Volume of revolution — shell method: Worked examples

4.2.15 · D3 · Maths › Calculus II — Integration › Volume of revolution — shell method

Kuch bhi karne se pehle, har shell ke teen pieces se milo. Neeche ke figure mein, radius $r$ (cyan, bottom par) axis se strip tak ki horizontal distance hai; height $h$ (amber, right par) cyan strip kitni lambi hai; aur thickness — jo ek vertical strip ke liye bilkul symbol $d x$ hai (ek horizontal axis ise $d y$ bana deti) — strip kitni wide hai (white, top par). Amber vertical line jis par "axis" likha hai woh $y$ -axis hai ( $c = 0$ ) — dekho radius arrow bilkul waheen se shuru hota hai. Is page par koi symbol inse aur $\int$ se aage nahi hai, jo simply matlab hai "sabhi strips ko jodo."

Figure — Volume of revolution — shell method

Figure s01 — ek cylindrical shell dissected: cyan radius $r = x$ amber axis se, amber height $h$ , white thickness labelled $d x$ .

Scenario matrix

Har shell-method problem in cells mein se ek hai (ya inka blend). Poori mushkil hamesha radius aur height mein hoti hai — inhe sahi karo aur integral routine ho jaata hai. (Yaad karo $c$ = axis ko name karne wala number, bilkul upar defined; $f (x)$ = top curve ki height.)

Cell	Kya badalta hai	Radius $r$	Height $h$	Thickness	Example
A	Axis = $y$ -axis ( $c = 0$ ), region uske right mein	$x$	$f (x)$	$d x$	Ex 1
B	Axis = vertical line $x = c$ , region uske left mein	$c - x$	top − bottom	$d x$	Ex 2
C	Axis = vertical line $x = c$ , region uske right mein	$x - c$	top − bottom	$d x$	Ex 3
D	Region axis $x = c$ ko straddle karti hai (radius ka sign flip hota hai)	$	x-c	$	$f (x)$
E	Axis = horizontal line $y = c$ → $d y$ use karna padta hai	$	y-c	$	right − left
F	Curve $x = g (y)$ ke roop mein diya gaya hai (inversion ki zaroorat nahi)	$y$ or $	y-c	$	$g (y)$
G	Degenerate: axis region ko touch karti hai (radius $= 0$ ek edge par)	$x$	$f (x)$	$d x$	Ex 7
H	Word problem with units	$x$ (distance in cm)	$f (x)$ (length in cm)	$d x$	Ex 8
I	Exam twist: do curves ke beech region + shifted axis	$	x-c	$	top − bottom

Cell A — axis $y$ -axis hai ( $c = 0$ )

Worked example Example 1 (Cell A) — region axis ke right mein

Region $y = 4 - x^{2}$ ke neeche, $y = 0$ ke upar, $0 \leq x \leq 2$ ke liye, $y$ -axis ( $c = 0$ ) ke around rotate kiya gaya.

Forecast: region $x = 0$ se $x = 2$ tak ek "hill" hai. Guess karo: kya volume $8 π$ ke kareeb hai ya $30 π$ ke? Apna guess likh lo.

Step 1 — axis ke parallel strips. Axis vertical hai, isliye main vertical strips use karta hoon (thickness $d x$ ). Yeh step kyun? Axis ke parallel strip ek clean tube sweep karta hai; perpendicular strip disks deta, jo yahan mujhe $x = 4 - y$ solve karne par majboor karta.

Step 2 — radius. $r = x$ (yahan $c = 0$ , isliye $∣ x - c ∣ = ∣ x - 0∣ = x$ ). Kyun? $y$ -axis (line $x = 0$ ) se position $x$ par strip tak ki distance bas $x$ hai.

Step 3 — height. $h = f (x) = 4 - x^{2}$ (yaad karo $f (x)$ = top curve ki height). Kyun? Strip floor $y = 0$ se upar curve $y = 4 - x^{2}$ tak jaati hai; height = top minus bottom.

