Exercises — Volume of revolution — disk method, washer method

KYA ho raha hai: region x-axis par baith raha hai (iska bottom edge $y=0$ hai, yaani axis khud). Jab yeh spin karta hai, har slice ek full solid coin hai — koi hole nahi. Toh yeh ek disk hai.

KYUN disk: hole tab hi banta hai jab region axis se door hota hai. Yahan region axis ko touch karta hai, toh coin ka centre filled hai.

Set-up: radius $=$ distance from axis to curve $=f(x)=x^2$. Thickness $=dx$. Toh $V=\pi {\int }_0^2(x^2{)}^{2}dx=\pi {\int }_0^2{x}^{4}dx.$ (Hum set-up par ruk jaate hain — sirf recognition.)

KYA: do curves region ko bound karti hain — ek upper line $y=4$ aur ek lower parabola $y=x^2$. $[0,2]$ par poora region axis se upar hai $y=0$ se ek gap ke saath... lekin jo gap matter karta hai woh region aur axis ke beech wala hai. Kyunki lower edge $y=x^2$ axis se upar hai (sirf $x=0$ ko chhod ke), spinning se ek hole ban jaata hai. Yeh ek washer hai.

KYUN: axis se door wali curve $y=4$ hai (bada $y$ = door), toh $R=4$. Paas wali curve $y=x^2$ hai, toh $r=x^2$.

Set-up: $V=\pi {\int }_0^2(4^2-(x^2{)}^2)dx=\pi {\int }_0^2(16-x^4)dx.$

$V=\pi {\int }_0^2{x}^{4}dx=\pi {\left[\frac{x^5}{5}\right]}_0^2=\pi \cdot \frac{32}{5}=\frac{32\pi }{5}.$ KYA kiya humne: power rule $\int x^{n}dx=\frac{x^{n+1}}{n+1}$ use karke $x^4$ integrate kiya, phir limits plug in ki. KYUN: definite integral hai hi coin-sum ka limit jab thickness $\to 0$ ho jaati hai (Definite Integral as a Limit of Sums). $\boxed{V=\frac{32\pi }{5}}$

KYUN disk: region axis par rest karta hai → solid coins. $V=\pi {\int }_1^3{\left(\frac{1}{x}\right)}^{2}dx=\pi {\int }_1^3{x}^{-2}dx.$ WHAT the square did: ${\left(\frac{1}{x}\right)}^2=x^{-2}$ — squaring the radius turned the fraction into a clean power we can integrate. $=\pi {\left[\frac{x^{-1}}{-1}\right]}_1^3=\pi {\left[-\frac{1}{x}\right]}_1^3=\pi \left(-\frac{1}{3}+1\right)=\frac{2\pi }{3}.$ $\boxed{V=\frac{2\pi }{3}}$

$V=\pi {\int }_0^2(16-x^4)dx=\pi {\left[16x-\frac{x^5}{5}\right]}_0^2.$ Compute: $16(2)=32$ aur $\frac{2^5}{5}=\frac{32}{5}$, toh $V=\pi \left(32-\frac{32}{5}\right)=\pi \cdot \frac{160-32}{5}=\frac{128\pi }{5}.$ $\boxed{V=\frac{128\pi }{5}}$

KYA decide karna hai pehle: $(0,1)$ par kaun si curve axis se door hai? $x=\frac{1}{4}$ test karo: $\sqrt{\frac{1}{4}}=\frac{1}{2}$ lekin $\frac{1}{4}=0.25$. Toh $\sqrt{x}>x$ yahan — root upar hai. Isliye $R=\sqrt{x}$ (outer) aur $r=x$ (inner).

KYUN yeh matter karta hai: agar inhe ulta likh do toh aap ek negative integrand integrate karoge aur negative "volume" milega. Door wali curve hi $R$ honi chahiye. $V=\pi {\int }_0^1((\sqrt{x}{)}^2-x^2)dx=\pi {\int }_0^1(x-x^2)dx.$ $=\pi {\left[\frac{x^2}{2}-\frac{x^3}{3}\right]}_0^1=\pi \left(\frac{1}{2}-\frac{1}{3}\right)=\pi \cdot \frac{1}{6}=\frac{\pi }{6}.$ $\boxed{V=\frac{\pi }{6}}$

KYA badla: axis ab $y=0$ nahi hai; yeh line $y=-1$ hai (figure mein dekho, region ke neeche dashed line). Radius woh doori hai us line se curve tak.

KYUN re-measure karna padega: radius ka matlab hamesha "axis-se-curve ki doori" hota hai. $y=-1$ se upar $y=x$ tak ki doori hai $x-(-1)=x+1$. Kyunki $[0,3]$ par $x+1>0$ hai, absolute value ki zaroorat nahi.

Kya hole hai? Region ka bottom $y=0$ hai, jo axis $y=-1$ se $1$ ki doori upar hai — toh region axis ko touch nahi karta. Iska matlab hai ek washer, inner radius $r=0-(-1)=1$ ke saath. $V=\pi {\int }_0^3((x+1{)}^2-1^2)dx=\pi {\int }_0^3(x^2+2x+1-1)dx=\pi {\int }_0^3(x^2+2x)dx.$ $=\pi {\left[\frac{x^3}{3}+x^2\right]}_0^3=\pi \left(\frac{27}{3}+9\right)=\pi (9+9)=18\pi .$ $\boxed{V=18\pi }$

STEP 1 — kahan milte hain? $2x=x^2$ set karo $\Rightarrow x^2-2x=0\Rightarrow x(x-2)=0$, toh $x=0$ aur $x=2$. Yahi limits hain.

