Worked examples — Volume of revolution — disk method, washer method

4.2.14 · D3 · Maths › Calculus II — Integration › Volume of revolution — disk method, washer method

Yeh page disk & washer topic ka drill floor hai. Parent note ne dono formulas build kiye. Yahan hum har tarah ki situation unpe throw karte hain, ek ek karke, aur har ek ko puri tarah solve karte hain. Agar exam mein rotation ka koi problem mile, toh woh neeche diye gaye kisi na kisi cell mein hoga.

Shuru karne se pehle, ek reminder seedhe shabdon mein. Ek radius simply ek distance hota hai: curve ka koi point spinning line (axis) se kitna door hai. Kyunki distance kabhi negative nahi ho sakti, radius hamesha axis aur curve ke beech ke gap ki absolute value hoti hai: $r=|f(x)-(\text{axis value})|.$ Jab curve axis ke neeche ho (yaani $f(x)$ axis value se chhota ho), toh yeh absolute value sign flip kar deti hai — isliye Example 4 mein aap $f(x)-2$ ki jagah $2-f(x)$ dekhenge. Yeh yaad rakho: radius ek length hai, kabhi signed number nahi. Volume thin coins ko add karne se aata hai, $\pi r^{2}dx$. Baaki sab kaunsi distance aur kaunsa variable ka bookkeeping hai.

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Scenario matrix

Is topic ke har problem ka type inhi cells mein se ek hai. Last column us worked example ka naam batata hai jo ise cover karta hai.

Cell	Kya change hota hai	Disk ya Washer	Variable	Example
A. Region x-axis ko touch karta hai	radius $=f(x)$, koi hole nahi	Disk	$dx$	Ex 1
B. Do curves ke beech region, axis ke upar	outer/inner radii	Washer	$dx$	Ex 2
C. y-axis ke baare mein rotate karo	$x=g(y)$ ke liye solve karo	Disk	$dy$	Ex 3
D. Horizontal line $y=k$ ke baare mein rotate karo	dono radii ko $k$ se shift karo	Washer	$dx$	Ex 4
E. Vertical line $x=k$ ke baare mein rotate karo	radii shift karo, $y$ mein integrate karo	Washer	$dy$	Ex 5
F. Degenerate / zero case (curves meet, ya region ka koi area nahi)	volume collapse ho jaata hai	—	—	Ex 6
G. Real-world word problem (ek physical vase/tank)	words → curve mein translate karo	Disk	$dx$	Ex 7
H. Exam twist: interval ke andar outer/inner swap ho jaate hain (curves cross karte hain)	integral split karo	Washer	$dx$	Ex 8
I. Axis region se guzarti hai	koi hole nahi, max distance use karo	Disk	$dx$	Ex 9

Neeche di gayi figure poori page ka map hai. Ise aise padho: gray solid lines x- aur y-axes hain; har coloured dashed line rotation ka ek alag possible axis hai jo cells use karte hain — orange for horizontal line $y=k$, red for $y=2$, green for vertical line $x=3$. Blue arrow woh ek idea dikhata hai jo har cell share karta hai: radius jis bhi axis ko aapne choose kiya usse curve tak ki distance hai. Jaise jaise aap examples work karo, wapas aao aur apne cell ka axis yahan dhundho.

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Example 1 — Cell A (region x-axis ko touch karta hai, pure disk)

Forecast: height tezi se badhti hai (yeh $x^2$ hai), toh squaring se $x^4$ milega — kaafi bada number expect karo, kuch $\pi$ jaisa. Aage padhne se pehle guess karo.

Figure region (green) aur ek representative red coin dikhata hai: kyunki region x-axis par rest karta hai, coin solid hai — koi hole nahi.

1. Radius identify karo. Region x-axis par baitha hai, toh har coin ka radius curve ki height hai, $r=f(x)=x^2$. Yeh step kyun? Region aur axis ke beech koi gap nahi ⇒ solid coins ⇒ disk method.
2. Disk integral likho. $V=\pi {\int }_0^2[x^2{]}^{2}dx=\pi {\int }_0^2{x}^{4}dx.$ Yeh step kyun? Coin area $=\pi r^2=\pi (x^2{)}^2=\pi x^4$; thickness $dx$ ke saath sum karo.
3. Integrate karo. $V=\pi {\left[\frac{x^5}{5}\right]}_0^2=\pi \cdot \frac{32}{5}=\frac{32\pi }{5}.$ Yeh step kyun? Power rule: $\int x^{n}dx=x^{n+1}/(n+1)$, toh $\int x^{4}dx=x^5/5$; phir $x=2$ plug in karo ($32/5$ milega) aur $x=0$ ($0$ milega) aur subtract karo.

