Double integrals in polar coordinates — Jacobian r: Question bank

4.4.19 · D5 · Maths › Multivariable Calculus › Double integrals in polar coordinates — Jacobian r

True ya false — justify karo

Polar coordinates mein area element

d r d θ

hota hai.

False. Yeh expression ek length aur ek dimensionless angle ka product hai — yeh ek length hai, area nahi. Sahi tile

r d r d θ

hai, kyunki curved side ek arc hai jiskI length

r d θ

hai (dekho Arc Length and Radian Measure).

Factor

r

isliye aata hai kyunki humne convention se

r \geq 0

choose kiya hai.

False.

r

geometric hai: tile ki arc side

r d θ

hai, isliye yeh kisi bhi convention mein exist karta hai.

r \geq 0

convention sirf yeh allow karta hai ki hum absolute value

∣ det J ∣ = ∣ r ∣ = r

drop kar sakein.

Agar koi region origin se door ek patla sa square hai, tab bhi polar coordinates use karne par

r

ka factor integral mein aayega.

True.

r

coordinate change se aata hai, region ki shape se nahi. Off-center square ke liye bhi

d A = r d r d θ

hold karta hai — bas limits messy ho jaati hain, isliye hum yahan polar avoid karte.

Gaussian integral mein,

r

drop karne ke baad bhi koi mushkil antiderivative leke kaam khatam ho sakta hai.

False.

r

ke bina tumhare paas

\int e^{- r^{2}} d r

hoga, jiska koi elementary antiderivative nahi hai.

r

bilkul wahi cheez hai jo substitution

u = r^{2}

ko power deti hai; uske bina method collapse ho jaata hai (dekho Gaussian Integral).

Polar coordinates ka Jacobian kuch quadrants mein negative hota hai.

False.

det J = r cos^{2} θ + r sin^{2} θ = r \geq 0

har jagah hota hai, kyunki

r

ek distance hai. Trig terms har

θ

ke liye

r

mein combine ho jaate hain, isliye koi bhi quadrant sign flip nahi karta.

Kyunki

θ

dimensionless hai, uska differential

d θ

koi units carry nahi karta.

True. Ek radian arc-length divided by radius hota hai, isliye yeh ek pure number hai. Isliye

r d θ

(length

\times

pure number) ki units length hain aur yeh area tile ki ek side ka kaam kar sakta hai.

r

ko ek fixed

d r

se badhane par tile mein har jagah equal amount of area add hoti hai, chahe aap kahaan bhi ho.

False. Tile ki area

r d r d θ

hai, isliye same

d r

aur

d θ

ke liye area $r$ ke saath linearly badhti hai — door wali tile paas wali tile se moti hoti hai, bilkul pizza-slice picture ki tarah.

Error pakdo

\iint_{R} x d A = \iint_{R^{'}} x r d r d θ

, aur

x

constant hai isliye isse bahar nikaalo."

Error yeh hai ki

x

unconverted chhod diya. Tumhe

x = r cos θ

substitute karna hoga, jisse

\iint (r cos θ) r d r d θ

milta hai;

x

(r, θ)

mein constant nahi hai.

"Disk ki area

= \int_{0}^{2 π} \int_{0}^{a} d r d θ = 2 π a

Jacobian missing hai. Answer

2 π a

ek length hai (dimensional alarm), area nahi.

r

restore karne par

\int_{0}^{2 π} \int_{0}^{a} r d r d θ = π a^{2}

milta hai.

x^{2} + y^{2} \leq a^{2}

ke liye main

\int_{0}^{a} \int_{0}^{2 π} r d θ d r

likhूnga same limits ke saath — order matter nahi karta."

Yahan yeh kaam karta hai kyunki dono limits constant hain, lekin stated reason dangerous hai. Order-swapping tabhi free hai jab koi bhi limit doosre variable par depend na kare; jaise hi

r = 2 cos θ

jaisi boundary aaye, swapping se sab toot jaata hai.

"Circle

r = 2 cos θ

ke liye,

r

0

2

tak integrate karo aur

θ

0

2 π

tak."

Do errors hain: outer

r

-limit

2 cos θ

hai (constant

2

nahi), aur yeh circle

θ \in [- \frac{π}{2}, \frac{π}{2}]

se trace hota hai, poora chakkar nahi — poora sweep disk ko double-cover kar deta hai.

d x d y = det J d r d θ

jahan

det J = r

hai, isliye general rule ko absolute value ki zaroorat nahi."

General Change of Variables Theorem mein

∣ det J ∣

chahiye hota hai. Yahan sirf isliye unnecessary lagta hai kyunki

r \geq 0

pehle se

det J

ko non-negative bana deta hai; doosre maps ke liye modulus zaroori hai.

"Kyunki

det J = r

hai aur origin par

r = 0

hai, substitution origin par invalid hai isliye disk integral galat hai."

Jacobian ka ek single point par (zero area ka set) vanish karna koi nuksaan nahi karta — yeh integral mein kuch contribute nahi karta. Origin ek coordinate singularity hai zero measure ka, koi genuine problem nahi.

