Volume of revolution — disk method, washer method: Visual walkthrough

4.2.14 · D2 · Maths › Calculus II — Integration › Volume of revolution — disk method, washer method

Hum ek flat region ko ek straight line ke around rotate karte hain aur poochte hain: usske andar kitna stuff hai jo solid sweep karta hai? Neeche har ek symbol pehle earn kiya jaata hai, phir appear hota hai.

Step 1 — "Rotate a region" ka matlab kya hota hai

WHAT. Ek flat shaded region paper par banaao. Ek straight line chuno — isse axis kaho. Ab socho ki paper ke us region ko us line ke around bahut tez ghuma rahe ho. Jo blur woh chhod jaata hai woh ek solid 3D object hai.

WHY. Humein koi bhi algebra se pehle ek mental picture chahiye. "Volume of revolution" word scary lagta hai; picture (ek shape jo ek skewer par ghoom raha ho) scary nahi hai. Baad mein sab kuch bas is blur ko measure karna hai.

PICTURE. Left panel mein ek flat region ek curve ke neeche dikh raha hai. Right panel mein wahi region horizontal line ke around spin hone ke baad ek bowl-jaisa solid ban jaata hai.

Figure — Volume of revolution — disk method, washer method

Step 2 — Ek thin slice (almost) ek cylinder hota hai

WHAT. Poore curvy solid ki jagah, horizontal position $x$ par ek thin slice kaato, axis ke perpendicular. Isko ek chhoti si thickness do jise hum $Δ x$ kahenge — padhte hain "a small change in $x$ ."

WHY. Poora solid curvy hai aur uske liye koi formula nahi hai. Lekin ek thin slice thoda hi curve karta hai, isliye woh almost ek perfect cylinder hai — aur cylinder ka volume hum jaante hain. Yahi poora trick hai: ek impossible shape ko kai easy shapes se replace karo.

PICTURE. Ek single upright coin solid se bahar nikali gayi. Uska round face axis ke along point karta hai; uski thickness $Δ x$ coin ki width hai.

Step 3 — Us circle ka radius curve ki height hai

WHAT. Humare coin ka round face ek circle hai. Uska radius $r$ axis se curve tak ki distance hai. Agar curve $y = f (x)$ hai aur axis x-axis hai, toh woh distance bas height $f (x)$ hai.

WHY. Ek circle poori tarah se uske radius se decide hota hai. Isliye coin ka face jaanne ke liye humein sirf ek number chahiye: curve axis se kitni door hai. Us gap ko ghoomane se poora circle sweep ho jaata hai.

PICTURE. Ek red arrow axis se seedha upar curve par point $(x, f (x))$ tak. Uski length $r = f (x)$ label hai. Woh arrow, ghoomne par, woh circle hai.

Ab tak ke saare symbols, plain words mein:

$r = height of the curve above the axis at position x f (x)$

Step 4 — Coin ka volume: face ka area × thickness

WHAT. Ek cylinder ka volume uske circular face ka area hai times uski thickness. Face ek circle hai radius $r$ ka, jiska area $π r^{2}$ hai. Thickness $Δ x$ hai. Toh:

$Δ V \approx area of the round face π r^{2} π [f (x)]^{2} \times thickness of the coin Δ x$

$π r^{2}$ kyun? Kyunki face ek genuine circle hai, aur radius $r$ ke circle ka area $π r^{2}$ hota hai — yahi reason hai ki tool " $π r^{2}$ " use hota hai na ki kuch aur: humara slice round hai, square nahi. Hum radius ko square karte hain kyunki area do directions mein grow karta hai (across aur upar), sirf ek mein nahi.

$\approx$ (almost-equals) kyun? Coin ki edge perfectly straight nahi hai — curve $Δ x$ width ke across thoda rise karta hai. Mote coin ke liye yeh error dikh sakti hai; thin coin ke liye yeh gayab ho jaati hai. Step 5 isko exactly zero kar deta hai.

PICTURE. Coin ko flat kiya gaya: radius $f (x)$ ka ek circle shaded, uska area $π [f (x)]^{2}$ naam ke saath, paas mein thin gap $Δ x$ .

Step 5 — Coins stack karo, phir shrink karo: integral appear hoti hai

WHAT. $x = a$ se $x = b$ tak saari coins line up karo aur unke volumes add karo:

$V \approx \sum π [f (x)]^{2} Δ x .$

Ab har coin ko thinner aur thinner banao ( $Δ x \to 0$ , matlab thickness kuch bhi nahi ho jaati). Infinitely many infinitely-thin coins — woh limiting sum exactly definite integral hai (yahi poori baat hai Definite Integral as a Limit of Sums ki):

$V = π \int_{a}^{b} [f (x)]^{2} d x$

Limit kyun, aur yeh integral kyun ban jaata hai? Step 4 ka $\approx$ sirf tab $=$ banta hai jab coins infinitely thin hoti hain, kyunki tab unka curved-edge error zero hota hai. Symbol $\int$ defined hi hai "limit of a sum of tiny pieces" ke roop mein — toh $\sum$ literally $\int$ mein badal jaata hai, aur $Δ x$ (ek small width) $d x$ (ek infinitesimal width) ban jaata hai.

PICTURE. Teen panels: kuch fat coins (rough), bahut si thin coins (better), smooth filled solid (exact). Rougher stack overshoot/undershoot karta hai; fine wala true solid ke saath chipak jaata hai.

Boxed result ka term-by-term:

$V = from π r^{2} π add all coins from x = a to b \int_{a}^{b} r^{2}, the face [f (x)]^{2} infinitesimal thickness d x$

Step 6 — Jab region mein gap ho: ek hole punch karo (the washer)

WHAT. Maano region do curves ke beech hai, ek outer (door wali) $y = f (x)$ aur ek inner (paas wali) $y = g (x)$ , dono axis ke upar gap ke saath. Isko spin karo. Ab har coin ke beech mein ek hole hai — yeh ek ring hai, jise washer kehte hain.

