Worked examples — Surface area of revolution

4.2.17 · D3 · Maths › Calculus II — Integration › Surface area of revolution

Pehli line se pehle, teen tools ki reminder jo hum use karte hain, simple words mein:

• $ds$ = curve ke tiny piece ki asli slanted length (dekho Arc length). Yeh us "rope" ki length hai jo wrap karti hai.
• $2\pi r$ = woh circumference jo ek point spin karte waqt trace karta hai; $r$ woh distance hai jitni door woh apne pole se baitha hai jiske around spin kar raha hai.
• substitution (Integration by substitution) = ek renaming trick jab square root ke bahar ka stuff, andar ke stuff ka (ek multiple mein) derivative ho.

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Scenario matrix

Har surface-of-revolution problem in cells mein se exactly ek mein hota hai. Neeche ke examples us cell ke saath tagged hain jo woh hit karte hain.

#	Case class	Kya alag hai	Covered by
A	$y=f(x)$ ko x-axis ke around rotate karo	radius $=y$	Ex 1
B	y-axis ke around rotate karo	radius $=x$, dhyan se integrate karo	Ex 2
C	Shifted line $y=k$ ke around rotate karo	radius $=|y-k|$ (sign matter karta hai!)	Ex 3
D	Parametric curve	$ds$ from $x^{\prime }(t),y^{\prime }(t)$	Ex 4
E	Degenerate: horizontal segment	$dy/dx=0$ → cylinder, ek sanity anchor	Ex 5
F	Degenerate: point on the axis (radius $\to 0$)	band zero ho jaata hai — limiting behaviour	Ex 6
G	Word problem (real object, units)	story → curve → integral translate karo	Ex 7
H	Exam twist: curve axis cross karta hai (radius sign change)	$|y|$ use karna zaroori hai, integral split karo	Ex 8
I	Pappus's theorem se cross-check	$S=2\pi \bar{r}L$ shortcut	Ex 9

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Example 1 — X-axis ke around rotate karo (Cell A)

Neeche ka figure setup dikhata hai: magenta curve $y=\sqrt{x}$, radius (x-axis tak height $y$) dene wala violet arrow, aur ek chhota orange segment — ek tiny slanted piece $ds$ jo wrap karta hai. Notice karo ki orange piece apni horizontal shadow se lamba hai: yahi wajah hai hum $ds$ use karte hain, $dx$ nahi.

Step 1 — Radius identify karo. Curve par har point $(x,y)$ x-axis se height $y$ par baitha hai. Uss axis ke around spin karne par, woh radius $y=\sqrt{x}$ ka circle banata hai (figure mein violet arrow). Yeh step kyun? Radius "distance to the axis" hai — yahan axis x-axis hai, toh vertical height $y$ exactly woh distance hai.

Step 2 — $ds$ banao. Humein slope chahiye: $\frac{dy}{dx}=\frac{1}{2\sqrt{x}}$, toh $ds=\sqrt{1+\frac{1}{4x}}dx.$ Yeh step kyun? Band tilted curve (orange segment) ke along wrap karta hai, uski flat shadow ke along nahi — isliye arc length chahiye, aur $ds$ ko slope chahiye.

Step 3 — Integral assemble karo. $S={\int }_0^{4}2\pi \sqrt{x}\sqrt{1+\frac{1}{4x}}dx.$ Yeh step kyun? Har band $2\pi \cdot (\text{radius})\cdot ds$ contribute karta hai; $x=0$ se $x=4$ tak sab bands ko sum (integrate) karne se poori surface milti hai.

Step 4 — Integrand simplify karo. $\sqrt{x}$ andar kheencho: $\sqrt{x}\sqrt{1+\frac{1}{4x}}=\sqrt{x+\frac{1}{4}}.$ Yeh step kyun? $x\cdot (1+\frac{1}{4x})=x+\frac{1}{4}$ — awkward $1/x$ gayab ho jaata hai, ek plain finite root bachta hai jise hum $x=0$ tak integrate kar sakte hain.

Step 5 — Integrate karo. $u=x+\frac{1}{4}$, $du=dx$ ke saath: $S=2\pi {\int }_0^4\sqrt{x+\frac{1}{4}}dx=2\pi \cdot \frac{2}{3}{\left[(x+\frac{1}{4}{)}^{3/2}\right]}_0^4=\frac{4\pi }{3}\left[{\left(\frac{17}{4}\right)}^{3/2}-{\left(\frac{1}{4}\right)}^{3/2}\right].$ Numerically $S\approx 36.18$.

