Worked examples — Double integrals over rectangles — Fubini's theorem

4.4.16 · D3 · Maths › Multivariable Calculus › Double integrals over rectangles — Fubini's theorem

Yeh page ek drill of cases hai. Parent note ne machine banai thi (freeze → fill → file). Yahan hum us machine ko har tarah ke input se guzarte hain jo ek rectangle de sakta hai: positive surfaces, plane ke neeche jaane wali surfaces, zero answers, degenerate rectangles, separable shortcuts, ek physics word problem, aur ek exam twist jahan integration ka order decide karta hai ki aap ek line mein finish karoge ya ek ghante mein.

Shuru karne se pehle, ek reminder seedhe alfazon mein. Ek double integral ${\iint }_{R}fdA$ woh signed volume hai jo surface $z=f(x,y)$ aur flat floor ($xy$-plane) ke beech rectangle $R$ ke upar trapped hai. "Signed" ka matlab hai: jahan surface floor ke upar hai woh volume positive count hota hai, jahan neeche jaati hai woh negative count hota hai. Yeh picture yaad rakho — neeche ke zyaadatar cases sirf konsa sign jeet ta hai ke baare mein hain.

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Scenario matrix

Har rectangular double integral jo aap dekhoge in cells mein se kisi ek mein aata hai. Worked examples ko uss cell se tag kiya gaya hai jo woh cover karte hain.

#	Case class	Kya tricky hai isme	Example
A	All-positive integrand	Surface poori tarah floor ke upar → answer ek genuine positive volume hai	Ex 1
B	Sign-changing integrand	Surface floor ko cross karti hai → signed cancellation, answer chhota ya zero ho sakta hai	Ex 2
C	Answer exactly zero hai (symmetry)	Positive lump exactly negative lump ko cancel karta hai	Ex 3
D	Separable shortcut	$f=g(x)h(y)$ → do 1-D integrals mein split karo	Ex 4
E	Degenerate rectangle ($a=b$ ya $c=d$)	Zero-width slab → volume must be $0$	Ex 5
F	Order matters for effort	Ek order ek line hai, doosre ko parts chahiye	Ex 6
G	Real-world word problem (units!)	Ek physical quantity ko $\iint$ mein translate karo; units carry karo	Ex 7
H	Exam twist: constant / average value	Integral ko average height ke roop mein reinterpret karo	Ex 8

Hum saaton aath cells cover karenge.

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Ex 1 — Cell A: all-positive integrand (ek real volume)

Pehle surface dekho — yeh ek valley floor hai jo dono directions mein upar curve karti hai.

Step 1 — $x$ ko freeze karo, $y$ ke upar integrate karo (inner integral fill karo). ${\int }_0^1(x^2+y^2)dy={\left[x^{2}y+\frac{y^3}{3}\right]}_{y=0}^{y=1}=x^2+\frac{1}{3}.$ Yeh step kyun? $x$ ko freeze karna inner job ko $y$ mein ek ordinary 1-D integral bana deta hai; term $x^2$ ek frozen constant hai, toh $y$ mein iski antiderivative $x^{2}y$ hai. Result $A(x)=x^2+\frac{1}{3}$ position $x$ par solid ka cross-sectional area hai.

Step 2 — $A(x)$ ko outer integral mein file karo. ${\int }_0^1\left(x^2+\frac{1}{3}\right)dx={\left[\frac{x^3}{3}+\frac{x}{3}\right]}_0^1=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}.$ Yeh step kyun? Saare thin slabs $A(x)dx$ ko add karna exactly Volume by slicing (single-variable) hai — ek dimension neeche.

Verify: Order swap karo. Symmetry se ($x\leftrightarrow y$ integrand ko unchanged chhod ta hai) inner-$x$ integral bhi $y^2+\frac{1}{3}$ deta hai, aur outer $\frac{2}{3}$ phir se deta hai. Answer $\frac{2}{3}\approx 0.667$ predicted range $[0,2]$ mein hai. ✅

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Ex 2 — Cell B: sign-changing integrand

Red patch (floor ke neeche, $x<y$) origin ke paas ek chhota triangle hai; blue region (floor ke upar) dominate karta hai.

Step 1 — $y$ ke upar inner integral ($x$ ko freeze karo). ${\int }_0^1(x-y)dy={\left[xy-\frac{y^2}{2}\right]}_0^1=x-\frac{1}{2}.$ Yeh step kyun? Constant $x$ integrate hokar $xy$ deta hai; genuinely $y$-dependent term $-y$ integrate hokar $-\frac{y^2}{2}$ deta hai. Note karo $A(x)=x-\frac{1}{2}$ khud negative hai $x<\frac{1}{2}$ ke liye — yahi below-floor part dikh raha hai.