Step 4 — yeh limits kyun, phir integrate karo. Region defined hai $x = 0$ (left edge, axis) se $x = 2$ tak (right edge, problem mein diya gaya), isliye $x$ , $0 \to 2$ sweep karta hai aur wahi integral bounds hain: $V = \int_{0}^{2} 2 π x (4 - x^{2}) d x = 2 π \int_{0}^{2} (4 x - x^{3}) d x = 2 π [2 x^{2} - \frac{x ^{4}}{4}]_{0}^{2} .$ $= 2 π (8 - 4) = 8 π .$

Verify: $x = 1$ par shell ka $r = 1$ , $h = 3$ hai, isliye $d V / d x = 2 π (3) = 6 π$ — ek plausible mid-value; width $2$ par integrate karo toh $8 π$ ke kareeb aata hai. ✓ Units: agar $x, y$ cm mein hain toh volume cm³ mein hai.

Cells B & C — vertical line $x = c$ as axis, region dono sides par

Yahan ek subtle cheez hai radius ke andar ka sign. Figure dekho: cyan strip kisi $x$ par baith ti hai $0$ aur $2$ ke beech, aur wahi strip do alag-alag axes tak maapi jaati hai. Right axis ( $c = 5$ ) tak distance $5 - x$ hai; left axis ( $c = - 1$ ) tak $x + 1$ hai. Dono cyan arrows positive aate hain — yahi is pair ka poora lesson hai.

Figure s02 — wahi cyan region do amber axes tak maapi gayi; cyan arrows $r = 5 - x$ (right axis) aur $r = x + 1$ (left axis) dikhate hain, dono positive.

Worked example Example 2 (Cell B) — region axis ke LEFT mein

Region $y = x^{2}$ ke neeche, $0 \leq x \leq 2$ , vertical line $x = 5$ ke around rotate kiya gaya (toh $c = 5$ ).

Forecast: axis region ke bahut right mein hai. Har jagah bada radius ⇒ kya tumhe expect hai ki volume $8 π$ (jo $x = 0$ ke around mila) se kaafi bada hoga? Ek number guess karo.

Step 1 — radius. Axis $x = 5$ har strip ke right mein hai ( $0 \leq x \leq 2$ ), isliye strip ki distance $r = ∣ x - 5∣ = 5 - x$ hai. Kyun? $5 - x$ , $[0, 2]$ par positive hai (5 se 3 tak jaata hai). $x - 5$ likhne par negative number aata — radius ke liye galat sign.

Step 2 — height. $h = x^{2}$ (unchanged; height ko axis ki parwah nahi).

Step 3 — yeh limits kyun, phir integrate karo. Region ki apni edges $x = 0$ aur $x = 2$ hain (dono statement mein diye gaye), isliye wahi bounds hain: $V = \int_{0}^{2} 2 π (5 - x) x^{2} d x = 2 π \int_{0}^{2} (5 x^{2} - x^{3}) d x = 2 π [\frac{5 x ^{3}}{3} - \frac{x ^{4}}{4}]_{0}^{2} .$ $= 2 π (\frac{40}{3} - 4) = 2 π \cdot \frac{28}{3} = \frac{56 π}{3} .$

Verify: $\frac{56 π}{3} \approx 58.6$ , sachchi mein kaafi bada $8 π \approx 25.1$ se — "far axis, bada radius" forecast se match karta hai. ✓

Worked example Example 3 (Cell C) — region axis ke RIGHT mein

Wahi region $y = x^{2}$ , $0 \leq x \leq 2$ , lekin vertical line $x = - 1$ ke around rotate kiya gaya (toh $c = - 1$ ).

Forecast: ab axis region ke left mein baith ti hai. $x = 5$ wale case se bada ya chhota? (Hint: $x = - 1$ sirf $1$ se $3$ units door hai; $x = 5$ $3$ se $5$ units door tha.)