Verify: ${\left(\frac{17}{4}\right)}^{3/2}=8.755$, ${\left(\frac{1}{4}\right)}^{3/2}=0.125$; difference $8.630$; $\times \frac{4\pi }{3}\approx 36.18$. Positive aur finite — theek hai, kyunki simplification ke baad integrand $[0,4]$ par bounded hai. ✅

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Example 2 — Y-axis ke around rotate karo (Cell B)

Step 1 — Radius ab $x$ hai. $(x,y)$ se y-axis tak distance horizontal distance $x$ hai. Yeh step kyun? "Distance to the axis" ne axis change kiya; ab horizontally measure karo.

Step 2 — $x$ mein integrate karte raho. Hum abhi bhi $x$ ko apna variable use kar sakte hain; $ds=\sqrt{1+(2x{)}^2}dx$ kyunki $dy/dx=2x$. Yeh step kyun? $ds$ ek length hai — isse koi fark nahi ki hum kis axis ke around spin kar rahe hain; sirf radius factor badalta hai.

Step 3 — Integral. $S={\int }_0^{1}2\pi x\sqrt{1+4x^2}dx.$ Yeh step kyun? Har band $2\pi \cdot (\text{radius }x)\cdot ds$ hai; $x\in [0,1]$ par integrate karne se sab bands stack ho kar poori surface banti hai.

Step 4 — Substitute $u=1+4x^2$, $du=8xdx$: $S=\frac{2\pi }{8}{\int }_1^5\sqrt{u}du=\frac{\pi }{4}\cdot \frac{2}{3}{\left[u^{3/2}\right]}_1^5=\frac{\pi }{6}\left(5^{3/2}-1\right)\approx \mathrm{5.33.}$ Substitution kyun? Bahar ka $x$ exactly $\frac{1}{8}du$ hai — ek gift.

Verify: $5^{3/2}=11.180$, minus $1=10.180$, $\times \frac{\pi }{6}\approx 5.33$. Parent note ke Example 3 se match karta hai. ✅

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Example 3 — Shifted line $y=k$ ke around rotate karo (Cell C)

Figure magenta segment $y=1$, uske upar dashed navy axis $y=3$, aur dono ke beech gap measure karte violet arrows dikhata hai. Har arrow ki length $2$ hai — woh constant gap radius hai. Kyunki strip flat hai aur gap kabhi nahi badlta, swept surface ek plain cylinder hai.

Step 1 — Radius $|y-k|$ hai. Point height $1$ par baitha hai, axis height $3$ par, isliye radius $=|1-3|=2$ (violet arrows). Yeh step kyun? Height $k$ par horizontal line tak distance vertical gap $|y-k|$ hai — absolute value isse positive rakhta hai chahe $y<k$ ho.

Step 2 — Flat line ke liye $ds$. $dy/dx=0$, isliye $ds=\sqrt{1+0}dx=dx$. Yeh step kyun? Horizontal segment tilted nahi hai, isliye uski slanted length uski flat length ke equal hai.

Step 3 — Integral (yeh cylinder hai!). $S={\int }_0^{3}2\pi \cdot 2\cdot dx=4\pi \cdot 3=12\pi \approx \mathrm{37.70.}$ Yeh step kyun? Constant radius $2$ aur $ds=dx$ ke saath, har band identical $2\pi (2)dx$ hai; $x=0$ se $3$ tak sum karna bas length $3$ se multiply karna hai — literally ek cylinder ko rectangle mein unroll karna.

Verify: Radius $2$, length $3$ ke cylinder ka lateral area $2\pi rL=2\pi (2)(3)=12\pi$. Bilkul wahi mila. ✅ Agar hum absolute value bhool jaate toh radius $=1-3=-2$ likhte aur negative area milta — nonsense; $|\cdot |$ humein bachata hai.

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Example 4 — Parametric curve (Cell D)

Step 1 — Radius $=y=t^2$. X-axis ke around rotate karte hue, axis tak distance height $y$ hai. Yeh step kyun? Same rule; bas $y$ ko $t$ ke through express karo.

Step 2 — Parametric $ds$. $x^{\prime }(t)=3t^2$, $y^{\prime }(t)=2t$, isliye $ds=\sqrt{(3t^2{)}^2+(2t{)}^2}dt=\sqrt{9t^4+4t^2}dt=t\sqrt{9t^2+4}dt(t\ge 0).$ Yeh step kyun? Parametric curve ke liye arc length $\sqrt{(x^{\prime }{)}^2+(y^{\prime }{)}^2}dt$ hai — dono coordinates move karte hain toh Pythagoras. Hum $t^2$ root se bahar factor karte hain (safe hai kyunki $t\ge 0$ se $\sqrt{t^2}=t$).