Step 2 — $x$ ke upar outer integral. ${\int }_0^2\left(x-\frac{1}{2}\right)dx={\left[\frac{x^2}{2}-\frac{x}{2}\right]}_0^2=(2-1)-0=1.$

Verify: Signed pieces mein split karo. $\iint xdA=\left({\int }_0^{2}xdx\right)\left({\int }_0^{1}dy\right)=2\cdot 1=2$ aur $\iint ydA=\left({\int }_0^{2}dx\right)\left({\int }_0^{1}ydy\right)=2\cdot \frac{1}{2}=1$. Difference $2-1=1$. ✅ Positive aur chhota, jaise forecast tha.

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Ex 3 — Cell C: answer exactly zero hai (symmetry)

Step 1 — Recognize karo ki yeh separable hai, split karo. ${\iint }_R\sin x\cos ydA=\left({\int }_{-\pi }^{\pi }\sin xdx\right)\left({\int }_0^{\pi }\cos ydy\right).$ Yeh step kyun? $f=g(x)h(y)$ hai jahan $g=\sin x$, $h=\cos y$; separable integrands ke liye product rule (parent note) ek 2-D job ko do 1-D jobs mein badal deta hai.

Step 2 — $x$-factor evaluate karo. ${\int }_{-\pi }^{\pi }\sin xdx=[-\cos x{]}_{-\pi }^{\pi }=-\cos \pi +\cos (-\pi )=-(-1)+(-1)=0.$ Yeh step kyun? Koi bhi ek factor ka zero hona poore product ko zero kar deta hai — $y$-integral finish karne ki zaroorat nahi. Yeh odd-function-over-symmetric-interval cancellation hai jo rigorously prove ho raha hai.

Verify: $y$-factor bhi ${\int }_0^{\pi }\cos ydy=[\sin y{]}_0^{\pi }=0-0=0$ hai — lekin hume sirf ek zero factor chahiye tha. Product $=0\times 0=0$. ✅

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Ex 4 — Cell D: separable shortcut with a twist

Step 1 — Do 1-D integrals ke product mein split karo. ${\iint }_{R}x{e}^{x}ydA=\left({\int }_0^{1}x{e}^{x}dx\right)\left({\int }_0^{2}ydy\right).$ Yeh step kyun? Variables kabhi mix nahi hote aur dono limits constants hain, toh double integral factor ho jaata hai — yahi Cell D ka poora point hai.

Step 2 — $x$-factor by parts. $u=x,dv=e^{x}dx$ choose karo, toh $du=dx,v=e^x$: ${\int }_0^{1}x{e}^{x}dx=[x{e}^x{]}_0^1-{\int }_0^1{e}^{x}dx=e-(e-1)=1.$ Yeh step kyun? Integration by parts "polynomial $\times$ exponential" product ke liye tool hai — $x$ ko differentiate karna polynomial factor ko ek step mein kill karta hai, yahi parts ka use karne ka point hai.

Step 3 — $y$-factor aur product. ${\int }_0^{2}ydy={\left[\frac{y^2}{2}\right]}_0^2=2,\text{answer}=1\times 2=2.$

Verify: Numerically ${\int }_0^{1}x{e}^{x}dx=e\cdot 1-(e-1)=1$ exactly; times $2$ gives $\boxed{2}$. ✅ Forecast se match karta hai.

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Ex 5 — Cell E: degenerate rectangle (zero width)

Step 1 — Outer limits padhо. ${\int }_2^2\left({\int }_0^5(x^2+3xy+7)dy\right)dx.$ Yeh step kyun? Inner integral koi function $A(x)$ produce karta hai — koi fark nahi kya — kyunki outer integral $2$ se $2$ tak chalta hai.

Step 2 — Zero-length rule apply karo. ${\int }_2^{2}A(x)dx=0\text{kisi bhi integrable }A\text{ ke liye.}$ Yeh step kyun? ${\int }_a^{a}g=0$ definite integrals ki fundamental property hai: koi interval nahi, koi accumulation nahi. Geometrically "solid" zero thickness ki flat sheet hai — volume $0$.

Verify: Brute force bhi agree karta hai. Inner: ${\int }_0^5(x^2+3xy+7)dy=5x^2+\frac{75}{2}x+35$, ek finite $A(x)$. Outer ${\int }_2^{2}A(x)dx=0$. ✅ Yeh degenerate limiting case hai — algebra grind karne se pehle hamesha check karo.

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Ex 6 — Cell F: order matters for effort

Step 1 — Pehle $y$ ke upar integrate karo (freeze $x$). ${\int }_0^{1}x{e}^{xy}dy=[e^{xy}{]}_{y=0}^{y=1}=e^x-1.$ Yeh step kyun? Kyunki ${\partial }_y{e}^{xy}=x{e}^{xy}$ integrand se exactly match karta hai — saamne wala $x$ precisely chain-rule factor hai. $dy$ pehle choose karna hi ise one-liner banata hai. Yeh Fubini in action hai: same solid, sasta slicing direction.

