Exercises — Volume of revolution — shell method

Axis = $y$-axis, toh strips vertical hain ($y$-axis ke parallel) thickness $dx$ ke saath.

• Radius: $y$-axis (line $x=0$) se $x$ position pe ek strip ki distance $r=x$ hai.
• Height: strip $y=0$ se upar $y=x^3$ tak jaati hai, toh $h=x^3$.
• Limits: $x$, $0$ se $2$ tak jaata hai. $V={\int }_0^{2}2\pi x\cdot x^{3}dx.$

• $y$-axis se radius: $r=x$.
• Strip ki height: $h=\sqrt{x}$.
• Limits: $x$, $1$ se $4$ tak. $V={\int }_1^{4}2\pi x\sqrt{x}dx.$ Dhyan do ki humne $x=y^2$ mein invert nahi kiya. Yahi wajah hai ki shells yahan convenient hain.

$r=x$, $h=x^2$, $x\in [0,3]$. $V={\int }_0^{3}2\pi x\cdot x^{2}dx=2\pi {\int }_0^3{x}^{3}dx=2\pi {\left[\frac{x^4}{4}\right]}_0^3=2\pi \cdot \frac{81}{4}=\frac{81\pi }{2}.$ Humne kya kiya: circumference $2\pi x$ ko height $x^2$ se multiply kiya, phir axis se baahir ki taraf shells integrate kiye.

$r=x$, $h=\sqrt{x}=x^{1/2}$, $x\in [0,4]$. $V={\int }_0^{4}2\pi x\cdot x^{1/2}dx=2\pi {\int }_0^4{x}^{3/2}dx=2\pi {\left[\frac{2}{5}{x}^{5/2}\right]}_0^4.$ Ab $4^{5/2}=(4^{1/2}{)}^5=2^5=32$, toh $V=2\pi \cdot \frac{2}{5}\cdot 32=\frac{128\pi }{5}.$

$r=x$, $h=x^2+1$, $x\in [0,2]$. $V={\int }_0^{2}2\pi x(x^2+1)dx=2\pi {\int }_0^2(x^3+x)dx=2\pi {\left[\frac{x^4}{4}+\frac{x^2}{2}\right]}_0^2.$ $x=2$ plug karo: $\frac{16}{4}+\frac{4}{2}=4+2=6$, toh $V=12\pi$.

Axis $x=-1$ poore region ($0\le x\le 2$) ke left mein hai. $x$ position pe ek strip ki line $x=-1$ se distance hai $r=x-(-1)=x+1(\text{poore mein positive, kyunki }x\ge 0).$ Height $h=x^2$ rehta hai. $V={\int }_0^{2}2\pi (x+1)x^{2}dx=2\pi {\int }_0^2(x^3+x^2)dx=2\pi {\left[\frac{x^4}{4}+\frac{x^3}{3}\right]}_0^2.$ $x=2$ pe: $\frac{16}{4}+\frac{8}{3}=4+\frac{8}{3}=\frac{20}{3}$, toh $V=2\pi \cdot \frac{20}{3}=\frac{40\pi }{3}.$

Axis $x=2$ region ke right mein hai. $x$ position pe strip ki $x=2$ se distance hai $r=2-x(\text{positive, kyunki }x\le 2).$ Height $h=x^2$. $V={\int }_0^{2}2\pi (2-x)x^{2}dx=2\pi {\int }_0^2(2x^2-x^3)dx=2\pi {\left[\frac{2x^3}{3}-\frac{x^4}{4}\right]}_0^2.$ $x=2$ pe: $\frac{2\cdot 8}{3}-\frac{16}{4}=\frac{16}{3}-4=\frac{4}{3}$, toh $V=2\pi \cdot \frac{4}{3}=\frac{8\pi }{3}.$

Axis horizontal hai, toh strips thickness $dy$ ke saath horizontal hain, radius $r=y$ ($x$-axis se distance), aur height = us $y$ pe region ki horizontal width.

• Kisi given $y$ ke liye, region left mein curve $x=\sqrt{y}$ se aur right mein line $x=2$ se bounded hai. Toh width $=x_R-x_L=2-\sqrt{y}$.
• $y$, $0$ (bottom) se $4$ tak jaata hai (kyunki $y=x^2$, $x=2$ pe $4$ tak pahunchta hai). $V={\int }_0^{4}2\pi y(2-\sqrt{y})dy=2\pi {\int }_0^4(2y-y^{3/2})dy=2\pi {\left[y^2-\frac{2}{5}{y}^{5/2}\right]}_0^4.$ $y=4$ pe: $16-\frac{2}{5}\cdot 32=16-\frac{64}{5}=\frac{80-64}{5}=\frac{16}{5}$, toh $V=2\pi \cdot \frac{16}{5}=\frac{32\pi }{5}.$

$(0,1)$ pe line $y=x$, parabola $y=x^2$ ke upar hai, toh height $h=x-x^2$. Axis $x=2$ region ke right mein hai, toh radius $r=2-x$. $V={\int }_0^{1}2\pi (2-x)(x-x^2)dx.$ Integrand expand karo: $(2-x)(x-x^2)=2x-2x^2-x^2+x^3=2x-3x^2+x^3.$ $V=2\pi {\int }_0^1(2x-3x^2+x^3)dx=2\pi {\left[x^2-x^3+\frac{x^4}{4}\right]}_0^1.$ $x=1$ pe: $1-1+\frac{1}{4}=\frac{1}{4}$, toh $V=2\pi \cdot \frac{1}{4}=\frac{\pi }{2}.$

