Exercises — Double integrals in polar coordinates — Jacobian r

$R$ radius $3$ ka disk hai. Ek poore disk mein har point ke liye $r$, $0$ se $3$ tak aur $\theta$ poora ek chakkar $0$ se $2\pi$ tak jhaakta hai. Integrand $1$ ke saath Jacobian $r$ lagao: ${\int }_0^{2\pi }{\int }_0^{3}1\cdot rdrd\theta .$ Bas itna hi kaam hai — recognize karna ki $dA$, $rdrd\theta$ ban jaata hai.

Factor hai ==$r$==. "Jacobian" ka matlab yahan. Jab tum variables change karte ho, chhota area box apna size badalta hai. Jacobian matrix $J$ un charon partial derivatives ki grid hai jo record karti hai ki $r$ aur $\theta$ mein nudge dene par $x$ aur $y$ kaise respond karte hain: $J=\left(\begin{array}{cc}\frac{\partial x}{\partial r}& \frac{\partial x}{\partial \theta }\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial \theta }\end{array}\right)=\left(\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right).$ Iska determinant $\det J$ us box ke area-scaling factor ko measure karta hai (yeh Jacobian Determinant hai, parent mein prove kiya gaya hai): $\det J=\cos \theta (r\cos \theta )-(-r\sin \theta )(\sin \theta )=r{\cos }^2\theta +r{\sin }^2\theta =r.$ Change-of-variables theorem iska absolute value $|\det J|=r$ use karta hai. Kyunki $r$ ek distance hai, $r\ge 0$ satisfy karta hai, isliye $|r|=r$. Factor kabhi negative nahi hota kyunki distance kabhi negative nahi hoti.

Kyunki $x^2+y^2=(r\cos \theta {)}^2+(r\sin \theta {)}^2=r^2({\cos }^2\theta +{\sin }^2\theta )=r^2$: $x^2+y^2=r^2,\sqrt{x^2+y^2}=r.$ Yeh polar integrals mein sabse zyada use hone wala simplification hai: $x^2+y^2$ se bana koi bhi expression sirf $r$ ki function mein collapse ho jaata hai.

$r$ mein inner integral: ${\int }_0^{3}rdr={\left[\frac{r^2}{2}\right]}_0^3=\frac{9}{2}.$ $\theta$ mein outer integral (inner result ek constant hai, toh sirf sweep se multiply karo): ${\int }_0^{2\pi }\frac{9}{2}d\theta =\frac{9}{2}\cdot 2\pi =9\pi .$ Check: yeh radius $3$ ke disk ka area hai, aur $\pi r^2=\pi \cdot 3^2=9\pi$. ✓

Integrand $x^2+y^2\to r^2$; disk radius $2$ matlab $r:0\to 2$, $\theta :0\to 2\pi$; $r$ attach karo: ${\int }_0^{2\pi }{\int }_0^2{r}^2\cdot rdrd\theta ={\int }_0^{2\pi }{\int }_0^2{r}^{3}drd\theta .$ Inner: ${\int }_0^2{r}^{3}dr={\left[\frac{r^4}{4}\right]}_0^2=\frac{16}{4}=4.$ Outer: ${\int }_0^{2\pi }4d\theta =4\cdot 2\pi =8\pi .$

Integrand ko poora convert karo (yeh L1 ka doosra trap ulta hai): $x\to r\cos \theta$. Upper half-disk matlab $\theta :0\to \pi$, radius $1$ isliye $r:0\to 1$. $r$ attach karo: ${\int }_0^{\pi }{\int }_0^1(r\cos \theta )rdrd\theta ={\int }_0^{\pi }\cos \theta d\theta {\int }_0^1{r}^{2}dr.$ Split karo kyunki integrand ek $r$-part aur ek $\theta$-part mein factor hota hai. ${\int }_0^1{r}^{2}dr=\frac{1}{3},{\int }_0^{\pi }\cos \theta d\theta =[\sin \theta {]}_0^{\pi }=0-0=0.$ Answer $=\frac{1}{3}\cdot 0=0$. Symmetry se yeh expected hai: upper half-disk $y$-axis ke baare mein symmetric hai, isliye positive $x$, negative $x$ ko cancel karta hai.