Step 2 — Result ko $x$ ke upar integrate karo. ${\int }_0^1(e^x-1)dx=[e^x-x{]}_0^1=(e-1)-(1-0)=e-2.$

Verify: $e-2\approx 0.718$ — positive, $1$ se neeche, jaise forecast tha. Doosra order sanity-check karo: ${\int }_0^{1}x{e}^{xy}dx$ ko integration by parts chahiye aur ek messier expression produce karta hai jo bhi $e-2$ tak integrate hota hai — same answer, kaafi zyaada kaam. ✅ Fubini yahan apni value prove karta hai.

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Ex 7 — Cell G: real-world word problem (units carry karo!)

Step 1 — Mass integral set up karo. $M={\iint }_R\rho (x,y)dA={\int }_0^3{\int }_0^2(2+x+y)dydx[{\text{kg/m}}^2\cdot {\text{m}}^2=\text{kg}].$ Yeh step kyun? Har tiny tile jiska area $dA$ (m²) hai, mass $\rho dA$ (kg) carry karta hai. Tiles ko sum karna exactly ek 2-D Riemann sum hai jiska limit double integral hai — $\iint$ ka physical meaning.

Step 2 — $y$ ke upar inner integral ($x$ ko freeze karo). ${\int }_0^2(2+x+y)dy=[(2+x)y+\frac{y^2}{2}{]}_0^2=2(2+x)+2=2x+6.$ Yeh step kyun? $2+x$ $y$ mein ek frozen constant hai, integrate hokar $(2+x)y$ deta hai; actual $y$-term $\frac{y^2}{2}$ deta hai.

Step 3 — $x$ ke upar outer integral. $M={\int }_0^3(2x+6)dx=[x^2+6x{]}_0^3=9+18=27\text{kg}.$

Verify: Units: ${\text{kg/m}}^2\times {\text{m}}^2=\text{kg}$ ✅. Value $27\text{kg}$ average-density estimate se exactly match karta hai (kyunki $\rho$ linear hai, mean density really centre $(1.5,1)$ par value hai: $2+1.5+1=4.5$, times area $6$ = $27$). ✅

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Ex 8 — Cell H: exam twist (average value / constant height)

Step 1 — Average value ki definition yaad karo. $\bar{f}=\frac{1}{\text{area}(R)}{\iint }_{R}fdA.$ Yeh step kyun? Average height = (total volume)$÷$(base area). Yeh single-variable mean $\frac{1}{b-a}{\int }_a^{b}f$ ka 2-D twin hai. Yeh jawab deta hai: "kauni flat height same volume deti hai?"

Step 2 — Volume compute karo (separable). ${\iint }_R{x}^{2}ydA=\left({\int }_0^2{x}^{2}dx\right)\left({\int }_0^{3}ydy\right)=\frac{8}{3}\cdot \frac{9}{2}=12.$ Yeh step kyun? $x^{2}y$ separable hai, toh do 1-D integrals mein split hota hai — sabse fast route.

Step 3 — Area se divide karo. $\bar{f}=\frac{12}{2\cdot 3}=\frac{12}{6}=2.$

Verify: Base area $6$ aur height $\bar{f}=2$ wale box ka volume $6\times 2=12$ hai — computed volume se match karta hai. ✅ Aur $\bar{f}=2$ range $[0,12]$ ke andar hai, jaise forecast tha.

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Recall Saare cells ke upar quick self-test

Konse cell ka answer compute kiye bina guaranteed zero hai? ::: Cell E (degenerate rectangle, $a=b$ ya $c=d$) — aur Cell C jab ek odd factor ek symmetric interval span kare. Ek plate ka mass kaunsa integral chahta hai? ::: ${\iint }_R\rho dA$ (density times area, summed) — Cell G. Integration order choose karna real kaam kab bachata hai? ::: Cell F — jab ek order ka integrand already ek recognisable derivative ho (yahan ${\partial }_y{e}^{xy}=x{e}^{xy}$). $R$ ke upar average value formula? ::: $\bar{f}=\frac{1}{\text{area}(R)}{\iint }_{R}fdA$.

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Connections

• Riemann sums — word-problem mass (Ex 7) ek 2-D Riemann sum limit hai.
• Volume by slicing (single-variable) — har "inner ko outer mein file karo" step (Ex 1).
• Change of order of integration — Cell F (Ex 6) sasta slicing direction choose karta hai.
• Double integrals over general regions — aage: non-constant inner limits (yahan woh hamesha numbers hain).
• Fubini–Tonelli theorem — guarantee karta hai ki upar use kiye gaye order-swaps legal hain.
• Triple integrals — ek teesra slicing direction add karo plate nahi, solid ka mass ke liye.