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Example 2 — Cell B (axis ke upar do curves, washer)

Forecast: $\sqrt{x}$, $(0,1)$ par $x$ ke upar baitha hai, toh woh far curve hai. Gap thin hai, toh chhota answer expect karo, shayad $\pi$ se kam.

1. Outer vs inner decide karo. $(0,1)$ par, $\sqrt{x}>x$ (test karo $x=\frac{1}{4}$: $\frac{1}{2}>\frac{1}{4}$). Toh $R=\sqrt{x}$, $r=x$. Yeh step kyun? Outer radius = far curve tak distance; washer ka hole near curve se aata hai.
2. Har radius ko alag square karo, phir subtract karo. $V=\pi {\int }_0^1([\sqrt{x}{]}^2-[x{]}^2)dx=\pi {\int }_0^1(x-x^2)dx.$ Yeh step kyun? Ring area $=\pi R^2-\pi r^2$, kabhi $\pi (R-r{)}^2$ nahi.
3. Integrate karo. $V=\pi {\left[\frac{x^2}{2}-\frac{x^3}{3}\right]}_0^1=\pi \left(\frac{1}{2}-\frac{1}{3}\right)=\frac{\pi }{6}.$ Yeh step kyun? Har power term alag integrate karo ($\int xdx=x^2/2$, $\int x^{2}dx=x^3/3$); $x=1$ par evaluate karo (lower limit $0$ kuch contribute nahi karta) aur common denominator $6$ par combine karo.

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Example 3 — Cell C (y-axis ke baare mein rotate karo)

Forecast: axis ab vertical hai, toh hum horizontally slice karte hain aur $y$ mein integrate karte hain. Coins y-axis par $y=0$ se $y=4$ tak stack hote hain.

Figure flipped picture dikhata hai: coins ab horizontal hain, aur radius y-axis se sideways curve $x=\sqrt{y}$ tak jaata hai.

1. Slicing variable switch karo. Axis vertical hai ⇒ thickness $dy$ ⇒ radius ek horizontal distance $x$ hona chahiye. Curve ko $x$ ke liye solve karo: $y=x^2$ se $x=\sqrt{y}$ milta hai. Yeh step kyun? Vertical axis ke perpendicular ek coin ki thickness $dy$ hoti hai; uska radius sideways jaata hai, toh hume $y$ ke function ke roop mein $x$ chahiye.
2. Radius = y-axis se curve tak distance $=\sqrt{y}$. Koi hole nahi (region y-axis ko touch karta hai), toh disk. $V=\pi {\int }_0^4[\sqrt{y}{]}^{2}dy=\pi {\int }_0^{4}ydy.$ Yeh step kyun? Limits ab $y$-values hain: region $y=0$ se $y=4$ tak span karta hai.
3. Integrate karo. $V=\pi {\left[\frac{y^2}{2}\right]}_0^4=\pi \cdot \frac{16}{2}=8\pi .$ Yeh step kyun? $\int ydy=y^2/2$; upper limit $y=4$ substitute karo ($16/2=8$ milta hai) aur lower limit ka $0$ subtract karo.

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Example 4 — Cell D (horizontal line $y=k$ ke baare mein rotate karo)

Forecast: axis $y=2$ dono curves ke upar baitha hai, toh dono curves us ke neeche hain. $y=2$ se door curve neeche waali ($x^2$) hai, aur paas waali curve upar waali ($x$) hai. Is reversal ko dekho — yahi is cell ka poora point hai.