Pehle boundary samjho. $r=2\cos \theta$ ko $r$ se multiply karo: $r^2=2r\cos \theta$, yani $x^2+y^2=2x$, yani $(x-1{)}^2+y^2=1$ — radius $1$ ka circle $(1,0)$ pe centred. Neecha figure yeh circle kaafi rays ke saath draw karta hai origin se; har ray $\theta$ ki ek value hai, aur $r$ origin se lavender boundary tak jaata hai. Angle range: yeh circle right half-plane mein hai aur origin ko touch karta hai. Jaise $\theta$, $-\pi /2$ se $+\pi /2$ tak swing karta hai, origin se ray poore disk mein sweep karti hai (figure mein dashed coral rays ko fan out karte dekho); us range ke bahar $2\cos \theta <0$ hai aur koi region nahi hai. Toh $\theta :-\frac{\pi }{2}\to \frac{\pi }{2}$. Radial range: har ray ke saath, $r$ origin $0$ se boundary $2\cos \theta$ tak jaata hai. Yeh key analysis step hai — outer $r$-limit ek function of $\theta$ hai. ${\int }_{-\pi /2}^{\pi /2}{\int }_0^{2\cos \theta }rdrd\theta ={\int }_{-\pi /2}^{\pi /2}\frac{(2\cos \theta {)}^2}{2}d\theta ={\int }_{-\pi /2}^{\pi /2}2{\cos }^2\theta d\theta .$ Half-angle identity kyun? Hum ${\cos }^2\theta$ ke saath stuck hain, jiska us raw squared form mein koi elementary antiderivative nahi hai. Identity ${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$ square ko un cheezoon ki sum mein rewrite karti hai jinhe hum term-by-term integrate kar sakte hain: constant $\frac{1}{2}$ aur plain cosine $\frac{1}{2}\cos 2\theta$. (Yeh seedha double-angle formula $\cos 2\theta =2{\cos }^2\theta -1$ se aata hai; use ${\cos }^2\theta$ ke liye solve karo aur exactly upar wali line milti hai.) Toh: ${\int }_{-\pi /2}^{\pi /2}2{\cos }^2\theta d\theta ={\int }_{-\pi /2}^{\pi /2}(1+\cos 2\theta )d\theta ={\left[\theta +\frac{1}{2}\sin 2\theta \right]}_{-\pi /2}^{\pi /2}=\left(\frac{\pi }{2}+0\right)-\left(-\frac{\pi }{2}+0\right)=\pi .$ Check: radius $1$ ke circle ka area $\pi \cdot 1^2=\pi$ hai. ✓ (Dekho Area of Regions Bounded by Polar Curves.)

Integrand $y\to r\sin \theta$; limits yahan constant hain (ek asli quarter ring). $r$ attach karo: ${\int }_0^{\pi /2}{\int }_1^2(r\sin \theta )rdrd\theta ={\int }_0^{\pi /2}\sin \theta d\theta {\int }_1^2{r}^{2}dr.$ ${\int }_1^2{r}^{2}dr={\left[\frac{r^3}{3}\right]}_1^2=\frac{8-1}{3}=\frac{7}{3},{\int }_0^{\pi /2}\sin \theta d\theta =[-\cos \theta {]}_0^{\pi /2}=0-(-1)=1.$ Answer $=\frac{7}{3}\cdot 1=\frac{7}{3}$.

Poora plane: $r:0\to \infty$, $\theta :0\to 2\pi$. Integrand $e^{-(x^2+y^2)}\to e^{-r^2}$; $r$ attach karo: ${\int }_0^{2\pi }{\int }_0^{\infty }{e}^{-r^2}rdrd\theta .$ Kyun $r$ hero hai: yeh exactly woh derivative-piece hai jo $u=r^2,du=2rdr$ ke liye chahiye: ${\int }_0^{\infty }{e}^{-r^2}rdr=\frac{1}{2}{\int }_0^{\infty }{e}^{-u}du=\frac{1}{2}{\left[-e^{-u}\right]}_0^{\infty }=\frac{1}{2}(0-(-1))=\frac{1}{2}.$ Phir ${\int }_0^{2\pi }\frac{1}{2}d\theta =\frac{1}{2}\cdot 2\pi =\pi .$ Ab 1-D integral se connect karo. Maano $I={\int }_{-\infty }^{\infty }{e}^{-x^2}dx$. Kyunki $e^{-(x^2+y^2)}=e^{-x^2}{e}^{-y^2}$ aur double integral factor hota hai, $I^2={\int }_{-\infty }^{\infty }{e}^{-x^2}dx{\int }_{-\infty }^{\infty }{e}^{-y^2}dy=\pi \Rightarrow I=\sqrt{\pi }.$ Yeh Gaussian Integral hai — polar $r$ ke bina closed form mein impossible.

