar aur top $a$ se neeche. Aur $\overset{x}{ˉ} = 0$ ✓ symmetry se, bilkul forecast jaisa.

Example 5 — C5: polar tiles $r$ kyun "pehante" hain (disc ka moment of inertia $I_{0}$ )

Worked example Statement

Origin par centre kiya hua ek uniform disc radius $a$ , density $ρ$ , centred at the origin. $I_{0}$ nikalo, yaani centre se guzarne wale axis (the $z$ -axis) ke baare mein moment of inertia.

Forecast: parent ke " $1, r, r^{2}$ " idea se, $I_{0} = \iint r^{2} d m$ , aur areas (length) $^{2}$ ki tarah scale karte hain. Toh $I_{0}$ ko $a^{4}$ ki tarah badhna chahiye. Exact constant guess karo.

Step 1 — $d A = r d r d θ$ justify karo. Figure mein shaded polar tile dekho: yeh ek chhoti radial width $d r$ aur ek chhota angle $d θ$ span karta hai. Uski do "lambi" sides arcs hain jinka length $r d θ$ hai (arc length = radius × angle). Toh tile almost ek rectangle jaisi hai jinki sides $d r$ aur $r d θ$ hain, area $= r d r d θ$ . Yeh step kyun? Yeh Jacobian factor hai. $r$ bhool jaana sabse common blunder hai — isse rim ke paas ke tiles centre ke paas ke tiles jitni chhoti lagti hain, jo geometrically galat hai.

Step 2 — $I_{0}$ set up karo. Axis se distance $r$ hai, toh lever arm $^{2} = r^{2}$ : $I_{0} = ρ \int_{0}^{2 π} \int_{0}^{a} r^{2} \cdot d A r d r d θ = ρ \int_{0}^{2 π} d θ \int_{0}^{a} r^{3} d r .$ $r^{2} \cdot r = r^{3}$ kyun? Ek $r^{2}$ physics hai (distance²); extra $r$ geometry hai (Jacobian). Yeh alag hain aur dono zaroori hain.

Step 3 — Integrate karo. $I_{0} = ρ \cdot 2 π \cdot \frac{a ^{4}}{4} = \frac{π ρ a ^{4}}{2} .$

Step 4 — Per-mass form. Mass $m = ρ π a^{2}$ , toh $I_{0} = \frac{π ρ a ^{4}}{2} = \frac{( ρ π a ^{2} ) a ^{2}}{2} = \frac{m a ^{2}}{2}$ .

Verify: $I_{0} \propto a^{4}$ ✓ forecast jaisa, constant $\frac{π ρ}{2}$ ; per-mass form $\frac{1}{2} m a^{2}$ woh famous "solid disc" result hai Rotational Kinetic Energy and Angular Momentum se.

Example 6 — C6: degenerate limit (thin rod as a shrinking rectangle)

Worked example Statement

Rectangle $0 \leq x \leq L$ , $0 \leq y \leq b$ , uniform, jisme $I_{x} = \frac{m b ^{2}}{3}$ aur $I_{y} = \frac{m L ^{2}}{3}$ hain (parent Example B). Thin rod ke saath kya hoga — height $b \to 0$ le lo? $I_{y}$ nikalo rod ka ek end ke baare mein.

Forecast: jaise plate ek needle ban jaati hai $x$ -axis ke saath, $I_{x}$ (lambi axis ke baare mein spin) aur $I_{y}$ (chhote end ke baare mein spin) ka kya hoga?

Step 1 — $I_{x}$ mein $b \to 0$ lo. $I_{x} = \frac{m b ^{2}}{3} \to 0$ . Kyun? Jab rod infinitely thin ho jaati hai, koi bhi mass $x$ -axis se koi distance par nahi hota, toh apni lambi length ke baare mein ghoomne mein zero resistance hoti hai. Degenerate hai lekin sensible.

Step 2 — $I_{y}$ survive karta hai. $I_{y} = \frac{m L ^{2}}{3}$ mein $b$ nahi hai, toh $b \to 0$ hone par bhi finite rehta hai. Yeh step kyun? $I_{y}$ chhote end ke baare mein spin measure karta hai; mass $x \in [0, L]$ ke saath phayela hua hai, phir bhi $y$ -axis se real distance par.

Verify: length $L$ , mass $m$ ki rod, ek end ke baare mein rotate ki jaaye toh classic moment of inertia $\frac{1}{3} m L^{2}$ hai ✓ (textbook value). Hamaara 2-D rectangle formula exactly 1-D rod formula mein degenerate hua — ek accha consistency check ki limit mein machinery sahi hai.

