Double integrals over rectangles — Fubini's theorem: Exercises

4.4.16 · D4 · Maths › Multivariable Calculus › Double integrals over rectangles — Fubini's theorem

Parent se teen-step ritual yaad karo, jo har problem ke liye hamara compass hai:

Level 1 — Recognition

Ye sirf ek cheez test karte hain: kya tum ek iterated integral padh sakte ho aur dekh sakte ho kaunsa variable frozen hai?

Exercise 1.1

Integral $\int_{0}^{2} \int_{1}^{4} (x^{2} y) d y d x$ mein, kaunsa variable pehle integrate hota hai, aur pehle step mein kise constant treat kiya jaata hai?

Recall Solution 1.1

Inner differential $d y$ hai, isliye hum $y$ ke upar pehle integrate karte hain. Us inner step ke dauran, $x$ frozen rehta hai — bilkul ek fixed number ki tarah treat hota hai (jaise $7$ ). Concretely inner step yeh hai: $\int_{1}^{4} x^{2} y d y = x^{2} [\frac{y ^{2}}{2}]_{1}^{4} = x^{2} \cdot \frac{16 - 1}{2} = \frac{15}{2} x^{2},$ jahaan $x^{2}$ bahar aa gaya kyunki woh $y$ ke respect mein constant hai. Answer: pehle $y$ ; $x$ constant.

Exercise 1.2

Kya $f (x, y) = x^{3} e^{2 y}$ separable hai (yaani $g (x) h (y)$ form ka)? Agar haan, toh $g$ aur $h$ batao.

Recall Solution 1.2

Haan. Yeh sirf $x$ ki function aur sirf $y$ ki function ka product hai: $g (x) = x^{3}, h (y) = e^{2 y} .$ Kyunki yeh separate hota hai, ek rectangle par hum double integral ko do 1-D integrals ke product mein split kar sakte hain (parent shortcut). $x^{3} e^{2 x y}$ se compare karo: exponent $x$ aur $y$ ko mix karta hai, isliye woh separable nahi hai.

Level 2 — Application

Ritual ke saath kaam karo. Apna arithmetic check karo.

Exercise 2.1

$\iint_{R} (3 x^{2} + 2 y) d A$ compute karo, jahaan $R = [0, 1] \times [0, 2]$ .

Recall Solution 2.1

$x$ ko FREEZE karo, $y$ par FILL karo (inner): $\int_{0}^{2} (3 x^{2} + 2 y) d y = [3 x^{2} y + y^{2}]_{0}^{2} = 6 x^{2} + 4.$ Yahaan $3 x^{2}$ , $y$ mein constant hai, isliye $3 x^{2} y$ milta hai; $2 y$ ka antiderivative $y^{2}$ hai. Outer $x$ -integral mein FILE karo: $\int_{0}^{1} (6 x^{2} + 4) d x = [2 x^{3} + 4 x]_{0}^{1} = 2 + 4 = 6.$ Answer: $6$ .

Exercise 2.2

$\iint_{R} x e^{y} d A$ , $R = [1, 3] \times [0, 1]$ compute karo, separable shortcut use karke, phir confirm karo ki yeh iterated integral ke barabar hai.

Recall Solution 2.2

$f = x \cdot e^{y}$ separable hai, isliye $(\int_{1}^{3} x d x) (\int_{0}^{1} e^{y} d y) = [\frac{x ^{2}}{2}]_{1}^{3} \cdot [e^{y}]_{0}^{1} = \frac{9 - 1}{2} \cdot (e - 1) = 4 (e - 1) .$ Iterate karke confirm karo (inner $y$ par, $x$ frozen): $\int_{1}^{3} x (\int_{0}^{1} e^{y} d y) d x = \int_{1}^{3} x (e - 1) d x = (e - 1) \cdot \frac{9 - 1}{2} = 4 (e - 1) . ✓$ Answer: $4 (e - 1) \approx 6.873$ .

Exercise 2.3

$\int_{0}^{π /2} \int_{0}^{1} y sin x d y d x$ compute karo.