Parabola $y=2x-x^2$, $0\le x\le 2$ ke liye $x$-axis ke upar hai, toh $h=2x-x^2$. Axis $y$-axis hai, toh $r=x$. $V={\int }_0^{2}2\pi x(2x-x^2)dx=2\pi {\int }_0^2(2x^2-x^3)dx=2\pi {\left[\frac{2x^3}{3}-\frac{x^4}{4}\right]}_0^2.$ $x=2$ pe: $\frac{2\cdot 8}{3}-\frac{16}{4}=\frac{16}{3}-4=\frac{4}{3}$, toh $V=2\pi \cdot \frac{4}{3}=\frac{8\pi }{3}.$

$0\le x\le 1$ pe, upper curve $y=2-x^2$ hai aur lower $y=x^2$ hai, toh height $h=(2-x^2)-x^2=2-2x^2.$ Axis $x=1$ is piece ke right mein hai, toh radius $r=1-x$ ($[0,1)$ pe positive, $x=1$ pe zero). $V={\int }_0^{1}2\pi (1-x)(2-2x^2)dx=2\pi {\int }_0^{1}2(1-x)(1-x^2)dx.$ $(1-x)(1-x^2)=1-x^2-x+x^3$ expand karo, times $2$: $2-2x-2x^2+2x^3$. $V=2\pi {\int }_0^1(2-2x-2x^2+2x^3)dx=2\pi {\left[2x-x^2-\frac{2x^3}{3}+\frac{x^4}{2}\right]}_0^1.$ $x=1$ pe: $2-1-\frac{2}{3}+\frac{1}{2}=\frac{12-6-4+3}{6}=\frac{5}{6}$, toh $V=2\pi \cdot \frac{5}{6}=\frac{5\pi }{3}.$

Radius $r=x$, height $h=x^n$, limits $0$ se $a$ tak. $V={\int }_0^{a}2\pi x\cdot x^{n}dx=2\pi {\int }_0^a{x}^{n+1}dx=2\pi {\left[\frac{x^{n+2}}{n+2}\right]}_0^a=\frac{2\pi a^{n+2}}{n+2}.$ L2 ke saath algebra check karo: $n=3$, $a=2$ ke saath yeh $\frac{2\pi \cdot 2^5}{5}=\frac{64\pi }{5}$ deta hai. Hmm — boxed claim kehta hai $a^{n+3}/(n+3)$, jo $\frac{2\pi \cdot 2^6}{6}$ dega. Boxed statement ek deliberate trap hai: correct exponent $n+3$ nahi, $n+2$ hai, kyunki $x\cdot x^n=x^{n+1}$ aur integrate karne se ek aur add hota hai, jo $x^{n+2}$ deta hai. Sahi general formula hai $\boxed{V=\frac{2\pi a^{n+2}}{n+2}}.$ (Sanity: $n=1$, $a=1$ se $\frac{2\pi }{3}$ milta hai, jo $y=x$ ko $y$-axis ke baare mein spin karne se bane cone-like solid ka volume hai — sahi hai.)

Shells (Exercise 2.1 pattern se, lekin $a=2$): $r=x$, $h=x^2$, $V_{\text{shell}}={\int }_0^{2}2\pi x\cdot x^{2}dx=2\pi {\left[\frac{x^4}{4}\right]}_0^2=8\pi .$ Disks/washers: $y$-axis ke perpendicular slice karo, toh slices horizontal hain, thickness $dy$. Height $y$ pe, solid ek radius $R=2$ ka disk hai jisme curve $x=\sqrt{y}$ ne ek bita cut kar diya hai (inner radius $r_{\text{in}}=\sqrt{y}$), kyunki region sirf $x=\sqrt{y}$ se $x=2$ tak fill hota hai. $y$, $0$ se $4$ tak jaata hai: $V_{\text{washer}}={\int }_0^4\pi (2^2-(\sqrt{y}{)}^2)dy=\pi {\int }_0^4(4-y)dy=\pi {\left[4y-\frac{y^2}{2}\right]}_0^4=\pi (16-8)=8\pi .$ Dono $\boxed{8\pi }$ dete hain. Shells ne humein $x$ mein rakha; washers ne $x=\sqrt{y}$ inversion force kiya. Dekho Volume of revolution — disk and washer method.

$r$ jab $x=c$ ke baare mein rotate karo ::: $|x-c|$, strip se axis ki distance Thickness jab axis horizontal ho ::: $dy$ (strips axis ke parallel chalti hain) Do curves ke beech region ke liye height ::: top curve minus bottom curve $y=x^2$, $0\le x\le 3$, $y$-axis ke baare mein volume ::: $\frac{81\pi }{2}$ $y=\sqrt{x}$, $0\le x\le 4$, $y$-axis ke baare mein volume ::: $\frac{128\pi }{5}$ General $y=x^n$, $0\le x\le a$, $y$-axis ke baare mein ::: $\frac{2\pi a^{n+2}}{n+2}$