Example 7 — C7: ek real-world word problem

Worked example Statement

Ek rectangular metal sheet, $0 \leq x \leq 4$ m, $0 \leq y \leq 2$ m, ko garam kiya gaya toh uski areal density (kg/m²) height ke saath linearly badhti hai: $ρ (x, y) = 2 + y$ (kg/m²). Uski total mass aur balance point ki height $\overset{y}{ˉ}$ nikalo.

Forecast: density top par zyada hai ( $y = 2$ deta hai $ρ = 4$ ) neeche se ( $y = 0$ deta hai $ρ = 2$ ), toh $\overset{y}{ˉ}$ geometric mid-height $1$ m se upar push hona chahiye.

Step 1 — Words ko region + density mein translate karo. Region: rectangle $[0, 4] \times [0, 2]$ . Density: $ρ = 2 + y$ , figure mein colour gradient se dikhaya gaya hai (neeche light, upar dark). Koi $x$ -dependence nahi, toh sare $x$ -integrals bas width $4$ se multiply kar dete hain. Yeh step kyun? Word problems sirf region + density hain chhupe hue; unhe naam dena prose ko integral mein badle deta hai.

Step 2 — Mass. $m = \int_{0}^{4} \int_{0}^{2} (2 + y) d y d x = 4 \int_{0}^{2} (2 + y) d y = 4 [2 y + \frac{y ^{2}}{2}]_{0}^{2} = 4 (4 + 2) = 24 kg .$

Step 3 — $M_{x} = \iint y ρ d A$ . $M_{x} = 4 \int_{0}^{2} y (2 + y) d y = 4 \int_{0}^{2} (2 y + y^{2}) d y = 4 [y^{2} + \frac{y ^{3}}{3}]_{0}^{2} = 4 (4 + \frac{8}{3}) = \frac{80}{3} kg\cdotm .$ Yeh step kyun? $\overset{y}{ˉ}$ $x$ -axis ke baare mein balance karta hai, jiska lever arm $y$ hai; toh hum $M_{x}$ use karte hain, $M_{y}$ nahi.

Step 4 — $\overset{y}{ˉ} = \frac{M _{x}}{m} = \frac{80/3}{24} = \frac{80}{72} = \frac{10}{9} \approx 1.11$ m.

Verify: units: $M_{x}$ kg·m hai, $m$ kg hai, toh $\overset{y}{ˉ}$ metres mein hai ✓. Value $1.11$ m $> 1$ m ✓ — denser top ki taraf jhukta hai, bilkul forecast jaisa.

Example 8 — C8: perpendicular-axis theorem & radius of gyration

Worked example Statement

Example 5 ke uniform disc ke liye ( $I_{0} = \frac{1}{2} m a^{2}$ ), $I_{x}$ aur $I_{y}$ alag alag nikalo, phir centre ke baare mein radius of gyration nikalo.

Forecast: disc bilkul same dikhti hai chahe $x$ ke baare mein ghuma lo ya $y$ ke baare mein (circular symmetry hai), toh $I_{x}$ vs $I_{y}$ ke baare mein kya sach hona chahiye?

Step 1 — Symmetry se $I_{x} = I_{y}$ . Disc ko $9 0^{\circ}$ rotate karne par $x$ -axis $y$ -axis par map ho jaati hai aur disc unchanged rehti hai (figure mein same disc se guzarte do axes dekho), toh dono moments of inertia equal hain. Yeh step kyun? Symmetry phir se ek doosra integral bachne deti hai.

Step 2 — Perpendicular-axis theorem. Flat lamina ke liye, $I_{0} = I_{x} + I_{y}$ . $I_{x} = I_{y}$ ke saath: $I_{0} = 2 I_{x} \Rightarrow I_{x} = I_{y} = \frac{1}{2} I_{0} = \frac{1}{2} \cdot \frac{1}{2} m a^{2} = \frac{1}{4} m a^{2} .$ Yeh theorem kyun? Yeh perpendicular ( $z$ ) axis ke baare mein spin ko do in-plane axes se link karta hai — lekin sirf flat plate ke liye (parent ka mistake box warn karta hai ki 3-D solids ke liye kaam nahi karta).

Step 3 — Centre ke baare mein radius of gyration. $R_{g} = I_{0} / m = \frac{1}{2} m a^{2} / m = \frac{a}{2}$ . Yeh quantity kyun? $R_{g}$ woh single radius hai jis par tum saari mass rakh sakte ho (ek thin ring ki tarah) aur same $I_{0}$ milega.

