Volume of revolution — disk method, washer method

4.2.14 · Maths › Calculus II — Integration

WHY does slicing work?

Hum ek curvy solid ka volume directly nahi nikal sakte. Lekin hum ek thin slice ka volume nikal sakte hain, kyunki ek thin slice approximately ek cylinder hota hai, aur hum cylinder ka volume jaante hain: $V_{\text{cyl}}=(\text{area of circular face})\times (\text{thickness})=\pi r^{2}h.$

Agar hum x-axis ke around spin karein aur uske perpendicular slice karein, toh har slice ki thickness $dx$ hoti hai aur circular face ka radius curve ki height hoti hai, $r=f(x)$.

Disk Method ko scratch se derive karna

KYA: Region bounded by $y=f(x)$, the x-axis, between $x=a$ and $x=b$, rotated about the x-axis.

KAISE: Position $x$ par cut karo, thickness $\Delta x$ ka ek thin slab.

• Woh slab (approximately) ek cylinder hai jo apni side par leta hua hai.
• Uska radius = axis se curve tak ki doori = $f(x)$.
• Uski thickness = $\Delta x$.

Toh slab volume: $\Delta V\approx \pi [f(x){]}^2\Delta x.$ Yeh step kyun? Circular face ka area $\pi r^2=\pi [f(x){]}^2$ hai; thickness se multiply karo.

Saare slabs add karo aur $\Delta x\to 0$ jaane do — woh limit hi definite integral hai: $\boxed{V=\pi {\int }_a^b[f(x){]}^{2}dx}$

Washer Method derive karna

KYA: Ab region do curves ke beech mein hai, $y=f(x)$ (outer, top) aur $y=g(x)$ (inner, bottom), x-axis ke around rotate ho raha hai. Kyunki region axis ke upar ek gap ke saath baitha hai, har slice mein ek hole hota hai.

KAISE: Slice ek bada disk hai radius $R=f(x)$ ka, jisme se ek chhota disk radius $r=g(x)$ ka nikala gaya hai: $\Delta V\approx (\pi R^2-\pi r^2)\Delta x=\pi ([f(x){]}^2-[g(x){]}^2)\Delta x.$ Yeh step kyun? Ring ka area = (outer circle area) − (inner circle area). Subtract karo, na ki difference ko square karo!

$\boxed{V=\pi {\int }_a^b([R(x){]}^2-[r(x){]}^2)dx}$

Worked Example 1 — Disk

Region under $y=\sqrt{x}$, from $x=0$ to $x=4$, rotated about the x-axis.

$V=\pi {\int }_0^4(\sqrt{x}{)}^{2}dx=\pi {\int }_0^{4}xdx.$ Yeh step kyun? Radius $=\sqrt{x}$ hai, toh $[f(x){]}^2=x$. Square root ko khatam kar deta hai — clean.

$=\pi {\left[\frac{x^2}{2}\right]}_0^4=\pi \cdot \frac{16}{2}=8\pi .$

Worked Example 2 — Washer

Region between $y=x$ and $y=x^2$ (for $0\le x\le 1$), rotated about the x-axis.

Outer kaun sa hai? $(0,1)$ par, $x>x^2$, toh $R=x$ (axis se door), $r=x^2$ (paas).

$V=\pi {\int }_0^1(x^2-(x^2{)}^2)dx=\pi {\int }_0^1(x^2-x^4)dx.$ Yeh step kyun? $R^2=x^2$, $r^2=(x^2{)}^2=x^4$. Har radius ko alag alag square karo.

$=\pi {\left[\frac{x^3}{3}-\frac{x^5}{5}\right]}_0^1=\pi \left(\frac{1}{3}-\frac{1}{5}\right)=\pi \cdot \frac{2}{15}=\frac{2\pi }{15}.$

Worked Example 3 — Rotation about a non-axis line $y=2$

Region under $y=x$, $0\le x\le 2$, rotated about $y=2$.

Yahan axis line $y=2$ hai. Axis se curve tak ki doori $=|x-2|=2-x$ on $[0,2]$. Yeh ek disk hai ($x=2$ par region axis ko touch karta hai): $V=\pi {\int }_0^2(2-x{)}^{2}dx.$ Yeh step kyun? Naya radius $y=0$ se nahi, $y=2$ se measure hota hai.

Let $u=2-x$: $V=\pi {\int }_0^2{u}^{2}du\cdot (\text{sign handled by symmetry})=\pi {\left[\frac{(x-2{)}^3}{3}\right]}_0^2=\pi \left(0-\frac{(-2{)}^3}{3}\right)=\frac{8\pi }{3}.$

Common Mistakes

Active Recall

Connections

• Definite Integral as a Limit of Sums — slicing → Riemann sum → integral.
• Area Between Curves — same setup, lekin integrand alag hota hai ($f-g$ vs $R^2-r^2$).
• Shell Method — revolution ka alternative, slices axis ke parallel hote hain.
• Volume by Cross-Sections — disk/washer special cases hain (circular cross-sections).
• u-substitution — shifted-axis integrals evaluate karne ke liye use hoti hai.

Concept Map