Triple integrals in Cartesian, cylindrical, spherical coordinates: Question bank

4.4.20 · D5 · Maths › Multivariable Calculus › Triple integrals in Cartesian, cylindrical, spherical coordi

Pehle, teen words jinhe hum neeche har jagah use karte hain, har ek tumhare dimag mein ek picture se juda hai:

Yahan $r$ $z$ -axis se distance hai, $ρ$ origin se distance hai, $θ$ $z$ -axis ke around ka angle hai, aur $ϕ$ $+ z$ -axis se neeche ka angle hai. In charon pictures ko dhyan mein rakho — neeche har trap in boxes mein se kisi ek ke baare mein hi hai.

True or false — justify

True or false: Agar

f = 1

ho, toh

∭_{E} f d V

E

ka surface area deta hai.

False — yeh

E

ka volume deta hai. Surface area ek 2D integral hai

E

ki boundary ke upar, naa ki uske interior par ek 3D integral.

True or false: Cylindrical coordinates mein tum

r

factor tab chhod sakte ho jab integrand mein koi

r

naa ho.

False —

r

volume element se aata hai (fan-shaped wedge jo

r

ke saath widening hoti hai), integrand se nahi. Yeh

f

kaisi bhi ho, rehta hai.

True or false: Jacobian factor hamesha positive hota hai.

True formula mein, kyunki hum absolute value

∣ J ∣

lete hain. Raw determinant negative ho sakta hai (orientation-reversing), lekin volume kabhi negative nahi hoti, isliye hum sign hata dete hain.

True or false: Ek ball ke liye, spherical coordinates region ko

(ρ, ϕ, θ)

-space mein ek rectangular box mein convert kar dete hain.

True — ball

ρ \leq a

ban jaata hai

0 \leq ρ \leq a, 0 \leq ϕ \leq π, 0 \leq θ < 2 π

, sab constant limits. Woh constant-limit "box" exactly yehi reason hai ki round regions ke liye spherical itna powerful kyun hai.

True or false: Spherical coordinates mein angle

ϕ

ka range

[0, 2 π)

ho sakta hai bilkul

θ

ki tarah.

False —

ϕ \in [0, π]

hi hota hai. Yeh north pole se south pole tak ka tilt measure karta hai;

π

ke baad jaane par woh points dobara cover ho jaate hain jo tum already alag

θ

se reach kar chuke ho.

True or false: Order

d z d y d x

aur

d x d y d z

hamesha ek fixed region ke liye same number dete hain.

True — ek convergent integral ki value order-independent hoti hai (Fubini). Lekin limits usually bilkul change ho jaate hain, aur sirf kuch orders set up karne mein easy hote hain.

True or false:

d V = ρ^{2} sin ϕ d ρ d ϕ d θ

ρ^{2} sin θ d ρ d ϕ d θ

se replace kiya ja sakta hai jab region symmetric ho.

False — yeh kabhi bhi

sin θ

nahi hota. Joh shrinking sine encode karta hai woh poles par hoti hai (

ϕ = 0, π

), jo

ϕ

se govern hoti hai, naa ki around-axis angle

θ

se.

True or false: Cylindrical coordinates bas polar coordinates hain jinke upar ek

z

stack kiya gaya hai.

True — exactly yehi picture hai. Dekho Double integrals & polar coordinates:

r d r d θ

polar area element hai, aur usse

d z

se multiply karne par 3D mein lift ho jaata hai.

Spot the error

Ek student likhta hai

∭_{E} d V = \int_{0}^{1} \int_{0}^{x} \int_{0}^{x y} d z d y d x

aur kehta hai outer limits

0

x

hain. Error?

Outer

x

-limits constants hone chahiye, lekin yahan woh

x

par hi depend kar rahe hain — yeh bakwaas hai. Jab tak tum outer integral tak pahuncho, saare inner variables gone ho chuke hain, isliye sirf ek number hi reh sakta hai.

Ek student

d V = r d r d θ d z

use karta hai lekin

r

[- a, a]

tak range karne deta hai puri disk cover karne ke liye. Error?