Integrand $\frac{1}{\sqrt{x^2+y^2}}=\frac{1}{r}$. Dekho Jacobian ke saath kya hota hai: $dA$ ka $r$, $\frac{1}{r}$ ko cancel karta hai: ${\int }_0^{2\pi }{\int }_1^4\frac{1}{r}\cdot rdrd\theta ={\int }_0^{2\pi }{\int }_1^{4}1drd\theta .$ Inner: ${\int }_1^{4}dr=3$. Outer: ${\int }_0^{2\pi }3d\theta =6\pi$. Annulus ke baare mein note, disk nahi: humne $r=1$ se shuru kiya, $r=0$ se bachte hue. Agar region poora disk hota origin tak, toh integrand $\frac{1}{r}$ wahan blow up karta aur hum improper integral ki convergence check karte. Yahan inner radius $1$ sab kuch finite rakhta hai.

$z\ge 0$ matlab $4-x^2-y^2\ge 0$, yani $x^2+y^2\le 4$ — radius $2$ ka disk. Volume $={\iint }_{R}zdA$. Integrand $z=4-x^2-y^2\to 4-r^2$; disk deta hai $r:0\to 2$, $\theta :0\to 2\pi$; $r$ attach karo: $V={\int }_0^{2\pi }{\int }_0^2(4-r^2)rdrd\theta ={\int }_0^{2\pi }{\int }_0^2(4r-r^3)drd\theta .$ Inner: ${\int }_0^2(4r-r^3)dr={\left[2r^2-\frac{r^4}{4}\right]}_0^2=(8-4)-0=4.$ Outer: ${\int }_0^{2\pi }4d\theta =8\pi .$ Toh $V=8\pi$. (Sanity check: positive hona chahiye aur cylinder $\pi \cdot 2^2\cdot 4=16\pi$ se kam — sach mein $8\pi <16\pi$. ✓)

Region: cardioid $r=1+\cos \theta$ ek baar trace hoti hai jaise $\theta :0\to 2\pi$ ($\theta =0$ pe, $r=2$; $\theta =\pi$ pe, $r=0$ — origin pe cusp). Har $\theta$ ke liye, $r$, $0\to 1+\cos \theta$ tak jaata hai. Integrand $x^2+y^2\to r^2$; $r$ attach karo taaki $r^3$ mile: ${\int }_0^{2\pi }{\int }_0^{1+\cos \theta }{r}^{3}drd\theta ={\int }_0^{2\pi }\frac{(1+\cos \theta {)}^4}{4}d\theta .$ Expand karo $(1+\cos \theta {)}^4=1+4\cos \theta +6{\cos }^2\theta +4{\cos }^3\theta +{\cos }^4\theta$. $[0,2\pi ]$ ke upar standard averages dete hain: ${\int }_0^{2\pi }1d\theta =2\pi ,{\int }_0^{2\pi }\cos \theta d\theta =0,{\int }_0^{2\pi }{\cos }^2\theta d\theta =\pi ,$ ${\int }_0^{2\pi }{\cos }^3\theta d\theta =0,{\int }_0^{2\pi }{\cos }^4\theta d\theta =\frac{3\pi }{4}.$ Andar sum: $2\pi +0+6\pi +0+\frac{3\pi }{4}=8\pi +\frac{3\pi }{4}=\frac{35\pi }{4}$. $4$ se divide karo: $\frac{1}{4}\cdot \frac{35\pi }{4}=\frac{35\pi }{16}$.

Pehli quadrant quarter disk: $r:0\to a$, $\theta :0\to \pi /2$. Convert karo $xy\to (r\cos \theta )(r\sin \theta )=r^2\cos \theta \sin \theta$; $r$ attach karo: ${\int }_0^{\pi /2}{\int }_0^a{r}^2\cos \theta \sin \theta \cdot rdrd\theta ={\int }_0^{\pi /2}\cos \theta \sin \theta d\theta {\int }_0^a{r}^{3}dr.$ ${\int }_0^a{r}^{3}dr=\frac{a^4}{4},{\int }_0^{\pi /2}\cos \theta \sin \theta d\theta ={\left[\frac{{\sin }^2\theta }{2}\right]}_0^{\pi /2}=\frac{1}{2}.$ Answer $=\frac{a^4}{4}\cdot \frac{1}{2}=\frac{a^4}{8}.$ Interpretation: L2.3 ke unlike, yahan kuch symmetry se cancel nahi hota kyunki poore region mein $x,y\ge 0$ hai, isliye $xy\ge 0$ har jagah hai aur integral genuinely positive hai.

Yahan har problem teen hi moves hai. Move 1 ::: Integrand convert karo: $x^2+y^2\to r^2$, $x\to r\cos \theta$, $y\to r\sin \theta$ replace karo — koi $x,y$ nahi bachna chahiye. Move 2 ::: $dxdy$ ko $rdrd\theta$ se replace karo ($r$ kabhi mat chhodho). Move 3 ::: Geometry se limits padho — full disks/annuli ke liye constant, $r=2\cos \theta$ ya cardioid jaise curves ke liye $\theta$-dependent.