Worked examples — Area between curves — horizontal and vertical slices

4.2.13 · D3 · Maths › Calculus II — Integration › Area between curves — horizontal and vertical slices

Do symbols jinhe hum poore page mein use karte hain, plain words mein dobara batate hain:

• ${\int }_a^b(\dots )dx$ ka matlab hai "infinitely many thin khade rectangles ko $x=a$ se $x=b$ tak add karo"; $dx$ har rectangle ki tiny width hai. (Scratch se Definite Integral as Riemann Sum mein banaya gaya hai.)
• ${\int }_c^d(\dots )dy$ ka matlab same hai lekin lying-down rectangles ke saath $y=c$ se $y=d$ tak; $dy$ tiny height hai.

Scenario matrix

Har area-between-curves problem in cells mein se kisi ek mein aata hai. Table ke neeche, ek example har cell ko clear karta hai.

Ex 1 — Cell A: clean vertical slice

Forecast: compute karne se pehle guess karo — kya answer $5$ se bada hai ya chota? (Region ek fat lens jaisi dikhti hai.)

1. Ye find karo ke wo kahan milti hain. $4-x^2=x+2$ set karo. $4-x^2=x+2\Rightarrow x^2+x-2=0\Rightarrow (x+2)(x-1)=0\Rightarrow x=-2,1.$ Ye step kyun? Limits $a,b$ exactly wahan hain jahan region pinch shut hoti hai — do curves wahan touch karti hain (dekho Solving Quadratic & Polynomial Equations).
2. Top vs bottom decide karo. $x=0$ test karo: parabola $=4$, line $=2$. To parabola top par hai $[-2,1]$ across. Ye step kyun? Ek interior test point puri order fix kar deta hai crossings ke beech — wo sirf endpoints par swap karti hain.
3. Top minus bottom integrate karo. $A={\int }_{-2}^1[(4-x^2)-(x+2)]dx={\int }_{-2}^1(2-x-x^2)dx.$ Subtract kyun? Har thin standing rectangle ki height (top $y$) − (bottom $y$) hoti hai; wo vertical gap parabola minus line hai.
4. Antiderivative (Fundamental Theorem of Calculus se): $[2x-\frac{x^2}{2}-\frac{x^3}{3}{]}_{-2}^1=(2-\frac{1}{2}-\frac{1}{3})-(-4-2+\frac{8}{3})=\frac{7}{6}-(-\frac{10}{3})=\frac{9}{2}.$

Verify: integrand $2-x-x^2$ at $x=0$ positive hai (value $2$) aur dono ends par zero — koi sign flip nahi, to cancellation ki koi chinta nahi. Answer $\frac{9}{2}=4.5$, humara "around 5" ka guess ke kareeb. ✔

Ex 2 — Cell B: curves interval ke andar cross karti hain

Forecast: agar tum naively ${\int }_0^{\pi /2}(\cos x-\sin x)dx$ karo to milega... kuch. Guess karo kya wo bahut chota hoga.

1. Interval ke andar crossing find karo. $\sin x=\cos x\Rightarrow x=\frac{\pi }{4}$. Ye step kyun? Agar curves $[a,b]$ ke andar top/bottom swap karti hain, to ek formula poori cheez cover nahi kar sakta — ye exactly Cell B hai.
2. Dekho kaun top par hai har piece mein. $[0,\frac{\pi }{4})$ par: $x=0$ par, $\cos 0=1>\sin 0=0$, to cosine top hai. $(\frac{\pi }{4},\frac{\pi }{2}]$ par: $x=\frac{\pi }{2}$ par, $\sin =1>\cos =0$, to sine top hai. Ye step kyun? "Top" literally crossing par flip hota hai; har region ko apna khud ka (top − bottom) chahiye.
3. Split karo aur add karo, hamesha top minus bottom. $A={\int }_0^{\pi /4}(\cos x-\sin x)dx+{\int }_{\pi /4}^{\pi /2}(\sin x-\cos x)dx.$ Pehla piece: $[\sin x+\cos x{]}_0^{\pi /4}=(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2})-(0+1)=\sqrt{2}-1$. Doosra piece: $[-\cos x-\sin x{]}_{\pi /4}^{\pi /2}=(0-1)-(-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2})=\sqrt{2}-1$. $A=(\sqrt{2}-1)+(\sqrt{2}-1)=2\sqrt{2}-2\approx \mathrm{0.828.}$

Verify: dono pieces equal nikle ($\sqrt{2}-1$), jo $\sin$ aur $\cos$ ki $x=\frac{\pi }{4}$ ke baare mein mirror symmetry se match karta hai. Positive answer, jaisa area hona chahiye. ✔

Ex 3 — Cell C: forced horizontal slicing

Forecast: kya tum $y$ ke liye solve karna prefer karoge (jisse $\pm \sqrt{\cdot }$ milega) ya cheezein $x=f(y)$ ke roop mein rakhoge?