WHY. Paas wali curve aur axis ke beech ka gap kabhi fill nahi hota, toh wahan kuch nahi hai — ek hole. Ring ka area bada circle minus chhota hole hota hai:

$Δ V \approx (big circle, radius R = f (x) π [f (x)]^{2} - hole, radius r = g (x) π [g (x)]^{2}) Δ x$

Step 5 ki tarah exactly sum karo aur shrink karo:

$V = π \int_{a}^{b} ([R (x)]^{2} - [r (x)]^{2}) d x$

PICTURE. Ek washer face: outer circle radius $R$ , inner hole radius $r$ , ring unke beech shaded. Centre se do arrows: ek lamba $R$ tak, ek chhota $r$ tak.

Step 7 — Edge & degenerate cases (taaki kuch surprise na kare)

WHAT. Disk sirf woh washer hai jisme koi hole nahi: inner radius $r = 0$ set karo aur $R^{2} - r^{2} = R^{2}$ — tum exactly Step 5 recover kar lete ho. Do aur edges:

Region axis ko touch karta hai (curve axis par ek point par milti hai). Wahan radius $0$ hai, toh coin ek point tak shrink ho jaati hai — solid ek tip par aa jaata hai. Bilkul theek hai: zero-radius coin ka area zero hai, kuch contribute nahi karta, aur integral isko automatically handle kar leta hai.
Axis ek shifted line $y = k$ hai, x-axis nahi. Radius ab height $f (x)$ nahi hai; yeh curve se line tak ki distance hai, $∣ f (x) - k ∣$ . Distance hamesha positive hoti hai, isliye absolute value aata hai. Lekin hum usse square karte hain, aur squaring koi bhi sign mita deta hai — toh formula mein $∣ f (x) - k ∣^{2} = (f (x) - k)^{2}$ aur tum bars hata sakte ho.

Yeh sab yahan kyun? Taaki tum kabhi aisi case se na milo jise derivation ne cover nahi kiya: hole/no-hole, touching/not-touching, on-axis/off-axis sab ek hi formula hai different radii ke saath.

PICTURE. Teen mini-panels: (a) $r = 0$ disk-as-washer, (b) curve axis par ek point par tip karta hua, (c) ek curve jisme axis raised line $y = k$ ki tarah draw hai aur radius arrow wahan se measure kiya gaya hai.

$r = 0 \Rightarrow R^{2} - r^{2} = R^{2} (washer collapses to disk)$ $axis y = k \Rightarrow r = distance to the line ∣ f (x) - k ∣, r^{2} = (f (x) - k)^{2}$

Ek-picture summary

Sab kuch ek canvas par: flat region (left) → ek coin bahar nikali gayi apne radius arrow ke saath (middle) → stacked coins solid ban rahi hain, boxed integral neeche (right). Arrows ko left se right follow karo aur tumne dono formulas re-derive kar liye.

Recall Feynman retelling — poori walkthrough plain words mein

Maine ek shape paper par banayi aur usse ek skewer par ghoomaya; woh blur ho kar ek solid ban gayi. Main poore curvy solid ko measure nahi kar sakta, toh main isse skewer ke across thin coins mein slice karta hoon. Har coin round hai, aur round face ka area $π r^{2}$ hai, jahan $r$ bas itna hai ki curve skewer se kitni door hai. Coin ka volume woh area hai times uski tiny thickness. Main saari coins add karta hoon; jab main unhe infinitely thin banata hoon toh sum ek integral ban jaata hai, $V = π \int_{a}^{b} [f (x)]^{2} d x$ . Agar spinning shape ka skewer se gap tha, toh har coin mein hole hoga — ek ring — toh main big-circle-area minus hole-area leta hoon, $R^{2} - r^{2}$ , jisse $V = π \int_{a}^{b} (R^{2} - r^{2}) d x$ milta hai. Disk sirf ek ring hai bina hole ke ( $r = 0$ ), aur agar skewer ek raised line $y = k$ hai toh main $r$ wahan se measure karta hoon. Same coins, same $π r^{2}$ , hamesha.

Active Recall

Woh single 3D shape kaun sa hai jis par har step built hai?

Ek thin cylinder (ek coin) radius

r

aur thickness

Δ x

ka.

Coin ke face ka area

π r^{2}

kyun hai?

Face ek circle hai, aur radius

r

ke circle ka area

π r^{2}

hota hai.

Coins ka sum integral mein kya convert karta hai?

Thickness

Δ x \to 0

ko let karna; limiting Riemann sum hi definite integral hai.

Disk ek special washer kaise hai?

Inner radius

r = 0

set karo, toh

R^{2} - r^{2} = R^{2}

y = k

ke around rotate karne par

r

kya hai?

Distance

∣ f (x) - k ∣

; squaring sign ko irrelevant bana deta hai.

R^{2} - r^{2}

kyun aur

(R - r)^{2}

kyun nahi?

R - r

ek length hai (ring thickness); area ek thickness ko square karke nahi aa sakta. Ring area = big disk − hole.

Connections

Definite Integral as a Limit of Sums — Step 5 yahi hai: sum of coins → integral.
Area Between Curves — same do-curve picture; wahan tum heights subtract karte ho, yahan tum squared radii subtract karte ho.
Volume by Cross-Sections — disk/washer is general slicing idea ka circular case hai.
Shell Method — alternative: axis ke parallel slice karo instead of perpendicular.
u-substitution — woh tool jo Step 7 ka shifted-axis integral evaluate karta hai.