Step 1 — radius. Axis $x = - 1$ har strip ke left mein hai, isliye $r = ∣ x - (- 1) ∣ = x + 1$ . Kyun? $[0, 2]$ par, $x + 1$ , $1$ se $3$ tak jaata hai, poori tarah positive. Strip ek left axis se utni hi door hoti hai jitna bada $x$ ho.

Step 2 — height. $h = x^{2}$ .

Step 3 — yeh limits kyun, phir integrate karo. Wahi region, wahi edges $x = 0$ se $x = 2$ tak: $V = \int_{0}^{2} 2 π (x + 1) x^{2} d x = 2 π \int_{0}^{2} (x^{3} + x^{2}) d x = 2 π [\frac{x ^{4}}{4} + \frac{x ^{3}}{3}]_{0}^{2} .$ $= 2 π (4 + \frac{8}{3}) = 2 π \cdot \frac{20}{3} = \frac{40 π}{3} .$

Verify: $\frac{40 π}{3} \approx 41.9$ — $x = 5$ wale case ( $58.6$ ) se chhota, kyunki yeh axis kareeb hai. Sign of radius positive nikla jaise chahiye tha. ✓

Cell D — region ek vertical axis $x = c$ ko straddle karti hai

Yahan khatra yeh hai ki radius ka sign axis cross karte hi badal jaata hai. Figure ek region dikhata hai jo $x = 1$ se $x = 4$ tak jaati hai aur axis $x = 2$ se split hoti hai: axis ke left mein strips ka radius $2 - x$ hai, right mein $x - 2$ hai, aur ek formula $∣ x - 2∣$ dono ko capture karta hai. Dekho cyan radius axis par exactly zero tak simat ta hai aur dusri taraf phir badhta hai.

Figure s04 — amber axis $x = 2$ ko straddle karta rectangle; cyan arrows left par radius $2 - x$ aur right par $x - 2$ dikhate hain, dono axis par vanish hote hain.

Worked example Example 4 (Cell D) — region line

$x = 2$ cross karti hai Region $y = 3$ ke neeche, $y = 0$ ke upar, $1 \leq x \leq 4$ ke liye (ek rectangle), vertical line $x = 2$ ke around rotate kiya gaya (toh $c = 2$ ). Axis region ke through jaati hai.

Forecast: rectangle ka ek part $x = 2$ ke left mein hai (width $1$ ), ek part right mein (width $2$ ). Dono parts real, positive volume sweep karte hain. Kya woh add honge, ya partly cancel? Guess karo.

Step 1 — sign flip pakdo. Axis ke left mein strips ( $1 \leq x < 2$ ) ke liye signed quantity $x - 2$ negative hai; uske right mein ( $2 < x \leq 4$ ) positive hai. Distance negative nahi ho sakti, isliye true radius $r = ∣ x - 2∣$ hai. Yeh step kyun matters? Agar tum $x - 2$ raw use karo, left part negative volume contribute karega aur galat tarike se right part cancel ho jaayega.

Step 2 — axis par split karo. Region ko wahaan tod do jahan radius formula badlti hai, yaani $x = 2$ par, aur har piece par absolute value hata do: $V = \int_{1}^{2} 2 π (2 - x) \cdot 3 d x + \int_{2}^{4} 2 π (x - 2) \cdot 3 d x .$ Kyun? $[1, 2]$ par, $∣ x - 2∣ = 2 - x$ ; $[2, 4]$ par, $∣ x - 2∣ = x - 2$ . Bounds $1, 2, 4$ region ki edges aur axis crossing se aate hain.

