Exercises — Surface area of revolution

(a) X-axis ke baare mein rotate karna matlab radius = point ki x-axis se doori = uski height $=y=\sqrt{r^2-x^2}$. (b) $\sqrt{r^2-x^2}=(r^2-x^2{)}^{1/2}$ par chain rule se: $y^{\prime }=\frac{1}{2}(r^2-x^2{)}^{-1/2}\cdot (-2x)=\frac{-x}{\sqrt{r^2-x^2}}.$ Chain rule kyun? Inside $r^2-x^2$ khud $x$ ka function hai, isliye hum outer power differentiate karte hain aur inner derivative $-2x$ se multiply karte hain.

$y^{\prime }=\frac{r}{h}$ — straight line ka ek fixed slope hota hai, toh haan, constant hai. $\sqrt{1+(y^{\prime }{)}^2}=\sqrt{1+\frac{r^2}{h^2}}=\sqrt{\frac{h^2+r^2}{h^2}}=\frac{\sqrt{h^2+r^2}}{h}.$ Geometrically iska matlab: ek straight generating line perfect cone banati hai, isliye slant har jagah same hota hai — frustum/cone geometry se match karta hai.

Y-axis ke baare mein rotate karna matlab points vertical line $x=0$ ke around circle karte hain. Radius = horizontal doori $=x=3$. Ek spin mein tay ki gayi doori $=2\pi r=2\pi (3)=6\pi$. Trap se bachao: height $7$ yahan radius se irrelevant hai — sirf axis tak ki doori matter karti hai.

Radius $=y=\frac{r}{h}x$; slant factor $=\frac{\sqrt{h^2+r^2}}{h}$ (1.2 se). $S={\int }_0^{h}2\pi \cdot \frac{r}{h}x\cdot \frac{\sqrt{h^2+r^2}}{h}dx=\frac{2\pi r\sqrt{h^2+r^2}}{h^2}{\int }_0^{h}xdx.$ ${\int }_0^{h}xdx=\frac{h^2}{2}$, toh $S=\frac{2\pi r\sqrt{h^2+r^2}}{h^2}\cdot \frac{h^2}{2}=\pi r\sqrt{h^2+r^2}=\pi rL.✓$

$y^{\prime }=3x^2$, toh $1+(y^{\prime }{)}^2=1+9x^4$. $S={\int }_0^{1}2\pi x^3\sqrt{1+9x^4}dx.$ Substitution kyun choose karein: kyunki outside factor $x^3$, inside $1+9x^4$ ke derivative ka (ek multiple) hai. Let $u=1+9x^4$, $du=36x^{3}dx\Rightarrow x^{3}dx=\frac{du}{36}$. Limits: $x=0\to u=1$; $x=1\to u=10$. $S=2\pi {\int }_1^{10}\sqrt{u}\frac{du}{36}=\frac{\pi }{18}\cdot \frac{2}{3}[u^{3/2}{]}_1^{10}=\frac{\pi }{27}\left(10^{3/2}-1\right).$ Numerically $\approx 3.563$.

$y=x^{1/2}$, $y^{\prime }=\frac{1}{2}{x}^{-1/2}$, toh $(y^{\prime }{)}^2=\frac{1}{4x}$ aur $1+(y^{\prime }{)}^2=\frac{4x+1}{4x}$. Phir $y\sqrt{1+(y^{\prime }{)}^2}=x^{1/2}\cdot \frac{\sqrt{4x+1}}{2\sqrt{x}}=\frac{\sqrt{4x+1}}{2}$. Kyun khoobsurat hai: $\sqrt{x}$ cancel ho jaata hai, ek clean root bachta hai. $S={\int }_0^{4}2\pi \cdot \frac{\sqrt{4x+1}}{2}dx=\pi {\int }_0^4\sqrt{4x+1}dx.$ Let $u=4x+1$, $du=4dx$; $x=0\to u=1$, $x=4\to u=17$: $S=\pi {\int }_1^{17}\sqrt{u}\frac{du}{4}=\frac{\pi }{4}\cdot \frac{2}{3}[u^{3/2}{]}_1^{17}=\frac{\pi }{6}\left(17^{3/2}-1\right)\approx \mathrm{36.18.}$

Axis = y-axis ⇒ radius = horizontal doori = $x$. Hum $y$ mein integrate kar sakte the ($x=\sqrt{y}$ use karke) lekin $x$ mein integrate karna zyada clean hai kyunki $y^{\prime }=2x$ simple hai aur radius literally $x$ hai. $S={\int }_0^{1}2\pi x\sqrt{1+(2x{)}^2}dx={\int }_0^{1}2\pi x\sqrt{1+4x^2}dx.$ Sub $u=1+4x^2$, $du=8xdx\Rightarrow xdx=\frac{du}{8}$; $x=0\to u=1$, $x=1\to u=5$: $S=2\pi {\int }_1^5\sqrt{u}\frac{du}{8}=\frac{\pi }{4}\cdot \frac{2}{3}[u^{3/2}{]}_1^5=\frac{\pi }{6}\left(5^{3/2}-1\right)\approx \mathrm{5.33.}$ Yeh parent note ke Example 3 se match karta hai. ✅

Curve $y$ ke terms mein $x$ diya gaya hai, aur hum y-axis ke baare mein rotate kar rahe hain (radius $=x$). $y$ mein integrate karna natural hai: limits $y$-limits hain aur $\frac{dx}{dy}$ easy hai. $\frac{dx}{dy}=3y^2,ds=\sqrt{1+(3y^2{)}^2}dy=\sqrt{1+9y^4}dy.$ Radius $=x=y^3$: $S={\int }_0^{1}2\pi y^3\sqrt{1+9y^4}dy.$ 2.2 jaisi hi integral shape! Sub $u=1+9y^4$, $du=36y^{3}dy$; $u:1\to 10$: $S=\frac{\pi }{27}\left(10^{3/2}-1\right)\approx \mathrm{3.563.}$ Insight: $y=x^3$ ko x-axis ke baare mein rotate karna (2.2) aur $x=y^3$ ko y-axis ke baare mein rotate karna mirror-images hain, isliye equal $S$ expected hai.