Cylindrical coordinates mein

r \geq 0

hamesha; negative

r

points ko double-count karega. Poori disk

r \in [0, a]

aur

θ \in [0, 2 π)

se cover hoti hai.

Ek student

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

ko spherical mein convert karta hai aur boundary ko

ρ^{2} = a^{2}

likhta hai, phir

ρ

- a

a

tak integrate karta hai. Error?

ρ

ek distance hai, isliye

ρ \geq 0

; yeh

0

a

tak jaata hai,

- a

a

tak nahi. Ball ki doosri side angles se reach hoti hai, negative

ρ

se nahi.

Ek student cylinder volume ko

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

(bina

r

ke) compute karta hai aur

2 π ah

paata hai. Error?

Unhone Jacobian

r

chhod diya. Correct integrand

r

hai, jo deta hai

2 π \cdot \frac{a ^{2}}{2} \cdot h = π a^{2} h

r

missing hone se galat units aur galat number milta hai.

Ek student Jacobian ko matrix ke determinant ki jagah partial derivatives ka sum likhta hai. Error?

Local volume stretch derivative matrix ka determinant hota hai (uske columns span karne wale parallelepiped ka signed volume), naa ki koi sum. Dekho Jacobian and change of variables.

Ek student

\int_{0}^{π} sin ϕ d ϕ

integrate karta hai aur "kyunki sine symmetric hai" kehke

0

paata hai. Error?

[0, π]

par,

sin ϕ \geq 0

poore interval mein, isliye integral

2

hai,

0

nahi. Cancellation wali intuition

[0, 2 π]

par

sin

ke liye apply hoti hai, jo ek alag range hai jo

ϕ

kabhi use nahi karta.

Ek student ek hemisphere ko

x y

-plane par project karta hai aur

z

0

a

(ek constant) tak integrate karta hai. Error?

Top surface curved sphere

z = a^{2} - x^{2} - y^{2}

hai, jo

x, y

par depend karti hai — constant

a

nahi. Sirf flat base constant height par hai.

Why questions

Cylindrical volume element mein extra

r

kyun hota hai lekin Cartesian mein kuch nahi?

Ek Cartesian brick har jagah same size ka hota hai, isliye koi correction nahi chahiye. Ek cylindrical wedge fan out karta hai, isliye uski arc-length

r d θ

r

ke saath badhti hai —

r

us growth ko record karta hai.

Spherical mein

ρ^{2}

(do powers) kyun hai lekin cylindrical mein sirf ek

r

Spherical mein do curved arc-edges hain (

ρ d ϕ

aur

ρ sin ϕ d θ

), har ek ek

ρ

contribute karta hai. Cylindrical mein sirf ek curved edge hai (

r d θ

), isliye sirf ek factor.

Poles par

sin ϕ \to 0

physically sense kyun karta hai?

ϕ = 0

ϕ = π

par latitude circles ek single point tak collapse ho jaate hain, isliye

θ

-arc ki length zero hoti hai.

sin ϕ

factor box ko exactly wahan zero tak shrink karta hai, over-counting rokta hai.

Inside-out integrate karna kyun zaroori hai (innermost limits outer variables par depend kar sakte hain, lekin kabhi ulta nahi)?

Har integration apna variable "use up" kar deta hai; integrate hone ke baad, woh variable gone hai. Isliye ek inner limit abhi bhi outer variables dekh sakta hai (jo abhi tak integrate nahi hue), lekin ek outer limit kisi aise inner variable ko nahi dekh sakta jo ab exist hi nahi karta.

x^{2} + y^{2} + z^{2}

ek region ya integrand mein dekhte hi spherical prefer kyun karte hain?

Kyunki

x^{2} + y^{2} + z^{2} = ρ^{2}

ek single clean variable mein collapse ho jaata hai, ugly nested square-root limits ko constant boundary

ρ = a

mein convert kar deta hai.

Local stretch ka sahi measure determinant kyun hai — say, edge lengths ka average kyun nahi?

Ek determinant hai hi column (edge) vectors se span hone wale parallelepiped ka signed volume, jo precisely woh warped-box volume hai jo humein chahiye. Lengths ko average karna edges ke beech ke angles ko ignore kar deta.