1. Har radius ko naye axis se re-measure karo. $y=2$ se neeche curve $y=f(x)$ tak ki distance $|f(x)-2|=2-f(x)$ hai (dono curves line ke neeche hain, toh $f(x)-2$ negative hai aur absolute value ise $2-f(x)$ mein flip kar deti hai). Yeh step kyun? Radius = axis se curve tak ki distance, aur axis $y=2$ par move ho gaya; distance hamesha $\ge 0$ hoti hai, jo exactly woh hai jo absolute value guarantee karti hai.
2. Outer vs inner dhundho. $y=2$ se sabse door point sabse neechi curve hai. $(0,1)$ par, $x^2<x$, toh $x^2$ sabse neecha hai ⇒ sabse door hai ⇒ $R=2-x^2$. Near curve $x$ hai ⇒ $r=2-x$. Yeh step kyun? "Outer" ka matlab axis se physically door hai, "bigger $y$" nahi. Axis ko region ke upar shift karne se kaun si curve outer hai woh flip ho jaati hai.
3. Washer integral. $V=\pi {\int }_0^1[(2-x^2{)}^2-(2-x{)}^2]dx.$ Expand karo: $(2-x^2{)}^2=4-4x^2+x^4$, $(2-x{)}^2=4-4x+x^2$. Subtract karo: $(4-4x^2+x^4)-(4-4x+x^2)=x^4-5x^2+4x.$ Yeh step kyun? Pehle squares, phir subtract; constant $4$'s cancel ho jaate hain, jo ek achha sign hai.
4. Integrate karo. $V=\pi {\int }_0^1(x^4-5x^2+4x)dx=\pi {\left[\frac{x^5}{5}-\frac{5x^3}{3}+2x^2\right]}_0^1=\pi \left(\frac{1}{5}-\frac{5}{3}+2\right).$ Combine karo: $\frac{1}{5}-\frac{5}{3}+2=\frac{3-25+30}{15}=\frac{8}{15}$. $V=\frac{8\pi }{15}.$ Yeh step kyun? Term by term integrate karo ($\int x^4=x^5/5$, $\int 5x^2=5x^3/3$, $\int 4x=2x^2$); $x=1$ par har power $1$ ke barabar hoti hai, toh bracket simply $\frac{1}{5}-\frac{5}{3}+2$ hai, aur $x=0$ par woh $0$ hai. Finish karne ke liye common denominator $15$ par rakho.

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Example 5 — Cell E (vertical line $x=k$ ke baare mein rotate karo)

Forecast: vertical axis ⇒ $y$ mein slice karo. Axis $x=3$ poore region ke daaye hai ($x$, $0$ se $2$ tak chalta hai), toh near edge curve $x=\sqrt{y}$ hai aur far edge y-axis line $x=0$ hai.

Figure region ko red vertical axis $x=3$ ke saath right side par aur ek horizontal washer slice dikhata hai: uska outer radius $x=0$ tak poora jaata hai, uska inner radius curve par ruk jaata hai.

1. Horizontally slice karo. Vertical axis ⇒ thickness $dy$, radius $x=3$ se horizontal distance hai. Boundary ko $x=\sqrt{y}$ ke roop mein express karo ($y=x^2$ se). Yeh step kyun? Vertical axis ke perpendicular coins $y$ mein stack hote hain.
2. $x=3$ se do radii. Height $y$ par, region $x=0$ (left edge) se $x=\sqrt{y}$ (curve) tak chalta hai. $x=3$ se distances (har ek absolute value, dono edges axis ke left):
   • far edge $x=0$: $R=|0-3|=3$,
   • near edge $x=\sqrt{y}$: $r=|\sqrt{y}-3|=3-\sqrt{y}$. Yeh step kyun? Farther point ($x=0$) $x=3$ se zyada door hai, outer radius deta hai; absolute value dono radii ko positive rakhti hai.
3. Washer integral $y$ mein, $y=0$ se $y=4$ tak. $V=\pi {\int }_0^4[3^2-(3-\sqrt{y}{)}^2]dy.$ $(3-\sqrt{y}{)}^2=9-6\sqrt{y}+y$ expand karo. Toh $9-(9-6\sqrt{y}+y)=6\sqrt{y}-y$. $V=\pi {\int }_0^4(6\sqrt{y}-y)dy.$ Yeh step kyun? Pehle squares; $9$'s cancel ho jaate hain.
4. Integrate karo ($\int \sqrt{y}dy=\frac{2}{3}{y}^{3/2}$): $V=\pi {\left[6\cdot \frac{2}{3}{y}^{3/2}-\frac{y^2}{2}\right]}_0^4=\pi {\left[4y^{3/2}-\frac{y^2}{2}\right]}_0^4.$ $y=4$ par: $y^{3/2}=8$, toh $4\cdot 8-\frac{16}{2}=32-8=24$. $V=24\pi .$ Yeh step kyun? $\sqrt{y}=y^{1/2}$ likho taaki power rule apply ho ($\int y^{1/2}=\frac{2}{3}{y}^{3/2}$); $y=4$ par $4^{3/2}=(\sqrt{4}{)}^3=8$ use karke evaluate karo, aur lower limit $0$ phir se kuch contribute nahi karta.

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Example 6 — Cell F (degenerate / zero volume)

Forecast: agar do curves coincide hoti hain, toh koi region nahi — zero area — toh zero volume. Degeneracy ko pehchanna aapko nonsense compute karne se bachata hai.