Step 3 — Integral. $S={\int }_0^{1}2\pi t^2\cdot t\sqrt{9t^2+4}dt=2\pi {\int }_0^1{t}^3\sqrt{9t^2+4}dt.$ Yeh step kyun? Radius $\times ds\times 2\pi$, ab sab kuch single variable $t$ mein likha hai aur limits $0$ se $1$ tak.

Step 4 — Substitute $u=9t^2+4$, $du=18tdt$, aur $t^2=\frac{u-4}{9}$: $S=2\pi {\int }_4^{13}\frac{u-4}{9}\sqrt{u}\cdot \frac{du}{18}=\frac{2\pi }{162}{\int }_4^{13}\left(u^{3/2}-4u^{1/2}\right)du.$ Yeh step kyun? $t^{3}dt=t^2\cdot tdt$, aur $tdt=\frac{1}{18}du$ jabki $t^2$ $u$ ka function ban jaata hai — poori cheez ab $\sqrt{u}$ mein polynomial hai.

Step 5 — Evaluate karo. ${\int }_4^{13}\left(u^{3/2}-4u^{1/2}\right)du={\left[\frac{2}{5}{u}^{5/2}-\frac{8}{3}{u}^{3/2}\right]}_4^{13}.$ $u=13$ par: $\frac{2}{5}(13^{5/2})-\frac{8}{3}(13^{3/2})=243.86-124.98=118.88$. $u=4$ par: $\frac{2}{5}(4^{5/2})-\frac{8}{3}(4^{3/2})=\frac{2}{5}(32)-\frac{8}{3}(8)=12.80-21.33=-8.53$. Bracket $=118.88-(-8.53)=127.41$, isliye $S=\frac{2\pi }{162}\cdot 127.41\approx \mathrm{4.94.}$

Verify: $\frac{2\pi }{162}\approx 0.03879$; $0.03879\times 127.41\approx 4.94$. Itne chhote curve ke liye small positive number — sensible hai. ✅

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Example 5 — Degenerate: horizontal segment → cylinder (Cell E)

Step 1 — Radius $=5$ (constant), $ds=dx$ kyunki slope $0$ hai. Yeh step kyun? Koi tilt nahi ⇒ arc length horizontal length ke equal hai; yeh sabse simple possible case hai, hamara "kya machine abhi bhi kaam karti hai?" test.

Step 2 — Integral. $S={\int }_0^{2}2\pi (5)dx=10\pi \cdot 2=20\pi \approx \mathrm{62.83.}$ Yeh step kyun? Har band identical $2\pi (5)dx$ hai; integrate karna bas length $2$ se multiply karna hai — yahi hai cylinder ko height $2\pi r$ aur width $L$ ke rectangle mein unroll karne ki definition.

Verify: Cylinder ka lateral area $=2\pi rL=2\pi (5)(2)=20\pi$. Jab koi tilt nahi hota toh calculus grade-school geometry tak reduce ho jaata hai — exactly jaisa hona chahiye. ✅

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Example 6 — Degenerate: radius zero ho jaata hai (Cell F)

Figure mein, magenta line $y=x$ generating curve hai aur neeche dotted mirror dikhata hai woh surface jo spin se banti hai — ek cone. Origin par orange dot tip hai: woh axis par baitha hai, isliye uska radius (violet arrow $x=0.8$ par) wahan zero ho jaata hai. Dekho ki arrow tip ki taraf left jaate waqt chhota hota jaata hai.

Step 1 — Radius $=y=x$, jo $x\to 0$ ke saath $\to 0$ ho jaata hai. Tip par woh jo circle trace karta hai uska radius $0$ hai — ek single point (orange dot). Yeh step kyun? Yeh limiting/degenerate check hai: jahan surface ek point tak pinch hoti hai wahan kya hota hai?

Step 2 — $ds=\sqrt{1+1}dx=\sqrt{2}dx$ (slope $=1$).

Step 3 — Integral. $S={\int }_0^{1}2\pi x\sqrt{2}dx=2\sqrt{2}\pi {\left[\frac{x^2}{2}\right]}_0^1=\sqrt{2}\pi \approx \mathrm{4.443.}$ Yeh step kyun? Integrand $2\pi x\sqrt{2}$ $x=0$ par $0$ hai — pinch point kuch add nahi karta, aur koi blow-up nahi hai. Integral cleanly converge karta hai.