Why questions

Tile ki arc side

r d θ

kyun use karti hai, sirf

d θ

kyun nahi?

Radian measure arc length ko radius times angle ke roop mein define karta hai: arc

= r d θ

. Angle akela nahi jaanta ki tum kitni door ho;

r

woh distance supply karta hai (dekho Arc Length and Radian Measure).

Polar mein jaane se pehle hum

I = \int e^{- x^{2}} d x

ko square kyun karte hain?

Square karne se ek 1-D integral ek 2-D integral mein convert ho jaata hai poore plane par integrand

e^{- (x^{2} + y^{2})}

ke saath. Tabhi

x^{2} + y^{2} = r^{2}

appear hota hai, jisse polar coordinates aur

r

-Jacobian ise rescue kar paate hain.

Geometric derivation mein second-order term

d r d θ \cdot d r

kyun discard kiya jaata hai?

Tile ki do edges mein fark hota hai arc

(r + d r) d θ

r d θ

; gap

d r^{2} d θ

ke proportional hota hai. Leading

r d r d θ

ke compare mein yeh increments

\to 0

hone par faster shrink karta hai, isliye limit mein vanish ho jaata hai.

Polar disk ki limits ko tame kyun karta hai lekin square ki nahi?

Disk ki boundary

r = a

hai (polar mein ek constant), jisse fixed limits milti hain; square ki sides

x =

const,

y =

const hoti hain, jo tangled

r

–

θ

relations ban jaati hain. Coordinate system ko boundary ki natural symmetry se match karo.

Sirf ek nahi, dono — integrand aur area element — kyun convert karne chahiye?

Yeh alag alag cheezein describe karte hain: integrand

f (x, y)

kya sum kar rahe ho, aur

d A

kitni space har sample cover karta hai. Sirf ek ko convert karna coordinate systems ko mix kar deta hai aur ek meaningless quantity deta hai.

r

same Jacobian kyun hai chahe tum geometry se derive karo ya determinant se?

Dono same physical cheez compute karte hain — coordinate map ka area-stretch. Chhota-box picture aur

∣ det J ∣

dono "yahan area kitna scale hota hai" ke liye do alag languages hain, isliye inhe agree karna hi hai (dekho Jacobian Determinant).

Edge cases

Origin par (

r = 0

), ek tile ki area kya hai, aur kya yeh koi problem hai?

Tile area

r d r d θ \to 0

: saare angles ek point par collapse ho jaate hain, isliye wahan genuinely koi area nahi hai. Yeh ek harmless singularity hai — ek akela point kisi bhi integral mein kuch contribute nahi karta.

Agar

θ

poore disk ke liye

[0, 2 π]

ki jagah

[0, 4 π]

tak jaaye, to kya hoga?

Tum disk ko do baar sweep karte ho, answer

2 π a^{2}

ho jaata hai. Polar coordinates unique nahi hain — tumhe ek

θ

-range choose karna hoga jo region ko exactly ek baar cover kare.

Annulus

1 \leq r \leq 2

ke liye,

θ

-limits abhi bhi

0

2 π

tak constant kyun hain?

Hole origin par centered hai, isliye har angle par same radial slice appear hota hai — boundary mein full rotational symmetry hai.

r

-limits mein saari "ring" information hai;

θ

ek full, constant sweep rehta hai.

Agar sirf first quadrant par integrate karo to Gaussian integral kya ban jaata hai?

Tab

θ

0 \to \frac{π}{2}

tak run karta hai, jisse

I_{quad}^{2} = \frac{π}{2} \cdot \frac{1}{2} = \frac{π}{4}

milta hai, yaani

π

ka ek quarter — consistent hai, kyunki ek quadrant plane ka ek quarter hota hai.

Origin se guzarne wale circle jaise

r = 2 cos θ

ke liye, point

θ = π /2

par tile ki area kya hai?

Wahan

r

0 \to 2 cos (π /2) = 0

tak jaata hai, isliye radial slice ki length zero hai aur koi area contribute nahi hoti — curve us angle par sirf origin ko graze karta hai.

Agar kisi polar-curve convention mein region

r < 0

mein jaata hai, to hum

r \geq 0

kaise rakhein?

Negative

r

ka matlab hai "opposite direction mein plot karo," yaani

θ

mein

π

add karo aur

r

ka sign flip karo. Isse rewrite karne par

r \geq 0

rehta hai aur clean

∣ det J ∣ = r

abhi bhi hold karta hai.

Connections

Double Integrals in Polar Coordinates — Jacobian r — parent jisko yeh traps stress-test karte hain.
Change of Variables Theorem — woh $∣ det J ∣$ jo $r$ ban jaata hai.
Jacobian Determinant — geometric vs algebraic agreement.
Arc Length and Radian Measure — jahan $r d θ$ janam leta hai.
Gaussian Integral — killer edge case.
Area of Regions Bounded by Polar Curves — variable $r$ -limits.