Yahan parametric $ds$ kyun? Yeh curve khud par loop karti hai (ek poora circle), isliye yeh ek single function $y=f(x)$ nahi hai — har $x$ ke liye do $y$ hain. Page ke top se general element $ds=\sqrt{(dx{)}^2+(dy{)}^2}$ hume bachaata hai: $dt$ se divide aur multiply karke parametric form milta hai $ds=\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt.$ Do derivatives compute karo: $\frac{dx}{dt}=-\sin t,\frac{dy}{dt}=\cos t,\sqrt{(x^{\prime }{)}^2+(y^{\prime }{)}^2}=\sqrt{{\sin }^{2}t+{\cos }^{2}t}=1.$ Yeh $1$ kyun hai: point unit circle par unit speed se move karta hai, isliye $ds=1dt$. X-axis se radius $=y=2+\sin t$: $S={\int }_0^{2\pi }2\pi (2+\sin t)\cdot 1dt=2\pi {\left[2t-\cos t\right]}_0^{2\pi }=2\pi (4\pi )=8{\pi }^2.$ (${\int }_0^{2\pi }\sin tdt=0$.) Toh $S=8{\pi }^2\approx 78.96$. Pappus se cross-check: $S=2\pi \bar{r}L$ jahan $L=2\pi$ (poore unit circle ki arc length, $2\pi \cdot 1$) aur $\bar{r}$ curve ke centroid ki height hai. Ek poore circle ka centroid uska geometric centre hota hai, jise humne height $2$ par rakha hai (exactly yahi matlab hai "$y=2+\sin t$, height $2$ par centred ek circle" ka — $2+\sin t$ ko ek poore turn par average karne par $2$ aata hai, kyunki $\sin t$ ka average $0$ hota hai). Toh $\bar{r}=2$ aur $S=2\pi (2)(2\pi )=8{\pi }^2$. ✅

Radius = line $y=-1$ tak ki doori, jo $|y-(-1)|=y+1=x+1$ hai (yahan $y=x\ge 0$, toh positive). $y^{\prime }=1\Rightarrow ds=\sqrt{1+1}dx=\sqrt{2}dx$. $S={\int }_0^{2}2\pi (x+1)\sqrt{2}dx=2\sqrt{2}\pi {\int }_0^2(x+1)dx.$ ${\int }_0^2(x+1)dx=[\frac{x^2}{2}+x{]}_0^2=2+2=4$. $S=2\sqrt{2}\pi \cdot 4=8\sqrt{2}\pi \approx \mathrm{35.54.}$

Parent note ke sphere derivation se, poora integrand ek constant par collapse ho jaata hai: $y\sqrt{1+(y^{\prime }{)}^2}=\sqrt{r^2-x^2}\cdot \frac{r}{\sqrt{r^2-x^2}}=r.$ Kyun collapse hota hai: height poles ke paas bilkul utni hi teezi se shrink hoti hai jitni teezi se slant grow karta hai, toh unka product flat ho jaata hai. $S={\int }_a^{b}2\pi rdx=2\pi r(b-a).$ Stunning conclusion: sphere par kahin bhi kati gayi equal width ki do bands ki surface area equal hoti hai — equator ke paas (chhoti, moti) ya pole ke paas (lambi, patli). $4\pi r^2$ recover karne ke liye $a=-r,b=r$ set karo.

2.3 ke antiderivative se, $S(\epsilon )=\frac{\pi }{6}\left((4\epsilon +1{)}^{3/2}-1\right)$ over $[0,\epsilon ]$. $\epsilon \to 0^{+}$ par: $(4\epsilon +1{)}^{3/2}\to 1$, toh $S\to \frac{\pi }{6}(1-1)=0$. ✅ Interpretation: origin ke paas vanishingly short curve ek vanishingly small surface sweep karta hai — formula degenerate end par sahi behave karta hai. Note karo curve $x=0$ par vertical hai (slope $\to \infty$), phir bhi kyunki radius $y=\sqrt{x}\to 0$ bhi hai, product finite rehta hai aur area phir bhi $0$ jaata hai.

Bug: radius axis tak ki doori honi chahiye, jo $|y|=|x^3|$ hai, kabhi negative nahi. Area negative nahi ho sakta; signed integrals cancel karne waali symmetry yahan illegal hai. Correct positive contributions mein split karo. Symmetry se dono halves ki area equal hai: $S=2{\int }_0^{1}2\pi x^3\sqrt{1+9x^4}dx.$ 2.2 se, ${\int }_0^{1}2\pi x^3\sqrt{1+9x^4}dx=\frac{\pi }{27}(10^{3/2}-1)$, toh $S=2\cdot \frac{\pi }{27}\left(10^{3/2}-1\right)=\frac{2\pi }{27}\left(10^{3/2}-1\right)\approx \mathrm{7.125.}$ Har point real, positive skin contribute karta hai — surface origin ke baare mein ek symmetric "double-horn" hai.

"Two-pie radius, slide along the rope." Radius = axis tak ki doori (shifts aur signs dhyaan mein rakho), rope = arc length $ds$. $ds$ ko uss variable se match karo jisme tumhare limits rehte hain. Distances kabhi negative nahi hoti.