Verify: Yeh base radius $r=1$, height $h=1$, slant $L=\sqrt{1^2+1^2}=\sqrt{2}$ wala cone hai. Formula $\pi rL=\pi (1)\sqrt{2}=\sqrt{2}\pi \approx 4.443$. ✅ Zero-radius tip harmless hai.

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Example 7 — Word problem: ek lampshade (Cell G)

Step 1 — Radius $=y=1/x$. X-axis tak distance height hai. Yeh step kyun? Story translate karo: "horizontal axis ke around spin" ⇒ radius vertical height hai.

Step 2 — Slope aur $ds$. $\frac{dy}{dx}=-\frac{1}{x^2}$, isliye $ds=\sqrt{1+\frac{1}{x^4}}dx.$

Step 3 — Integral. $S={\int }_1^{2}2\pi \cdot \frac{1}{x}\sqrt{1+\frac{1}{x^4}}dx=2\pi {\int }_1^2\frac{\sqrt{x^4+1}}{x^3}dx.$ Yeh step kyun? $\frac{1}{x}$ aur root combine karo: $\frac{1}{x}\sqrt{1+\frac{1}{x^4}}=\frac{1}{x}\cdot \frac{\sqrt{x^4+1}}{x^2}=\frac{\sqrt{x^4+1}}{x^3}$.

Step 4 — Simpson's rule se numerically evaluate karo. Integrand $g(x)=\frac{\sqrt{x^4+1}}{x^3}$ ka koi clean elementary antiderivative nahi hai, isliye approximate karte hain. $[1,2]$ ko $4$ equal strips mein tod lo ($h=0.25$) aur sample karo:

$g(1)=1.41421$, $g(1.25)=0.83328$, $g(1.5)=0.55372$, $g(1.75)=0.39725$, $g(2)=0.30208$.

Simpson's rule ${\int }_1^{2}gdx\approx \frac{h}{3}[g_0+4g_1+2g_2+4g_3+g_4]$: $\frac{0.25}{3}[1.41421+4(0.83328)+2(0.55372)+4(0.39725)+0.30208]=\frac{0.25}{3}(9.59775)\approx \mathrm{0.7998.}$ Simpson's rule kyun? Yeh sample points ke through parabolas fit karta hai — smooth curve ke liye trapezoids se kahin zyada accurate. Yahan yeh true value se $0.001$ ke andar aata hai (finer grid answer hardly move karta hai), aur story ko sirf kuch significant figures of cm² chahiye.

Step 5 — $2\pi$ se multiply karo. $S=2\pi (0.7998)\approx 5.03{\text{cm}}^2.$

Verify: Units: radius (cm) × $ds$ (cm) × $2\pi$ (dimensionless) ⇒ cm² ✅. Magnitude: shade ka average radius $\approx 0.7$ cm, slanted curve ki length $\approx 1.13$ cm, roughly $2\pi (0.7)(1.13)\approx 4.97$ — $5.03$ ke same ballpark mein. ✅

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Example 8 — Exam twist: curve axis cross karta hai (Cell H)

Step 1 — Radius $=|y|=|x^3|$. Axis ke neeche ka point abhi bhi positive distance par hai. Yeh step kyun? Agar hum naively $y=x^3$ use karte, toh $x<0$ wala part negative contribution deta — lekin area cancel nahi ho sakta. Humein $|y|$ use karna hi hoga.

Step 2 — Crossing point $x=0$ par split karo. Curve exactly $x=0$ par axis cross karti hai, jahan $|x^3|$ apna formula change karta hai. $[-1,0]$ par $|x^3|=-x^3$ likhte hain (kyunki wahan $x^3\le 0$ hai) aur $[0,1]$ par $|x^3|=x^3$: $S={\int }_{-1}^{0}2\pi (-x^3)\sqrt{1+9x^4}dx+{\int }_0^{1}2\pi x^3\sqrt{1+9x^4}dx.$ Yeh step kyun? Jahan bhi $y$ sign flip kare, integral wahan split karo taaki $|y|$ har piece par ek clean formula ho — warna antiderivative galat hogi.

Step 3 — Symmetry use karke combine karo. $ds=\sqrt{1+9x^4}dx$ ke saath, $|x^3|$ aur $\sqrt{1+9x^4}$ dono $x$ mein even hain, isliye dono pieces equal hain: $S=2{\int }_0^{1}2\pi x^3\sqrt{1+9x^4}dx.$ Yeh step kyun? Equal halves se sirf $[0,1]$ compute karke double kar sakte hain — yeh shortcut hamesha haath se split karne se behtar hai.