Recall Solution 2.3

$g (x) = sin x$ , $h (y) = y$ ke saath separable hai: $(\int_{0}^{π /2} sin x d x) (\int_{0}^{1} y d y) = [- cos x]_{0}^{π /2} \cdot [\frac{y ^{2}}{2}]_{0}^{1} = (0 - (- 1)) \cdot \frac{1}{2} = \frac{1}{2} .$ Answer: $\frac{1}{2}$ .

Level 3 — Analysis

Ab integration ka order ek choice hai. Woh choose karo jo extra kaam se bachaye.

Exercise 3.1

$\iint_{R} y e^{x y} d A$ , $R = [0, 1] \times [0, 1]$ compute karo. Woh order choose karo jo integration by parts se bachaye.

Recall Solution 3.1

Notice karo ki $\frac{\partial}{\partial x} e^{x y} = y e^{x y}$ — bilkul wohi integrand hai. Toh pehle $x$ par integrate karna (with $y$ frozen) trivial hai: $\int_{0}^{1} y e^{x y} d x = [e^{x y}]_{x = 0}^{x = 1} = e^{y} - 1.$ $y$ -integral mein FILE karo: $\int_{0}^{1} (e^{y} - 1) d y = [e^{y} - y]_{0}^{1} = (e - 1) - (1 - 0) = e - 2.$ Agar hum pehle $y$ par integrate karte, toh $\int y e^{x y} d y$ ke liye integration by parts chahiye hota — same answer, zyada mehnat. Answer: $e - 2 \approx 0.718$ .

Exercise 3.2

$\iint_{R} x cos (x y) d A$ , $R = [0, π /2] \times [0, 1]$ (yahaan $x \in [0, π /2]$ , $y \in [0, 1]$ ) compute karo. Clean order choose karo.

Recall Solution 3.2

Kyunki $\frac{\partial}{\partial y} sin (x y) = x cos (x y)$ , pehle $y$ par integrate karna (freeze $x$ ) directly derivative ko undo karta hai: $\int_{0}^{1} x cos (x y) d y = [sin (x y)]_{y = 0}^{y = 1} = sin x - 0 = sin x .$ FILE karo: $\int_{0}^{π /2} sin x d x = [- cos x]_{0}^{π /2} = 0 - (- 1) = 1.$ Answer: $1$ . (Pehle $x$ par integrate karne par parts ki zaroorat padti.)

Level 4 — Synthesis

Slicing, separability, aur geometry combine karo.

Exercise 4.1

$R = [- 1, 1] \times [- 1, 1]$ par $z = 4 - x^{2} - y^{2}$ ke neeche ke solid ka volume nikalo. (Figure dekho.)

Figure — Double integrals over rectangles — Fubini's theorem

Recall Solution 4.1

Volume $= \iint_{R} (4 - x^{2} - y^{2}) d A$ . Sum ko split karo: $\iint_{R} 4 d A - \iint_{R} x^{2} d A - \iint_{R} y^{2} d A .$ Pehla piece: $4 \cdot area (R) = 4 \cdot (2 \cdot 2) = 16.$ Doosra piece (separable, $x^{2} \cdot 1$ ): $(\int_{- 1}^{1} x^{2} d x) (\int_{- 1}^{1} 1 d y) = \frac{2}{3} \cdot 2 = \frac{4}{3} .$ Teesra piece symmetry se same hai: $\frac{4}{3} .$ $V = 16 - \frac{4}{3} - \frac{4}{3} = 16 - \frac{8}{3} = \frac{40}{3} .$ Surface $R$ par $x y$ -plane ke upar rehti hai (corners par minimum value: $4 - 1 - 1 = 2 > 0$ ), isliye yeh ek sach mein positive volume hai — koi cancellation nahi. Answer: $\frac{40}{3} \approx 13.33$ .

Exercise 4.2

$R = [0, 2] \times [0, 3]$ par $f (x, y) = x y$ ki average value nikalo. (Average value $= \frac{1}{area ( R )} \iint_{R} f d A$ .)

