Worked examples — Double integrals in polar coordinates — Jacobian r

4.4.19 · D3 · Maths › Multivariable Calculus › Double integrals in polar coordinates — Jacobian r

Yeh page parent topic ka drill floor hai. Parent ne prove kiya tha ki $dA=rdrd\theta$ kyun hota hai. Yahan hum ensure karte hain ki tum har tarah ke problem mein survive kar sako — chahe exam ho ya real world — full-circle sweeps, angle-dependent boundaries, integrands jo vanish ho jaate hain, integrals jo infinity tak jaate hain, aur ek word problem bhi.

Koi bhi symbol aane se pehle, sirf teen facts yaad karo jo tumhe chahiye:

Recall Teen facts jo neeche har example mein use hote hain

• $x=r\cos \theta$ aur $y=r\sin \theta$ (ek point ke Cartesian coordinates uski distance $r$ aur angle $\theta$ se).
• $x^2+y^2=r^2$ (Pythagoras, us chote triangle par jo point axes ke saath banata hai).
• $dA=rdrd\theta$ (chota curved box ka area = seedha side $dr$ × arc side $rd\theta$).

Yahan $==r==$ ka matlab hai "point origin se kitna door hai" aur $==\theta ==$ ka matlab hai "positive $x$-axis se counter-clockwise sweep kiya gaya angle". Dono pehle figure mein dikhaye gaye hain.

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Scenario matrix

Har polar double-integral problem in cells mein se kisi ek mein aata hai. Neeche ke examples in mein se har cell ko kam se kam ek baar cover karte hain.

Cell	Kya cheez ise alag banati hai	Example
A. Full sweep, constant radius	$\theta :0\to 2\pi$, $r:0\to a$ (ek disk)	Ex 1
B. Annulus (beech mein hole)	$r$ $0$ ke upar se shuru hota hai	Ex 2
C. Single quadrant / sector	$\theta$ limited range mein, $\cos \theta ,\sin \theta$ ka sign matter karta hai	Ex 3
D. Angle-dependent $r$-limit	boundary jaise $r=2\cos \theta$	Ex 4
E. Degenerate / zero integrand	integrand origin par ya symmetry se vanish ho jaata hai	Ex 5
F. Infinite / limiting region	$r\to \infty$, improper integral	Ex 6
G. Real-world word problem	mass, average, physical units	Ex 7
H. Exam twist (region "do curves ke beech")	$r$ mein outer minus inner	Ex 8

Hum A → H chalte hain.

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Ex 1 — Cell A: disk (sanity anchor)

Forecast: tum pehle se jaante ho ki disk ka area $\pi r^2$ hota hai. Radius $3$ ke saath yeh $9\pi \approx 28.27$ hona chahiye. Dekho yeh machinery ise reproduce karti hai ya nahi.

1. Region set up karo. $r$ centre se rim tak jaata hai: $r:0\to 3$. Angle poora circle sweep karta hai: $\theta :0\to 2\pi$. Yeh step kyun? "Full sweep, constant radius" exactly Cell A hai — dono limits constants hain.
2. Jacobian ke saath $1$ ka integral likho. $\text{Area}={\int }_0^{2\pi }{\int }_0^{3}1\cdot rdrd\theta .$ $r$ kyun? Kyunki $dA=rdrd\theta$ hota hai. Isko drop karo toh tum lengths add kar rahe ho, areas nahi.
3. Inner integral. ${\int }_0^{3}rdr={\left[\frac{r^2}{2}\right]}_0^3=\frac{9}{2}.$ Inner pehle kyun? Limits constant hain, isliye $r$-integral independently kiya ja sakta hai.
4. Outer integral. ${\int }_0^{2\pi }\frac{9}{2}d\theta =\frac{9}{2}\cdot 2\pi =9\pi .$

Verify: $9\pi \approx 28.27$, $\pi \cdot 3^2$ se match karta hai. Units: $rdr$ hai (length)(length) = area — sahi hai.

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Ex 2 — Cell B: annulus (beech mein hole)

Forecast: yeh ek badi disk (radius 5) minus ek choti disk (radius 2) hai: $\pi (25-4)=21\pi \approx 65.97$.