Ball-volume integral teen alag 1D integrals mein factor kyun ho jaata hai?

Saari limits constants hain aur integrand

ρ^{2} sin ϕ

(

ρ

ka function)×(

ϕ

ka function)×

1

mein split hota hai. Fubini phir separable factors ke product ko independent integrals ke product banne deta hai.

Edge cases

∭_{E} f d V

kya hoga jab

E

ka volume zero ho (jaise 3D mein baithti ek flat disk)?

Yeh

0

hoga — ek 2D region mein koi 3D volume nahi hota, isliye har

Δ V_{k}

collapse ho jaata hai aur sum vanish ho jaata hai, chahe

f

kuch bhi ho.

Origin

ρ = 0

par, spherical coordinates degenerate hain — yeh integral ke liye harmless kyun hai?

ρ = 0

par volume element

ρ^{2} sin ϕ

zero hai, isliye woh single degenerate point (aur poora polar axis) kuch contribute nahi karta. Wahan

θ, ϕ

mein ambiguity answer ko kabhi affect nahi karti.

z

-axis ke around kholne wale ek cone ke liye, cylindrical ya spherical use karoge?

Dono kaam kar sakte hain, lekin agar cone ki boundary axis se ek fixed angle par hai, toh spherical usse constant

ϕ

banata hai; agar yeh

z

r

se describe hai (jaise

z = r

), toh cylindrical cleaner hai. Woh coordinate choose karo jiske constant surfaces region ke faces hain.

Agar

θ

sirf

[0, π]

tak range kare

[0, 2 π)

ki jagah, toh tumne physically kya restrict kiya?

Tumne

z

-axis ke around sirf aadha rotation rakha hai — ek axis se guzarne wale plane ka ek side. Half-cylinders ya half-balls ke liye useful hai, lekin ise bhoolne par tumhara answer silently aadha ho jaata hai.

Tetrahedron volume

\frac{1}{6}

example mein kya hoga agar tum innermost limit

z

1 - x - y

ki jagah constant set kar do?

Tum ek box par integrate karo ge, tetrahedron par nahi, aur galat region aur galat volume milega. Slanted face

x + y + z = 1

z

-upper-limit

1 - x - y

ke roop mein aana hi chahiye.

Kya density

f = ρ (x, y, z)

(mass density)

∭_{E} ρ d V

mein legitimately negative ho sakti hai?

Physical mass ke liye nahi — density

\geq 0

hai, isliye mass

\geq 0

hai. Mathematically negative "

f

" theek hai (net charge, signed quantities), lekin tab result ek signed total hai, mass nahi; dekho Center of mass and moments of inertia.

Active recall

Recall One-line self-test

Answers cover karo aur har ek ko ek full sentence mein defend karo.

Cylindrical $d V$ mein $r$ kahan se aata hai? ::: Fan-shaped wedge ki arc-length $r d θ$ se, jo axis se door hone ke saath wider hoti hai.
$sin θ$ nahi, $sin ϕ$ kyun? ::: Kyunki latitude circles poles par ( $ϕ = 0, π$ ) shrink hote hain, jise $ϕ$ control karta hai, $θ$ nahi.
Outer limits constants kyun hone chahiye? ::: Kyunki inner variables already integrate ho chuke hain, isliye bahar sirf ek number hi reh sakta hai.
Spherical par sight mein kab switch karo? ::: Jab $x^{2} + y^{2} + z^{2}$ dikhao, kyunki yeh single clean variable $ρ^{2}$ ban jaata hai.

Connections

Triple integrals in Cartesian, cylindrical, spherical coordinates — woh parent jise yeh page stress-test karta hai.
Jacobian and change of variables — yahan har "extra factor" trap ka source.
Double integrals & polar coordinates — cylindrical $r$ ki 2D root.
Spherical coordinates geometry — $ϕ \in [0, π]$ kyun hai aur $sin ϕ$ geometrically kahan se aata hai.
Center of mass and moments of inertia — jahan signed aur weighted integrands appear hote hain.