Figure degeneracy ko visible banata hai: do "curves" ek doosre ke exactly upar hain, toh shaded region ka literally zero width hai.

1. Pehla case: identical curves. $R=x$ aur $r=x$, toh $R^2-r^2=x^2-x^2=0$. $V=\pi {\int }_0^{1}0dx=0.$ Yeh step kyun? Curves ke beech koi gap nahi ⇒ koi washer thickness nahi ⇒ koi solid nahi.
2. Doosra case: zero width ka ek interval. $x=1$ se $x=1$ tak integrate karna: $V=\pi {\int }_1^1[(x^2+3{)}^2-(x^2{)}^2]dx=0.$ Yeh step kyun? ${\int }_a^a(\text{kuch bhi})dx=0$; ek slice ko volume hone ke liye width $\Delta x>0$ chahiye.

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Example 7 — Cell G (real-world word problem: ek bowl)

Forecast: yeh disguise mein Cell C hai (y-axis ke baare mein rotate karo). $\pi$ aur clean fraction wala answer expect karo; units cm³ hain.

Figure bowl ka cross-section (shaded parabola) aur yeh kaise y-axis ke baare mein spin karke paraboloid vessel carve karta hai woh dikhata hai.

1. Words ko ek curve aur axis mein translate karo. "y-axis ke baare mein rotate karo" ⇒ $y$ mein slice karo, radius $=x=\sqrt{y}$ ($y=x^2$ se). Bowl $y=0$ (bottom) se $y=9$ tak chalta hai (kyunki $x=3\Rightarrow y=9$). Yeh step kyun? Physical rim $x=3$ par hai, jo top height $y=9$ deta hai.
2. $y$ mein disk method (solid, axis ko touch karta hai): $V=\pi {\int }_0^9[\sqrt{y}{]}^{2}dy=\pi {\int }_0^{9}ydy.$ Yeh step kyun? Har horizontal coin ka radius $\sqrt{y}$ hai, toh uska area $\pi [\sqrt{y}{]}^2=\pi y$ hai; coins ko bottom $y=0$ se rim $y=9$ tak thickness $dy$ ke saath stack karo.
3. Integrate karo. $V=\pi {\left[\frac{y^2}{2}\right]}_0^9=\pi \cdot \frac{81}{2}=\frac{81\pi }{2}{\text{cm}}^3.$ Yeh step kyun? $\int ydy=y^2/2$; rim height $y=9$ plug in karo ($81/2$ milta hai) aur bottom $y=0$ ($0$ milta hai). Units cm³ rehti hain kyunki humne cm² area ko cm height par integrate kiya.

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Example 8 — Cell H (exam twist: curves cross karte hain, integral split karo)

Forecast: do curves $x=1$ par cross karti hain. $(0,1)$ par, $x>x^2$; $(1,2)$ par, $x^2>x$. Toh outer curve har piece par alag hai — aapko crossing par integral split karna hoga aur har part par sahi $R$ choose karna hoga. Guess: do washer integrals, add kiye gaye.

Figure dono curves $[0,2]$ par, $x=1$ par crossing marked, aur do colours mein shading dikhata hai jo flag karta hai ki "top" curve change hoti hai.