Step 4 — Substitute $u=1+9x^4$, $du=36x^{3}dx$: $S=4\pi \cdot \frac{1}{36}{\int }_1^{10}\sqrt{u}du=\frac{\pi }{9}\cdot \frac{2}{3}{\left[u^{3/2}\right]}_1^{10}=\frac{2\pi }{27}\left(10^{3/2}-1\right)\approx \mathrm{7.13.}$ Yeh step kyun? Bahar ka factor $x^{3}dx$ exactly $\frac{1}{36}du$ hai — substitution poori cheez ko $[1,10]$ par clean $\sqrt{u}$ integral mein badal deta hai.

Step 5 — Answer state karo. $S=\frac{2\pi }{27}\left(10^{3/2}-1\right)\approx 7.13$.

Verify: $10^{3/2}=31.623$, minus $1=30.623$; $\frac{2\pi }{27}\approx 0.23271$; $0.23271\times 30.623\approx 7.13$. Agar hum $|y|$ use nahi karte, toh $[-1,1]$ par $2\pi x^3\sqrt{1+9x^4}$ ka naive signed integral $0$ hota (odd function) — obviously ek real surface ke liye galat. $|\cdot |$ fix (aur $x=0$ par split) essential hai. ✅

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Example 9 — Pappus se cross-check (Cell I)

Step 1 — Length $L$. $(0,1)$ se $(3,5)$ tak: $L=\sqrt{3^2+4^2}=5$. Yeh step kyun? Pappus ko generating curve ki arc length chahiye; yahan yeh plain segment hai.

Step 2 — Centroid height $\bar{r}$. Segment ka midpoint $\left(\frac{0+3}{2},\frac{1+5}{2}\right)=(1.5,3)$ hai, isliye $\bar{r}=3$. Yeh step kyun? Uniform straight segment ke liye centroid midpoint hota hai; x-axis tak distance uski $y$-value hai.

Step 3 — Pappus. $S=2\pi \bar{r}L=2\pi (3)(5)=30\pi \approx \mathrm{94.25.}$ Yeh step kyun? Pappus ka shortcut kehta hai swept surface area, arc length aur ek full turn mein uske centroid ki travel ki distance $2\pi \bar{r}$ ka product hai — koi integral nahi chahiye.

Step 4 — Integral se confirm karo. Line $y=1+\frac{4}{3}x$ hai $[0,3]$ par, slope $\frac{4}{3}$, isliye $ds=\sqrt{1+\frac{16}{9}}dx=\frac{5}{3}dx$: $S={\int }_0^{3}2\pi \left(1+\frac{4}{3}x\right)\frac{5}{3}dx=\frac{10\pi }{3}{\left[x+\frac{2}{3}{x}^2\right]}_0^3=\frac{10\pi }{3}(3+6)=30\pi .$ Yeh step kyun? Do independent methods agree karen — yeh sabse strong sanity check hai.

Verify: Dono $30\pi \approx 94.25$ dete hain. Yeh radii $1$ aur $5$, slant $5$ wala frustum hai: parent ka frustum formula $\pi (r_1+r_2)\ell =\pi (1+5)(5)=30\pi$ bhi — triple agreement. ✅

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Recall Kaun sa cell kaun sa hai? (self-test)

Har situation ko uske matrix cell se match karo. $x=g(y)$ ko y-axis ke around rotate karna ::: Cell B (radius $=x$, lekin $y$ mein integrate karo) Ek curve x-axis ke neeche jaati hai, uske around rotate karna ::: Cell H ($|y|$ use karo, crossings par split karo) $y=2$ ko $y=7$ ke around rotate karna ::: Cell C (radius $=|2-7|=5$) $(x(t),y(t))$ ke roop mein di gayi curve ::: Cell D (parametric $ds$) Generating curve ka woh point jo axis ko touch karta hai ::: Cell F (radius $\to 0$, kuch contribute nahi karta)

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Connections

• Surface area of revolution — parent formula jise har example apply karta hai.
• Arc length — $ds$ factor, yahan identical hai.
• Integration by substitution — Ex 1, 2, 4, 8 mein use hua.
• Parametric curves — Ex 4 ka arc-length form.
• Frustum and cone geometry — Ex 5, 6, 9 ke liye sanity anchor.
• Pappus's theorem — Ex 9 mein shortcut cross-check.
• Volume of revolution (disk & shell) — same "distance to axis" idea, $r$ ki alag power.