1. Region. $r:2\to 5$ (note: $2$ se shuru hota hai, $0$ se nahi — yahi hole hai), $\theta :0\to 2\pi$. Yeh step kyun? Cell B, A se sirf lower $r$-limit mein alag hai; origin excluded hai.
2. Integral. ${\int }_0^{2\pi }{\int }_2^{5}rdrd\theta .$
3. Inner. ${\int }_2^{5}rdr={\left[\frac{r^2}{2}\right]}_2^5=\frac{25-4}{2}=\frac{21}{2}.$ Yeh step kyun? Do $r^2/2$ values subtract karna hi "big disk minus small disk" idea hai, jo limits se automatically ho jaata hai.
4. Outer. ${\int }_0^{2\pi }\frac{21}{2}d\theta =21\pi .$

Verify: $21\pi \approx 65.97=\pi (5^2-2^2)$. ✓ (General subtraction idea ke liye Area of Regions Bounded by Polar Curves dekho.)

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Ex 3 — Cell C: single quadrant, signs matter karte hain

Forecast: $x$ first quadrant mein har jagah positive hai, isliye answer positive hona chahiye. Rough size: average $x$ nearly $1$, area $=\frac{1}{4}\pi (2^2)=\pi$, toh kuch $2$–$3$ ke aaspaas guess karo.

1. Integrand convert karo. $x=r\cos \theta$. Yeh step kyun? Parent ka Mistake 2 — tumhe $x$ convert karna hi hoga, as-is nahi chhod sakte.
2. Jacobian ke saath assemble karo. $\iint xdA={\int }_0^{\pi /2}{\int }_0^2(r\cos \theta )rdrd\theta ={\int }_0^{\pi /2}{\int }_0^2{r}^2\cos \theta drd\theta .$ $r^2$ kyun? Ek $r$ $x=r\cos \theta$ se, ek $r$ Jacobian se.
3. Separate karo. Integrand ek $r$-part aur ek $\theta$-part mein factor hota hai: $\left({\int }_0^2{r}^{2}dr\right)\left({\int }_0^{\pi /2}\cos \theta d\theta \right).$ Kyun allowed hai? Kyunki limits constant hain aur integrand ek function of $r$ aur ek function of $\theta$ ka product hai.
4. Dono evaluate karo. ${\int }_0^2{r}^{2}dr=\frac{8}{3}$, aur ${\int }_0^{\pi /2}\cos \theta d\theta =[\sin \theta {]}_0^{\pi /2}=1$. $\cos \theta >0$ yahan kyun? First quadrant mein $\theta \in [0,\pi /2]$, isliye $\cos \theta \ge 0$ — $x$ ka sign kabhi flip nahi hota.
5. Multiply karo. $\frac{8}{3}\cdot 1=\frac{8}{3}$.

Verify: $\frac{8}{3}\approx 2.67>0$, forecast ke anusaar positive hai. Agar hum poore disk par integrate karte, toh symmetry ($x$ right mein positive, left mein negative) $0$ tak cancel ho jaati — Ex 5 dekho.

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Ex 4 — Cell D: angle-dependent radius $r=2\cos \theta$

Boundary $r=2\cos \theta$ ek full circle hai radius $1$ ki, $(1,0)$ par centered — yeh origin se guzarti hai. Yahan inner $r$-limit $\theta$ ka ek function hai, jo sabse tricky exam case hai.

Forecast: yeh radius $1$ ka circle hai, isliye iska area $\pi (1{)}^2=\pi \approx 3.14$ hona chahiye.