Step 3 — har piece integrate karo. Height $h = 3$ poori jagah hai (top $y = 3$ minus bottom $y = 0$ ). $\int_{1}^{2} 2 π (2 - x) 3 d x = 6 π [2 x - \frac{x ^{2}}{2}]_{1}^{2} = 6 π ((4 - 2) - (2 - \frac{1}{2})) = 6 π \cdot \frac{1}{2} = 3 π .$ $\int_{2}^{4} 2 π (x - 2) 3 d x = 6 π [\frac{x ^{2}}{2} - 2 x]_{2}^{4} = 6 π ((8 - 8) - (2 - 4)) = 6 π \cdot 2 = 12 π .$ $V = 3 π + 12 π = 15 π .$

Verify: rectangle ka har aadha ek solid cylinder sweep karta hai (koi hole nahi, kyunki strip axis tak pahunchi), isliye check ke liye disk method use karo. Left half → solid cylinder of radius $1$ , height $3$ : $π (1)^{2} (3) = 3 π$ ✓. Right half → solid cylinder of radius $2$ , height $3$ : $π (2)^{2} (3) = 12 π$ ✓. Total $15 π$ . Agar tum absolute value bhool jaate, toh $\int_{1}^{4} 2 π (x - 2) 3 d x = 6 π [\frac{x ^{2}}{2} - 2 x]_{1}^{4} = 6 π (0 - (- \frac{3}{2})) = 9 π$ — galat, jo confirm karta hai split zaroori tha. ✓

Cell E — horizontal axis $y = c$ $d y$ force karti hai

Jab axis horizontal ho, strips bhi horizontal honi chahiye (axis ke parallel), isliye unki thickness $d y$ hai aur unki length left-to-right maapi jaati hai. Figure mein, amber length ab ek width hai ( $x_{right} - x_{left}$ ) aur cyan radius axis $y = c$ se strip tak vertically mapa jaata hai. Yahan axis line $y = 1$ ke roop mein draw ki gayi hai (isliye $c = 1$ ), $x$ -axis nahi — dekho radius arrow $y = 1$ se shuru hota hai, deta hai $r = ∣ y - 1∣$ .

Figure s03 — horizontal strip (thickness $d y$ ) with cyan vertical radius $r = ∣ y - 1∣$ amber axis $y = 1$ se aur amber width $h =$ right $-$ left.

Worked example Example 5 (Cell E) — shifted line

$y = 1$ ke around rotation, $d y$ use karke Region $x = y$ , $x = 0$ , aur $y = 3$ se bounded (ek triangle with vertices $(0, 0), (0, 3), (3, 3)$ ), horizontal line $y = 1$ ke around rotate kiya gaya (toh $c = 1$ ).

Forecast: axis triangle ke vertical extent ke andar hai ( $y$ , $0$ se $3$ tak jaata hai, aur $1$ unke beech hai). Toh $y = 1$ ke paas radius tiny hai. $y = 0$ ke around rotate karne se chhota ya bada (jo $18 π$ deta)? Guess karo.

Step 1 — strips horizontal. Axis horizontal hai ⇒ thickness $d y$ . Region ka vertical extent $y = 0$ (bottom vertex) se $y = 3$ (given top edge) tak hai, toh wahi bounds hain. Yeh limits kyun? $y = 0$ triangle ka sabse neeche ka point hai, $y = 3$ uska top edge; strips region ko sweep karti hain jab $y$ , $0 \to 3$ jaata hai.

Step 2 — radius. $r = ∣ y - 1∣$ (axis $y = 1$ se strip tak ki vertical distance). Absolute value kyun? Axis ke neeche ( $0 \leq y < 1$ ) strip $1 - y$ door hai; uske upar ( $1 < y \leq 3$ ) $y - 1$ door hai. Region axis cross karti hai, isliye $y = 1$ par split karo.

Step 3 — height (ek width!). Height $y$ par, strip $x = 0$ se left par $x = y$ tak right par jaati hai, isliye $h = y - 0 = y$ . Kyun? Jab strips horizontal hon toh height = right minus left — curve ki $y$ -value nahi.