Verify: $I_{x} + I_{y} = \frac{1}{4} m a^{2} + \frac{1}{4} m a^{2} = \frac{1}{2} m a^{2} = I_{0}$ ✓ — perpendicular-axis check close ho gaya. Aur $R_{g} = a / 2 \approx 0.707 a$ $0$ aur $a$ ke beech hai ✓ — ek plausible "average radius."

Example 9 — C9: ek triple integral (3-D solid)

Worked example Statement

Ek solid cube $0 \leq x \leq 1$ , $0 \leq y \leq 1$ , $0 \leq z \leq 1$ mein volume density $ρ = 1$ hai (mass per unit volume). Uski mass aur $\overset{z}{ˉ}$ nikalo. (Ab $d m = ρ d V$ , ek extra dimension.)

Forecast: uniform cube → balance point dead centre par $(\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$ . Toh $\overset{z}{ˉ} = ?$

Step 1 — Triple integral se mass. $m = ∭_{V} ρ d V = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} 1 d z d y d x = 1.$ Triple integral kyun? Solid ko volume element chahiye $d V = d x d y d z$ ; recipe 2-D se bilkul same hai lekin teen nested integrals ke saath. Dekho Triple Integrals.

Step 2 — $x y$ -plane ke baare mein moment, $M_{x y} = ∭ z ρ d V$ . $M_{x y} = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} z d z d y d x = \int_{0}^{1} \int_{0}^{1} [\frac{z ^{2}}{2}]_{0}^{1} d y d x = \frac{1}{2} .$ Yeh step kyun? 3-D mein horizontal $x y$ -plane tak ka lever arm height $z$ hai; $\overset{z}{ˉ} = M_{x y} / m$ .

Step 3 — $\overset{z}{ˉ} = M_{x y} / m = \frac{1/2}{1} = \frac{1}{2}$ .

Verify: $\overset{z}{ˉ} = \frac{1}{2}$ ✓, uniform cube ka geometric centre, bilkul forecast jaisa. Identical symmetry se $\overset{x}{ˉ} = \overset{y}{ˉ} = \frac{1}{2}$ bhi.

Example 10 — C10: exam twist (density origin par blow up ho jaaye)

Worked example Statement

Ek plate quarter disc $r \leq 1$ , first quadrant, mein hai jisme density $ρ = \frac{1}{r}$ hai (centre ke paas zyada heavy, $r = 0$ par singular). Kya mass finite rehti hai? Nikalo.

Forecast: $ρ \to \infty$ jab $r \to 0$ . Autopilot chillata hai "infinite mass!" Lekin Jacobian $r$ shayad bacha le. Guess karo: finite ya infinite?

Step 1 — Polar mein set up karo, Jacobian rakho. $m = \int_{0}^{π /2} \int_{0}^{1} \frac{1}{r} \cdot d A r d r d θ = \int_{0}^{π /2} \int_{0}^{1} 1 d r d θ .$ Yahi trap hai: density ka $\frac{1}{r}$ Jacobian ke $r$ ko cancel kar deta hai! Jo singular lag raha tha woh $1$ ka integral ban jaata hai — bilkul finite. Jo Jacobian $r$ bhool jaata hai use $\int_{0}^{1} \frac{1}{r} d r = \infty$ milta hai aur galat answer.

Step 2 — Integrate karo. $m = \int_{0}^{π /2} [r]_{0}^{1} d θ = \int_{0}^{π /2} 1 d θ = \frac{π}{2} .$

Verify: $m = \frac{π}{2} \approx 1.571$ , finite ✓. Singular density polar area element se tame ho gayi — yahi exam ka poora point tha. Isliye "polar tiles wear an $r$ " koi cosmetic rule nahi hai balki woh cheez hai jo finiteness decide karti hai.

Recall Self-test: har example ko uski matrix cell se match karo

Ex 1 :::: C1 constant density / centroid Ex 2 :::: C2 variable density Ex 3 :::: C3 charon quadrants mein signed lever arms Ex 4 :::: C4 symmetry ⇒ moment $= 0$ Ex 5 :::: C5 polar Jacobian Ex 6 :::: C6 degenerate/limiting shape Ex 7 :::: C7 word problem Ex 8 :::: C8 perpendicular-axis + radius of gyration Ex 9 :::: C9 triple integral (3-D) Ex 10 :::: C10 exam twist (singular density)