1. $\theta$-range nikalo. $r=2\cos \theta \ge 0$ ke liye $\cos \theta \ge 0$ chahiye, yaani $\theta :-\frac{\pi }{2}\to \frac{\pi }{2}$. Yeh step kyun? $r$ ek distance hai aur negative nahi ho sakta; jab $\cos \theta <0$ hoga toh formula negative $r$ maangega, isliye woh angles exclude hain (woh same circle ko re-trace karenge).
2. $r$-limits set karo. Har fixed $\theta$ ke liye, origin se ek ray $r=0$ par enter karti hai aur circle ko $r=2\cos \theta$ par leave karti hai. Isliye $r:0\to 2\cos \theta$. Yeh step kyun? Cell D: outer boundary $\theta$ ke saath bend karti hai, isliye uski value $2\cos \theta$ hai, constant nahi. Figure mein red radius ko dekho — woh $\theta$ sweep karne par bada aur chota hota hai.
3. Integral. ${\int }_{-\pi /2}^{\pi /2}{\int }_0^{2\cos \theta }rdrd\theta .$
4. Inner. ${\int }_0^{2\cos \theta }rdr={\left[\frac{r^2}{2}\right]}_0^{2\cos \theta }=\frac{(2\cos \theta {)}^2}{2}=2{\cos }^2\theta .$ Upper limit squared kyun aata hai? Kyunki $\int rdr=r^2/2$ hota hai aur hum $\theta$-dependent top plug karte hain.
5. Outer. ${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$ use karo: ${\int }_{-\pi /2}^{\pi /2}2{\cos }^2\theta d\theta ={\int }_{-\pi /2}^{\pi /2}(1+\cos 2\theta )d\theta =[\theta +\frac{\sin 2\theta }{2}{]}_{-\pi /2}^{\pi /2}=\pi .$ Identity kyun? ${\cos }^2$ ka by inspection koi elementary antiderivative nahi hai; double-angle identity ise kuch aisa banaa deti hai jo tum integrate kar sako.

Verify: $\pi \approx 3.14=\pi (1{)}^2$, radius-$1$ circle ka area. ✓

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Ex 5 — Cell E: degenerate / vanishing integrand (symmetry se $0$ milta hai)

Forecast: disk $y$-axis ke baare mein symmetric hai. Har point jahan $x>0$ hai, uska ek mirror point hai jahan $x<0$ hai. Woh cancel ho jaate hain. Exactly $0$ predict karo — koi computation nahi chahiye.

1. Convert & assemble karo. ${\int }_0^{2\pi }{\int }_0^4(r\cos \theta )rdrd\theta =\left({\int }_0^4{r}^{2}dr\right)\left({\int }_0^{2\pi }\cos \theta d\theta \right).$ Factor out kyun? Same product structure Ex 3 jaisi.
2. $\theta$ integral isse khatam kar deta hai. ${\int }_0^{2\pi }\cos \theta d\theta =[\sin \theta {]}_0^{2\pi }=\sin 2\pi -\sin 0=0.$ Zero kyun? Ek poore chakkar mein $\cos \theta$ equally positive aur negative rehta hai — ek full cosine cycle ke neeche signed area zero hota hai. Yeh mirror-symmetry ka analytical roop hai.
3. Poora product = 0, chahe $r$-integral ki value kuch bhi ho ($\frac{64}{3}$).

Verify: kuch bhi $\times 0=0$. Symmetry forecast se match karta hai. Yeh degenerate case hai: Jacobian $r$ present aur correct hai, lekin integrand ka sign structure answer ko vanish hone par majboor karta hai.

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Ex 6 — Cell F: infinite region (Gaussian, generalised)

Forecast: parent ne dikhaya tha ki ${\left({\int }_{-\infty }^{\infty }{e}^{-x^2}dx\right)}^2=\pi$, aur yeh double integral wahi square hai. Isliye $\pi \approx 3.14$ predict karo.

1. Polarise karo. $x^2+y^2=r^2$, $dA=rdrd\theta$, region $r:0\to \infty$, $\theta :0\to 2\pi$: ${\int }_0^{2\pi }{\int }_0^{\infty }{e}^{-r^2}rdrd\theta .$ Polar kyun? Integrand sirf origin se distance par depend karta hai — perfect circular symmetry (Cell F meets Cell A but with infinite radius).
2. Substitute $u=r^2$, $du=2rdr$. ${\int }_0^{\infty }{e}^{-r^2}rdr=\frac{1}{2}{\int }_0^{\infty }{e}^{-u}du=\frac{1}{2}[-e^{-u}{]}_0^{\infty }=\frac{1}{2}(0-(-1))=\frac{1}{2}.$ Improper integral converge kyun karta hai? Jab $r\to \infty$, $e^{-r^2}\to 0$ itni tezi se ki $r$ se bhi fast, isliye tail kuch contribute nahi karta — "infinite" region ka finite integral hota hai. Jacobian $r$ exactly woh piece hai jo substitution ko chahiye.
3. Angle. ${\int }_0^{2\pi }\frac{1}{2}d\theta =\pi .$

Verify: $\pi \approx 3.14$, forecast aur Gaussian Integral se match karta hai. ✓

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Ex 7 — Cell G: real-world word problem (disk ka mass)

Forecast: density baahir ki taraf badhti hai, isliye zyaadatar mass rim ke paas hoga. Agar density constant average hoti, toh mass $\bar{\sigma }\cdot \pi a^2$ hota; $k{a}^3$ ka kuch guna expect karo. "Tens of kg" guess karo.