Step 4 — axis par split karo aur integrate karo. $V = \int_{0}^{1} 2 π (1 - y) y d y + \int_{1}^{3} 2 π (y - 1) y d y .$ $\int_{0}^{1} 2 π (y - y^{2}) d y = 2 π [\frac{y ^{2}}{2} - \frac{y ^{3}}{3}]_{0}^{1} = 2 π (\frac{1}{2} - \frac{1}{3}) = 2 π \cdot \frac{1}{6} = \frac{π}{3} .$ $\int_{1}^{3} 2 π (y^{2} - y) d y = 2 π [\frac{y ^{3}}{3} - \frac{y ^{2}}{2}]_{1}^{3} = 2 π ((9 - \frac{9}{2}) - (\frac{1}{3} - \frac{1}{2})) = 2 π (\frac{9}{2} + \frac{1}{6}) = 2 π \cdot \frac{14}{3} = \frac{28 π}{3} .$ $V = \frac{π}{3} + \frac{28 π}{3} = \frac{29 π}{3} .$

Verify: $\frac{29 π}{3} \approx 30.4$ , $y = 0$ ke answer $18 π \approx 56.5$ se bada? Nahi — $18 π \approx 56.5$ bada hai, jo sahi hai: region ke through ek axis average par chhote radii deti hai bajaay ek axis ke jo usse $1$ unit neeche ho. Forecast confirm hua (chhota $18 π$ se). ✓

Cell F — curve pehle se $x = g (y)$ form mein hai

Worked example Example 6 (Cell F) — koi inversion needed nahi,

$x$ -axis ke around rotate karo Region $x = y^{2}$ , $x = 0$ , aur $y = 2$ se bounded (with $y \geq 0$ ), $x$ -axis ( $c = 0$ ) ke around rotate kiya gaya.

Forecast: curve $x = y^{2}$ rightward open hoti hai. Guess karo ki $V$ , $10 π$ se zyada hai ya nahi.

Step 1 — horizontal strips. Axis horizontal ⇒ $d y$ . Region vertically $y = 0$ (jahan $x = y^{2}$ , $x = 0$ se milti hai) se $y = 2$ (given cap) tak jaati hai, isliye bounds $0 \to 2$ hain. Yeh limits kyun? $x = y^{2}$ aur $x = 0$ , $y = 0$ par intersect karte hain; line $y = 2$ top close karti hai.

Step 2 — radius. $r = ∣ y - 0∣ = y$ ( $x$ -axis se strip tak ki distance; $y \geq 0$ toh koi sign issue nahi).

Step 3 — height (width). $x = 0$ se $x = y^{2}$ tak, isliye $h = g (y) = y^{2}$ (yahan $g (y)$ = right-boundary curve as an $x$ -value). Yeh cell kyun easy hai: curve pehle se $y$ ki function ke roop mein $x$ hai, isliye invert karne ki koi algebra nahi — exactly woh situation jise shell method pasand karta hai.

Step 4 — integrate karo. $V = \int_{0}^{2} 2 π y \cdot y^{2} d y = 2 π \int_{0}^{2} y^{3} d y = 2 π [\frac{y ^{4}}{4}]_{0}^{2} = 2 π \cdot 4 = 8 π .$

Verify: $y = 2$ par strip ka $r = 2$ , $h = 4$ hai, isliye $d V / d y = 16 π$ top ke paas aur $0$ bottom par — average kuch $16 π \cdot 2 = 32 π$ se kaafi kum; $8 π$ consistent hai. ✓

Cell G — degenerate case (radius ek edge par zero)

Worked example Example 7 (Cell G) — axis region ko touch karti hai

Region $y = x$ , $y = 0$ , $0 \leq x \leq 4$ ke neeche, $y$ -axis ( $c = 0$ ) ke around rotate kiya gaya.

Forecast: $x = 0$ par strip axis par hi baith ti hai, isliye uska radius $0$ hai — us shell ka volume zero hai. Kya integral phir bhi theek behave karti hai? Guess karo ki answer finite hai ya nahi.