1. Dhundho kahan ordering swap hoti hai. Set karo $x=x^2\Rightarrow x(1-x)=0\Rightarrow x=0$ ya $x=1$. Toh $x=1$ $[0,2]$ ke andar crossing hai. Yeh step kyun? $R$ aur $r$ iss baat se define hote hain ki kaun si curve axis se zyada door hai; yeh sirf wahan change ho sakta hai jahan curves milti hain. Pehle woh points dhundho.
2. Har piece par $R,r$ assign karo.
   • $[0,1]$ par: $x\ge x^2$, toh $R=x$, $r=x^2$.
   • $[1,2]$ par: $x^2\ge x$, toh $R=x^2$, $r=x$. Yeh step kyun? Outer = x-axis se zyada door = zyada height (dono curves yahan $\ge 0$ hain). Sure hone ke liye har piece par ek point test karo ($x=\frac{1}{2}$: $\frac{1}{2}>\frac{1}{4}$; $x=\frac{3}{2}$: $\frac{9}{4}>\frac{3}{2}$).
3. Do washer integrals, add kiye gaye. $V=\pi {\int }_0^1(x^2-x^4)dx+\pi {\int }_1^2(x^4-x^2)dx.$ Yeh step kyun? Har piece apna $R^2-r^2$ use karta hai (hamesha outer squared minus inner squared); galat $R$ ke saath ek single integral ek piece par ek nonsensical negative integrand dega.
4. Pehla piece evaluate karo (yeh Example 2 ka integrand hai): $\pi {\int }_0^1(x^2-x^4)dx=\pi {\left[\frac{x^3}{3}-\frac{x^5}{5}\right]}_0^1=\pi \left(\frac{1}{3}-\frac{1}{5}\right)=\frac{2\pi }{15}.$
5. Doosra piece evaluate karo. $\pi {\int }_1^2(x^4-x^2)dx=\pi {\left[\frac{x^5}{5}-\frac{x^3}{3}\right]}_1^2.$ $x=2$ par: $\frac{32}{5}-\frac{8}{3}=\frac{96-40}{15}=\frac{56}{15}$. $x=1$ par: $\frac{1}{5}-\frac{1}{3}=-\frac{2}{15}$. Subtract karo: $\pi \left(\frac{56}{15}-\left(-\frac{2}{15}\right)\right)=\pi \cdot \frac{58}{15}=\frac{58\pi }{15}.$ Yeh step kyun? Piece 1 jaisi hi antiderivative lekin roles swap ho gaye, toh sign flip hota hai; naye limits $1$ aur $2$ ke beech evaluate karo.
6. Pieces add karo. $V=\frac{2\pi }{15}+\frac{58\pi }{15}=\frac{60\pi }{15}=4\pi .$ Yeh step kyun? Total volume do sub-solids ka sum hai; common denominator addition ko clean banata hai.

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Example 9 — Cell I (axis region ke andar se guzarti hai)

Forecast: kyunki axis region se guzarti hai, top half aur bottom half same cylinder sweep karte hain — woh overlap karte hain, stack nahi hote. Sahi radius maximum distance hai, top-minus-bottom nahi. Ek plain cylinder expect karo.

Figure square ko axis ko cross karte hue aur woh single cylinder jo woh sweep karta hai dikhata hai — note karo koi hole nahi hai, kyunki region axis ko touch karta hai.

1. Trap spot karo. Far edge axis ke $1$ upar hai; near edge axis khud hai ($y=0$) kyunki region us tak pahunchta hai. Toh $R=1$, $r=0$ — ek disk, washer nahi. Yeh step kyun? Jab axis region ke andar ho, toh solid ek full disk hota hai jiska radius = kisi edge tak sabse badi distance; koi hole nahi hota.
2. Disk integral. $V=\pi {\int }_{-1}^1[1{]}^{2}dx=\pi {\int }_{-1}^{1}1dx.$ Yeh step kyun? Har coin ka radius $1$ (constant) hai, thickness $dx$, $x=-1$ se $x=1$ tak.
3. Integrate karo. $V=\pi [x{]}_{-1}^1=\pi (1-(-1))=2\pi .$ Yeh step kyun? $\int 1dx=x$; constant integrand evaluate karna sirf interval ki length measure karta hai, $1-(-1)=2$, toh $V=2\pi$.

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Recall

Recall Kaun sa cell kaun sa hai?

X-axis ke baare mein rotate karo, region axis par (Cell A) ::: Disk in $dx$, radius $f(x)$. Y-axis ke baare mein rotate karo (Cell C) ::: Disk/washer in $dy$, $x=g(y)$ solve karo. $y=k$ ke baare mein rotate karo (Cell D) ::: Radii ko $|f(x)-k|$ par shift karo; sabse door curve outer hai. $x=k$ ke baare mein rotate karo (Cell E) ::: $dy$ mein slice karo, radius horizontal distance $|x-k|$ hai. Do identical curves (Cell F) ::: Zero volume — koi region nahi. Curves interval ke andar cross karti hain (Cell H) ::: Har crossing par integral split karo; har piece par outer²−inner² use karo. Axis region se guzarti hai (Cell I) ::: Max distance ka disk, washer nahi.

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Connections

• Definite Integral as a Limit of Sums — har "coins sum karo, $\Delta x\to 0$ hone do" step yahi hai.
• Area Between Curves — Cells B, D, E, H ek area-between-curves picture se shuru hote hain.
• Shell Method — ek alternative jab $x=g(y)$ ke liye solve karna ugly ho (Cell C/E compare karo).
• Volume by Cross-Sections — disks aur washers circular cross-section case hain.
• u-substitution — shifted-axis integrands expand karne ke liye handy (Cells D, E).