1. Mass = density ka area par integral. $M={\iint }_R\sigma dA={\int }_0^{2\pi }{\int }_0^2(kr)rdrd\theta .$ Yeh step kyun? Mass = density times area, plate par summed. Density $\sigma =kr$ already polar form mein hai.
2. Phir se do $r$'s. $\sigma =kr$ ek $r$ contribute karta hai, Jacobian doosra, giving $k{r}^2$: $M=k{\int }_0^{2\pi }{\int }_0^2{r}^{2}drd\theta .$
3. Inner. ${\int }_0^2{r}^{2}dr=\frac{8}{3}.$
4. Outer & constant. $M=k\cdot \frac{8}{3}\cdot 2\pi =\frac{16\pi k}{3}=\frac{16\pi \cdot 3}{3}=16\pi \approx 50.27\text{kg}.$

Verify — units: $[k]={\text{kg/m}}^3$, $\int r^{2}dr$ ke units hain m³, times dimensionless $\int d\theta$ ⇒ kg. ✓ Numeric: $16\pi \approx 50.27\text{kg}$, forecast ke anusaar "tens of kg".

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Ex 8 — Cell H: exam twist, region do curves ke beech

Forecast: $y=r\sin \theta \ge 0$ first quadrant mein poori jagah, isliye answer positive hai. $r$-integral $1\to 3$ run karta hai (Cell B/H "outer minus inner"), $\theta$-integral $\sin \theta$ pick up karta hai.

1. Integrand convert karo & assemble karo. $y=r\sin \theta$, Jacobian $r$: ${\int }_0^{\pi /2}{\int }_1^3(r\sin \theta )rdrd\theta =\left({\int }_1^3{r}^{2}dr\right)\left({\int }_0^{\pi /2}\sin \theta d\theta \right).$ $r^2$ kyun? Ek $r$ $y$ se, ek Jacobian se.
2. Radial part (outer minus inner). ${\int }_1^3{r}^{2}dr={\left[\frac{r^3}{3}\right]}_1^3=\frac{27-1}{3}=\frac{26}{3}.$ Yeh step kyun? Subtraction hi "do curves ke beech" mechanic hai — bada radius cubed minus chota radius cubed.
3. Angular part. ${\int }_0^{\pi /2}\sin \theta d\theta =[-\cos \theta {]}_0^{\pi /2}=-\cos \frac{\pi }{2}+\cos 0=0+1=1.$ Positive kyun? $\sin \theta \ge 0$ first quadrant mein, isliye $y\ge 0$ har jagah — koi cancellation nahi, Ex 5 ke unlike.
4. Multiply karo. $\frac{26}{3}\cdot 1=\frac{26}{3}.$

Verify: $\frac{26}{3}\approx 8.67>0$, forecast ke anusaar positive hai. ✓

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Case-coverage self-check

Recall Kya humne har cell hit ki?

A (disk) ::: Ex 1 B (annulus) ::: Ex 2 C (sector, signs) ::: Ex 3 D (angle-dependent $r$-limit) ::: Ex 4 E (degenerate / zero by symmetry) ::: Ex 5 F (infinite region) ::: Ex 6 G (word problem, units) ::: Ex 7 H (between two curves, exam twist) ::: Ex 8

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Connections

• Double Integrals in Polar Coordinates — Jacobian r — woh parent jisko yeh drill support karta hai.
• Change of Variables Theorem — sab kuch convert karna (Ex 3, 5, 8) mandatory kyun hai.
• Jacobian Determinant — woh source jahan se har example mein $r$ aata hai.
• Arc Length and Radian Measure — arc side $rd\theta$ kyun hoti hai.
• Gaussian Integral — Ex 6 full mein.
• Area of Regions Bounded by Polar Curves — Ex 2, 4, 8 area-style setups hain.
• Triple Integrals in Cylindrical and Spherical Coordinates — dimensions mein agla step.