Step 1 — degenerate edge check karo. $x = 0$ par: $r = 0$ , isliye $d V = 2 π (0) h d x = 0$ . Koi blow-up nahi, koi problem nahi — integrand wahaan simply vanish karta hai. Yeh step kyun matters? Jab bhi axis region ko graze kare, students panic karte hain; yahan $r = 0$ wala shell ek genuine "line" hai, kuch contribute nahi karta, aur integral finite rehti hai.

Step 2 — radius aur height. $r = x$ , $h = f (x) = x = x^{1/2}$ .

Step 3 — yeh limits kyun, phir integrate karo. Region $x = 0$ (jahan $y = x$ , $y = 0$ se milti hai) se $x = 4$ (given) tak hai, isliye bounds $0 \to 4$ : $V = \int_{0}^{4} 2 π x \cdot x^{1/2} d x = 2 π \int_{0}^{4} x^{3/2} d x = 2 π [\frac{2}{5} x^{5/2}]_{0}^{4} .$ Kyunki $4^{5/2} = (4)^{5} = 2^{5} = 32$ : $V = 2 π \cdot \frac{2}{5} \cdot 32 = \frac{128 π}{5} .$

Verify: $\frac{128 π}{5} \approx 80.4$ , forecast ke anusaar finite; $r = 0$ endpoint ne koi trouble nahi kiya kyunki integrand $x^{3/2} \to 0$ smoothly. ✓

Cell H — units wala word problem

Worked example Example 8 (Cell H) — ek machined bushing

Ek engineer region ko $y = 6 - x$ aur $y = 0$ ke beech, $2 \leq x \leq 6$ ke liye (sab cm mein), $y$ -axis ( $c = 0$ ) ke around spin karta hai ek tapered ring banane ke liye. Uska volume cm³ mein nikalo.

Forecast: region ek triangle hai jo $x = 2$ aur $x = 6$ ke beech baith ta hai, beech mein ek hole ke saath (kuch axis tak nahi pahunchta). Guess karo: tens ya hundreds of cm³?

Step 1 — radius. $r = x$ cm ( $y$ -axis se strip tak ki distance). Kyun? Cell A jaisa hi, lekin region $x = 2$ se shuru hoti hai, isliye kisi shell ka radius $2$ se kum nahi — woh ek hollow core chhod ta hai, jo ek "ring" ke liye sahi hai.

Step 2 — height. $h = f (x) = 6 - x$ cm (floor $y = 0$ se line tak).

Step 3 — yeh limits kyun, phir integrate karo. Strip ki left edge $x = 2$ hai aur right edge $x = 6$ (dono diye gaye), aur note karo $y = 6 - x$ exactly $x = 6$ par $0$ hit karta hai, isliye wahaan region close hoti hai: $V = \int_{2}^{6} 2 π x (6 - x) d x = 2 π \int_{2}^{6} (6 x - x^{2}) d x = 2 π [3 x^{2} - \frac{x ^{3}}{3}]_{2}^{6} .$ $x = 6$ par: $3 (36) - \frac{216}{3} = 108 - 72 = 36$ . $x = 2$ par: $3 (4) - \frac{8}{3} = 12 - \frac{8}{3} = \frac{28}{3}$ . $V = 2 π (36 - \frac{28}{3}) = 2 π \cdot \frac{80}{3} = \frac{160 π}{3} cm^{3} .$

Verify: $\frac{160 π}{3} \approx 167.6 cm^{3}$ . Units: (cm)(cm)(cm) → cm³ ✓. Empty core ( $x < 2$ ) sahi tarike se kuch contribute nahi karta. ✓

Cell I — exam twist (do curves + shifted axis)

Worked example Example 9 (Cell I) — curves ke beech,

$x = - 1$ ke around Region $y = x$ aur $y = x^{2}$ ke beech, vertical line $x = - 1$ ke around rotate kiya gaya (toh $c = - 1$ ).

Forecast: parent note mein $y$ -axis ke around yahi region $\frac{π}{6}$ deti thi. Axis ab $1$ unit aur left mein hai, isliye har radius $1$ se badh jaata hai. Obviously bada — lekin kitna? Guess karo.