1. $y$ ke terms mein intersect karo. $y^2-3=-y^2+5\Rightarrow 2y^2=8\Rightarrow y=\pm 2$. Ye step kyun? Dono curves already $x=(\text{function of }y)$ ke roop mein likhi hain, isliye sweep karne ka natural variable $y$ hai — crossings $y=-2,2$ par hain.
2. Kaun si curve right par hai? $y=0$ par: right-facing candidate $x=-y^2+5=5$ hai, doosri $x=y^2-3=-3$ hai. To $x=-y^2+5$ right curve hai, $x=y^2-3$ left hai. Ye step kyun? Horizontal rectangles ki width (right $x$) − (left $x$) hoti hai; humein pata hona chahiye kaun kya hai.
3. $y$ mein right minus left integrate karo. $A={\int }_{-2}^2[(-y^2+5)-(y^2-3)]dy={\int }_{-2}^2(8-2y^2)dy.$
4. Evaluate karo (integrand even hai, to $0\to 2$ part double karo): $A=2{\int }_0^2(8-2y^2)dy=2[8y-\frac{2y^3}{3}{]}_0^2=2(16-\frac{16}{3})=2\cdot \frac{32}{3}=\frac{64}{3}.$

Horizontal kyun aur vertical kyun nahi? $x$ ke function ke roop mein har parabola do $y$-values deti hai (top aur bottom halves). Vertical slicing splitting aur inverting maangti; horizontal slicing isse puri tarah dodge karti hai (ye wahi jagah hai jahan Inverse Functions otherwise bite karta).

Verify: $\frac{64}{3}\approx 21.3$; region roughly $2$-tall by up-to-$8$-wide lens hai, plausibly around $20$. Symmetry (even integrand) ne doubling confirm kiya. ✔

Ex 4 — Cell D: top curve mid-sweep mein badal jaati hai

Forecast: tumhare hisab se isko kitne alag $dx$ integrals chahiye?

1. Teeno pairwise crossings find karo.
   • $y=x$ milta hai $y=3-\frac{x}{2}$ se: $x=3-\frac{x}{2}\Rightarrow \frac{3x}{2}=3\Rightarrow x=2$ (point $(2,2)$).
   • $y=\frac{x}{4}$ milta hai $y=3-\frac{x}{2}$ se: $\frac{x}{4}=3-\frac{x}{2}\Rightarrow \frac{3x}{4}=3\Rightarrow x=4$ (point $(4,1)$).
   • $y=x$ milta hai $y=\frac{x}{4}$ se origin $(0,0)$ par. Ye step kyun? Region ke corners wahan hain jahan boundaries milti hain; top boundary sirf aise corner par switch ho sakti hai.
2. Switch spot karo. Bottom hamesha $y=\frac{x}{4}$ hai. Top $y=x$ hai $x=0$ se $x=2$ tak, phir $x=2$ se $x=4$ tak $y=3-\frac{x}{2}$ ban jaata hai. Ye step kyun? $x=2$ ke baad line $y=x$ region se bahar chali gayi hai — ceiling doosri line ko hand off ho gaya. Ye split force karta hai (Cell D).
3. Do integrals, har ek top − bottom. $A={\int }_0^2(x-\frac{x}{4})dx+{\int }_2^4(3-\frac{x}{2}-\frac{x}{4})dx.$ Pehla: ${\int }_0^2\frac{3x}{4}dx=\frac{3}{4}\cdot \frac{x^2}{2}{|}_0^2=\frac{3}{4}\cdot 2=\frac{3}{2}$. Doosra: ${\int }_2^4(3-\frac{3x}{4})dx=[3x-\frac{3x^2}{8}{]}_2^4=(12-6)-(6-\frac{3}{2})=6-\frac{9}{2}=\frac{3}{2}$. $A=\frac{3}{2}+\frac{3}{2}=3.$

Verify: region ke teen corners hain $(0,0),(2,2),(4,1)$. Us triangle ka Shoelace area: $\frac{1}{2}|0(2-1)+2(1-0)+4(0-2)|=\frac{1}{2}|2-8|=3$. Match karta hai. ✔

Ex 5 — Cell E: degenerate / zero area

Forecast: ye ek parabola aur line jaisi lagti hain — surely ek normal region trap karti hain?

1. Intersect karo. $x^2=2x-1\Rightarrow x^2-2x+1=0\Rightarrow (x-1{)}^2=0\Rightarrow x=1$ (double root). Ye step kyun? Repeated root ka matlab hai curves bina cross kiye touch karti hain — line parabola ki tangent hai.
2. Interpret karo. Sirf ek intersection point hai, isliye $a=b=1$. Limits same value par pinch ho jaati hain. Ye step kyun? Ek region ko width ke liye do alag crossings chahiye; yahan koi nahi hai.
3. Compute karo. $A={\int }_1^1(x^2-(2x-1))dx=0.$

Verify: $(x-1{)}^2\ge 0$ hamesha, isliye $x^2-(2x-1)\ge 0$ sirf $x=1$ par equality ke saath; "region" ek single point hai, area $0$. ✔

Ex 6 — Cell F: limiting behaviour

Forecast: jaise $k$ $0$ ki taraf shrink karta hai, parabola axis par flatten hoti hai — kya area smoothly vanish hoga, ya jump karega?