Recall Solution 4.2

$area (R) = 2 \cdot 3 = 6.$ Integral (separable): $\iint_{R} x y d A = (\int_{0}^{2} x d x) (\int_{0}^{3} y d y) = \frac{4}{2} \cdot \frac{9}{2} = 2 \cdot \frac{9}{2} = 9.$ Average value: $\frac{1}{6} \cdot 9 = \frac{3}{2} .$ Answer: $\frac{3}{2}$ .

Level 5 — Mastery

Ek aisa problem jo tumhe pakadne ke liye design kiya gaya hai. Slow down karo.

Exercise 5.1

$\int_{0}^{1} \int_{0}^{1} \frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}} d y d x$ aur $\int_{0}^{1} \int_{0}^{1} \frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}} d x d y$ compute karo. Kya dono orders agree karte hain? Explain karo ki yeh Fubini ke baare mein kya dikhata hai.

Recall Solution 5.1

Key antiderivative fact: $\frac{\partial}{\partial y} (\frac{- y}{x ^{2} + y ^{2}}) = \frac{- ( x ^{2} + y ^{2} ) + y \cdot 2 y}{( x ^{2} + y ^{2} ) ^{2}} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}} = - \frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}} .$ Toh $\frac{\partial}{\partial y} (\frac{y}{x ^{2} + y ^{2}}) = \frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}}$ , exactly hamara integrand.

Order $d y$ first (freeze $x > 0$ ): $\int_{0}^{1} \frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}} d y = [\frac{y}{x ^{2} + y ^{2}}]_{0}^{1} = \frac{1}{x ^{2} + 1} .$ Phir $\int_{0}^{1} \frac{1}{x ^{2} + 1} d x = [arctan x]_{0}^{1} = \frac{π}{4} .$

Order $d x$ first: integrand $x \leftrightarrow y$ swap karne par antisymmetric hai (sign flip ho jaata hai). $x$ mein same antiderivative se, $\int_{0}^{1} \frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}} d x = [\frac{- x}{x ^{2} + y ^{2}}]_{0}^{1} = \frac{- 1}{1 + y ^{2}},$ phir $\int_{0}^{1} \frac{- 1}{1 + y ^{2}} d y = - \frac{π}{4} .$

Dono orders $+ \frac{π}{4}$ aur $- \frac{π}{4}$ dete hain — woh DISAGREE karte hain.

Yahan Fubini humein kyun nahi bachata: integrand corner $(0, 0)$ ke paas blow up ho jaata hai; actually $\iint_{R} \frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}} d A = \infty$ . Fubini require karta hai ki $f$ integrable ho (closed rectangle par continuous ho, ya $\iint ∣ f ∣ < \infty$ ho). Yeh $f$ unbounded aur absolutely integrable nahi hai, isliye orders ki equality guarantee nahi hai — aur waakai mein fail hoti hai. Yeh woh textbook example hai jiske baare mein parent ke third mistake callout ne warn kiya tha. Answer: $+ \frac{π}{4}$ (order $d y d x$ ), $- \frac{π}{4}$ (order $d x d y$ ); Fubini fail hota hai kyunki $f$ , $(0, 0)$ ke paas integrable nahi hai.

Connections

Riemann sums — woh limit jinhe ye integrals secretly hain.
Volume by slicing (single-variable) — Exercise 4.1 ke peeche ka engine.
Change of order of integration — Level 3 rectangles par iska warm-up hai.
Double integrals over general regions — jahaan limits swap karna subtle ho jaata hai.
Fubini–Tonelli theorem — woh exact hypothesis jise Exercise 5.1 violate karta hai.
Triple integrals — same slicing idea, ek dimension aur gehri.

Flashcards

\int_{a}^{b} \int_{c}^{d} f d y d x

mein, pehle kaunsa integral kiya jaata hai?

Inner wala,

y

par (integrand ke sabse paas wala differential).

R

par

f

ki average value kya hoti hai?

\frac{1}{area ( R )} \iint_{R} f d A

\iint_{[0, 1] \times [0, 1]} y e^{x y} d A = ?

e - 2

\int\int \frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}}

ke dono orders kyun differ kar sakte hain?

Integrand

(0, 0)

ke paas unbounded / absolutely integrable nahi hai, isliye Fubini ki hypothesis fail hoti hai.