Step 1 — curves kahan milti hain (limits). Set karo $x = x^{2} \Rightarrow x^{2} - x = 0 \Rightarrow x (x - 1) = 0$ , isliye $x = 0$ aur $x = 1$ . Yeh intersection points hi integral bounds hain — region sirf inke beech exist karti hai. Yeh step kyun? Do curves $[0, 1]$ par bilkul region enclose karti hain; usse bahar kuch bound nahi hota.

Step 2 — height (top − bottom). $(0, 1)$ par, $x = \frac{1}{2}$ test karo: $y = x$ deta hai $\frac{1}{2}$ , $y = x^{2}$ deta hai $\frac{1}{4}$ , isliye $x > x^{2}$ aur $h = x - x^{2}$ . Kyun? Do curves ⇒ height unke beech ka gap hai, top minus bottom.

Step 3 — $x = - 1$ ke around radius. $r = ∣ x - (- 1) ∣ = x + 1$ , $[0, 1]$ par positive (1 se 2 tak jaata hai). Kyun? Axis left mein hai, isliye distance $x + 1$ hai.

Step 4 — integrate karo. Pehle product carefully expand karo: $(x + 1) (x - x^{2}) = x \cdot x - x \cdot x^{2} + 1 \cdot x - 1 \cdot x^{2} = x^{2} - x^{3} + x - x^{2} = x - x^{3} .$ Dono $x^{2}$ terms cancel ho jaate hain, bacha $x - x^{3}$ . Toh $V = \int_{0}^{1} 2 π (x - x^{3}) d x = 2 π [\frac{x ^{2}}{2} - \frac{x ^{4}}{4}]_{0}^{1} = 2 π (\frac{1}{2} - \frac{1}{4}) = 2 π \cdot \frac{1}{4} = \frac{π}{2} .$

Verify: $\frac{π}{2} = 3 \cdot \frac{π}{6}$ , exactly $3 \times$ $y$ -axis wala answer. Sanity: " $+ 1$ " shift ne $\int_{0}^{1} 2 π (x - x^{2}) d x = 2 π (\frac{1}{2} - \frac{1}{3}) = \frac{π}{3}$ original $\frac{π}{6}$ mein add kiya, deta hai $\frac{π}{6} + \frac{π}{3} = \frac{π}{2}$ . ✓

Recall checkpoint

Recall Which cell is each situation?

Axis is $x = 4$ , region lies in $0 \leq x \leq 3$ — which radius? ::: Region is left of the axis (Cell B): $r = 4 - x$ ; here $c = 4$ . Rotating about the horizontal line $y = 2$ — what is the strip thickness? ::: $d y$ (Cell E); strips are horizontal, radius $∣ y - 2∣$ . Region symmetric across the $y$ -axis, rotated about it — what safeguard? ::: Use $r = ∣ x ∣$ or exploit symmetry (Cell D); a signed $x$ would wrongly cancel. Region crosses the axis $x = 2$ — how do you set up the integral? ::: Use $r = ∣ x - 2∣$ and split at $x = 2$ into $2 - x$ (left) and $x - 2$ (right). A strip sits exactly on the axis — is the integral broken? ::: No (Cell G); that shell has $r = 0$ , contributing $0$ , integral stays finite. Curve given as $x = g (y)$ , axis horizontal — why is shell ideal? ::: No inversion needed (Cell F); integrate in $y$ directly.

Connections

shell method — the parent formula this page stress-tests.
Volume of revolution — disk and washer method — the perpendicular-slice alternative; Cell D's check used solid cylinders (the disk method).
Definite integral as a limit of Riemann sums — why summing infinitely many shells is exact.
Area between two curves — the "top − bottom" height reused in Cells D, E, I.
Choosing dx vs dy strips — decides between Cells A–D ( $d x$ ) and Cells E–F ( $d y$ ).
Arc length and surfaces of revolution — revolve the boundary instead of the region.

Worked examples — Volume of revolution — shell method