1. Parabola axis ko kahan hit karti hai? $k-x^2=0\Rightarrow x=\pm \sqrt{k}$. Ye step kyun? Region in do roots ke beech rehta hai; limits $k$ par depend karti hain.
2. Area (even integrand, half ko double karo). $A(k)={\int }_{-\sqrt{k}}^{\sqrt{k}}(k-x^2)dx=2{\int }_0^{\sqrt{k}}(k-x^2)dx=2[kx-\frac{x^3}{3}{]}_0^{\sqrt{k}}.$ $=2(k\sqrt{k}-\frac{k\sqrt{k}}{3})=2\cdot \frac{2}{3}{k}^{3/2}=\frac{4}{3}{k}^{3/2}.$ Ye step kyun? Height ($\sim k$) aur width ($\sim \sqrt{k}$) dono shrink karte hain, isliye area $k^{3/2}$ ki tarah scale karta hai.
3. Limit lo. ${\lim }_{k\to 0^{+}}\frac{4}{3}{k}^{3/2}=0.$

Verify: $k=1$ par, $A(1)=\frac{4}{3}$ — classic $y=1-x^2$ arch area se match karta hai. Jaise $k\to 0^{+}$ arch ka area continuously vanish hota hai (koi jump nahi), jo confirm karta hai region smoothly collapse hoti hai. ✔

Ex 7 — Cell G: real-world word problem (units ke saath)

Forecast: bed ek downward dip hai $2$ m deep aur $4$ m wide — integrate karne se pehle area guess karo.

1. Top aur bottom identify karo. Top curve: surface $y=0$. Bottom curve: bed $y=-(4-x^2)=x^2-4$. Ye step kyun? Paani flat surface aur curved bed ke beech fill hota hai — ek clean vertical-slice setup.
2. Limits banks hain. Bed surface se milta hai jahan $x^2-4=0\Rightarrow x=\pm 2$ (paani ke edges). Ye step kyun? $x=\pm 2$ se aage bed $y=0$ se upar uth jaata hai — wahan paani nahi.
3. Top − bottom integrate karo. $A={\int }_{-2}^2[0-(x^2-4)]dx={\int }_{-2}^2(4-x^2)dx=2[4x-\frac{x^3}{3}{]}_0^2=2(8-\frac{8}{3})=\frac{32}{3}.$

Verify: $\frac{32}{3}\approx 10.67{\text{m}}^2$. Units check: (metres) × (metres) integrate karke ${\text{m}}^2$ milta hai, cross-sectional area ke liye sahi. $4\text{ m}\times 2\text{ m}$ bounding box ka area $8{\text{ m}}^2$ hai; parabolic dip box ka $\frac{2}{3}$ fill karta hai $=\frac{16}{3}\times 2=\frac{32}{3}$. ✔

Ex 8 — Cell H: exam twist (axis ke upar aur neeche region)

Forecast: is region ka ek part $x$-axis ke neeche hai aur ek part upar. Kya ek single integral isko sahi karega?

1. $[-1,1]$ mein saari crossings find karo. $x^3=x\Rightarrow x^3-x=0\Rightarrow x(x-1)(x+1)=0\Rightarrow x=-1,0,1$. Ye step kyun? Teen crossings matlab top/bottom order flip ho sakta hai — ye really Cell B plus exam trap hai axis cross karne ka.
2. Har piece par top determine karo.
   • $(-1,0)$ par: $x=-\frac{1}{2}$ par, $x^3=-\frac{1}{8}$, $x=-\frac{1}{2}$. Kyunki $-\frac{1}{8}>-\frac{1}{2}$, $x^3$ top par hai.
   • $(0,1)$ par: $x=\frac{1}{2}$ par, $x^3=\frac{1}{8}$, $x=\frac{1}{2}$. Kyunki $\frac{1}{2}>\frac{1}{8}$, $x$ top par hai. Ye step kyun? Curves $x=0$ par swap karti hain; har piece ko apna khud ka (top − bottom) chahiye.
3. Do integrals, hamesha top − bottom. $A={\int }_{-1}^0(x^3-x)dx+{\int }_0^1(x-x^3)dx.$ Pehla: $[\frac{x^4}{4}-\frac{x^2}{2}{]}_{-1}^0=0-(\frac{1}{4}-\frac{1}{2})=\frac{1}{4}$. Doosra: $[\frac{x^2}{2}-\frac{x^4}{4}{]}_0^1=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$. $A=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}.$

Verify: region ki odd symmetry origin ke baare mein se, do pieces equal hone chahiye — dono ne $\frac{1}{4}$ diya. Total $\frac{1}{2}>0$. ✔

Connections

• Definite Integral as Riemann Sum — upar har $\int$ thin-rectangle sums ka limit hai.
• Fundamental Theorem of Calculus — har integral ko antiderivatives ke zariye evaluate karta hai.
• Inverse Functions — wo tool jo Ex 3 mein horizontally slice karke avoid kiya gaya.
• Solving Quadratic & Polynomial Equations — Ex 1, 4, 5, 8 mein intersection limits find karta hai.
• Volumes by Slicing & Disks — same slice-measure-integrate pattern, ek dimension upar.
• Hinglish version.