STEP 2 — kaun outer hai? $x=1$ test karo: $y=2x$ gives $2$, $y=x^2$ gives $1$. Line upar hai: $R=2x$, $r=x^2$. $V=\pi {\int }_0^2((2x{)}^2-(x^2{)}^2)dx=\pi {\int }_0^2(4x^2-x^4)dx.$ $=\pi {\left[\frac{4x^3}{3}-\frac{x^5}{5}\right]}_0^2=\pi \left(\frac{32}{3}-\frac{32}{5}\right).$ $=\pi \cdot 32\left(\frac{1}{3}-\frac{1}{5}\right)=\pi \cdot 32\cdot \frac{2}{15}=\frac{64\pi }{15}.$ $\boxed{V=\frac{64\pi }{15}}$

KYUN $y$ par switch karein: hum y-axis ke baare mein spin karte hain, toh slices horizontal coins hain jinka thickness $dy$ hai; unka radius ek horizontal doori hai, $x$ mein measured. Toh humein $x$ ko $y$ ke function ke roop mein likhna hoga.

STEP 1 — invert karo: $y=x^2\Rightarrow x=\sqrt{y}$ (first quadrant, $x\ge 0$). Har coin ka radius $x=\sqrt{y}$ hai. Jaise $y$, $0$ (bottom) se $4$ (top line) tak run karta hai, region ek solid sweep karta hai — yeh y-axis ko touch karta hai, toh disk. $V=\pi {\int }_0^4(\sqrt{y}{)}^{2}dy=\pi {\int }_0^{4}ydy=\pi {\left[\frac{y^2}{2}\right]}_0^4=\pi \cdot 8=8\pi .$ $\boxed{V=8\pi }$

Shell Method se sanity check (optional cross-check): vertical shells ke roop mein, radius $x$, height $4-x^2$, $V=2\pi {\int }_0^{2}x(4-x^2)dx=2\pi [2x^2-\frac{x^4}{4}{]}_0^2=2\pi (8-4)=8\pi$. ✓ Do methods, same number — confidence.

KYA solid hai: ek semicircle ko uske diameter ke baare mein spin karna radius $R$ ki ek full ball sweep karta hai. Har slice ek solid disk hai (region $x=\pm R$ par axis ko touch karta hai).

Ek slice ka radius: $f(x)=\sqrt{R^2-x^2}$, toh $[f(x){]}^2=R^2-x^2$ — square karne se square root cleanly erase ho jaata hai (yahi wajah hai ki disk-method problems circles se pyaar karte hain). $V=\pi {\int }_{-R}^R(R^2-x^2)dx.$

Symmetry use karo: integrand even hai ($x$ aur $-x$ par same), toh $0\to R$ integrate karo aur double karo: $V=2\pi {\int }_0^R(R^2-x^2)dx=2\pi {\left[R^{2}x-\frac{x^3}{3}\right]}_0^R.$ $=2\pi \left(R^3-\frac{R^3}{3}\right)=2\pi \cdot \frac{2R^3}{3}=\frac{4}{3}\pi R^3.$ $\boxed{V=\frac{4}{3}\pi R^3}$ KYUN yeh finish line hai: answer ek formula in $R$ hai, har sphere ke liye valid — tumne sirf coin-slicing idea use karke ek geometry fact re-derive kar di.

STEP 1 — axis $y=2$ se distances. $y=2$ se door wali curve woh hai jo neeche hai, kyunki neeche $y$ = $y=2$ se zyada doori. $[-1,1]$ par compare karo: $y=x^2$ ki range $0$ se $1$ hai, aur $y=1$ top edge hai. Parabola $x^2\le 1$ line se neeche hai, toh parabola $y=2$ se door hai: outer radius $R=2-x^2$; inner radius $r=2-1=1$. $V=\pi {\int }_{-1}^1((2-x^2{)}^2-1^2)dx.$

STEP 2 — expand karo. $(2-x^2{)}^2=4-4x^2+x^4$, toh integrand hai $4-4x^2+x^4-1=3-4x^2+x^4$. $V=\pi {\int }_{-1}^1(3-4x^2+x^4)dx.$

STEP 3 — even integrand, $0\to 1$ double karo: $V=2\pi {\int }_0^1(3-4x^2+x^4)dx=2\pi {\left[3x-\frac{4x^3}{3}+\frac{x^5}{5}\right]}_0^1.$ $=2\pi \left(3-\frac{4}{3}+\frac{1}{5}\right).$ Common denominator $15$: $3=\frac{45}{15}$, $\frac{4}{3}=\frac{20}{15}$, $\frac{1}{5}=\frac{3}{15}$, deta hai $\frac{45-20+3}{15}=\frac{28}{15}$. $V=2\pi \cdot \frac{28}{15}=\frac{56\pi }{15}.$ $\boxed{V=\frac{56\pi }{15}}$ (Direct u-substitution $u=2-x^2$ possible hai lekin expand karne se zyada messy hai — expansion yahan jeetta hai.)