4.4.19
Jab hum kisi region ko chhote polar boxes mein kaatten hain (do radii aur do angles ke beech ek wedge), to har box ka area d r d θ NAHI hota. Jitna door jaate hain center se, utna wider hota hai box. Yeh extra stretch factor exactly == r == d A = r d r d θ . r nahi hai — yeh us chhote curved box ka area hai, jo neeche derive kiya gaya hai.
Cartesian boxes d x d y x y disk, annulus, sector, ya kuch bhi circular symmetry wala hai, to yeh rectangles geometry se ladte hain: boundary x 2 + y 2 = a 2 ( r , θ ) constant limits ke saath describe karte hain, isliye integral trivial ho jaata hai.
Iska price: area element badal jaata hai. Yahi poora subtopic hai.
Definition Polar substitutionPlane mein ek point ko hum likhte hain
x = r cos θ , y = r sin θ , r ≥ 0 origin se distance hai aur θ x r = x 2 + y 2 θ = arctan ( y / x )
Derivation First principles: ek polar box ka areaEk chhoti region lo jo radii r r + d r θ θ + d θ
Radial side ki length d r Radius r arc side ki length = r d θ = ×
Chhote increments ke liye box almost ek rectangle hota hai sides d r r d θ d A ≈ ( d r ) ( r d θ ) = r d r d θ . ( r + d r ) d θ d r d θ ⋅ d r Isliye Jacobian factor r ∎
Derivation First principles: determinantGeneral change-of-variables theorem kehta hai, ( x , y ) = T ( u , v ) d x d y = det ∂ ( u , v ) ∂ ( x , y ) d u d v . ( u , v ) = ( r , θ ) x = r cos θ y = r sin θ
=\begin{vmatrix} \cos\theta & -r\sin\theta\\ \sin\theta & \;\;r\cos\theta\end{vmatrix}.$$
Expand karo:
$$\det J = \cos\theta\,(r\cos\theta) - (-r\sin\theta)(\sin\theta) = r\cos^2\theta + r\sin^2\theta = r.$$
To $|\det J| = r$ (kyunki $r\ge 0$), jo confirm karta hai $dx\,dy = r\,dr\,d\theta$. ∎
∬ R 1 d A x 2 + y 2 ≤ a 2
∫ 0 2 π ∫ 0 a 1 ⋅ r d r d θ .
r d A = r d r d θ Limits r : 0 → a θ : 0 → 2 π Disk mein saare radii a
= ∫ 0 2 π [ 2 r 2 ] 0 a d θ = ∫ 0 2 π 2 a 2 d θ = 2 a 2 ⋅ 2 π = π a 2 . ✓
Forecast Pehle forecast karo, phir verify karoCompute karne se pehle answer predict karo. Tum jaante ho ki disk ka area π a 2 r r ∫ 0 2 π ∫ 0 a d r d θ = 2 π a
Maano I = ∫ − ∞ ∞ e − x 2 d x I 2 = ∫ − ∞ ∞ ∫ − ∞ ∞ e − ( x 2 + y 2 ) d x d y .
Ise square kyun kiya? Do 1-D integrals ka product poore plane par ek 2-D integral ban jaata hai, aur x 2 + y 2
Poore R 2 r : 0 → ∞ θ : 0 → 2 π x 2 + y 2 = r 2 d A = r d r d θ I 2 = ∫ 0 2 π ∫ 0 ∞ e − r 2 r d r d θ .
Yeh ab easy kyun hai: extra r u = r 2 d u = 2 r d r
∫ 0 ∞ e − r 2 r d r = 2 1 ∫ 0 ∞ e − u d u = 2 1 . I 2 = 2 π ⋅ 2 1 = π ⇒ I = π .
Jacobian r closed form mein impossible hai. Iske saath , answer teen lines mein nikal aata hai. r
∬ R ( x 2 + y 2 ) d A 1 ≤ r ≤ 2 0 ≤ θ ≤ π /2
x 2 + y 2 = r 2 d A = r d r d θ ∫ 0 π /2 ∫ 1 2 r 2 ⋅ r d r d θ = ∫ 0 π /2 ∫ 1 2 r 3 d r d θ .
r 3 r 2 r
∫ 1 2 r 3 d r = [ 4 r 4 ] 1 2 = 4 16 − 1 = 4 15 , ∫ 0 π /2 d θ = 2 π . Answer = 4 15 ⋅ 2 π = 8 15 π .
Common mistake Mistake 1:d A = d r d θ r Sahi lagta kyun hai: Cartesian mein tum bas do differentials multiply karte ho, d x d y d r d θ universal lagta hai.
Galat kyun hai: r θ different units hain (r θ d r d θ length hai, area nahi. Arc length r Fix: Hamesha Jacobian saath rakho: d A = r d r d θ π a 2
Common mistake Mistake 2:r ∬ x d A = ∬ x r d r d θ x Fix: Sab kuch convert karo: x → r cos θ ∬ x d A = ∬ ( r cos θ ) r d r d θ
Common mistake Mistake 3: Galat order / limits jo ek-doosre par depend karti hainAgar r = 2 cos θ inner r θ Fix: Pehle region sketch karo; r θ θ
Recall Feynman: ek 12-saal ke bacche ko explain karoPizza kaatne ki imagine karo. Center ke paas slices patli chhoti triangles hoti hain; crust ke paas same-angle wali slice ek badi moti piece hoti hai. To ek "slice" ka fixed size nahi hota — jitna door jaao center se, utni badi hoti hai. Jab hum circles mein area add karte hain, to center se dur wali har tiny tile badi hoti hai, aur "kitni badi" bas kitni door hai woh hai — wahi r r
r "Arc grows with radius: r d θ Box ki curved side ek arc hai, arc = × d r r "Polar area Really needs r
Polar coordinates mein area element kya hai? d A = r d r d θ
Polar double integrals mein extra factor r Box ki curved side ki arc length r d θ d r ⋅ r d θ = r d r d θ
x y x = r cos θ , y = r sin θ
( x , y ) = ( r cos θ , r sin θ ) det ( cos θ sin θ − r sin θ r cos θ ) = r
d r d θ r r θ d r d θ r
∫ 0 2 π ∫ 0 a r d r d θ π a 2
Jacobian r Yeh woh r u = r 2 d u = 2 r d r ∫ e − r 2 r d r 2 1 ∫ e − u d u
∫ − ∞ ∞ e − x 2 d x π (squaring aur polar coordinates ke zariye).
Circle r = 2 cos θ r Nahi — outer r θ r 0 → 2 cos θ
area approx dr times r d theta
Circular symmetry disk annulus sector
Polar coordinates r theta
x = r cos theta, y = r sin theta
Constant integration limits
Tiny polar box, sides dr and r d theta
Master formula for polar integral
Disk area gives pi a squared
