Triple integrals in Cartesian, cylindrical, spherical coordinates: Visual walkthrough

4.4.20 · D2 · Maths › Multivariable Calculus › Triple integrals in Cartesian, cylindrical, spherical coordi

Hum assume karte hain ki tum triple integral ka idea samajhte ho (value × little volume ko add karna) aur yeh bhi ki ek warped little box ko ek stretch factor chahiye — Jacobian. Baaki sab hum scratch se banayenge.

Step 1 — Har coordinate ko point karke naam do

KYA. Kisi bhi algebra se pehle, hum agree karenge ki teeno numbers $(ρ, ϕ, θ)$ ka physical distances aur turns ke roop mein kya matlab hai. Space mein ek point in teeno se locate hota hai:

$ρ$ (Greek "rho") $=$ point ki origin se kitni door hai (ek length). $ρ \geq 0$ .
$ϕ$ (Greek "phi") $=$ polar angle, $+ z$ axis se neeche ki taraf measure kiya jaata hai ("north pole" direction). $0 \leq ϕ \leq π$ .
$θ$ (Greek "theta") $=$ azimuth, yani tum $z$ -axis ke around kitna ghoom gaye, flat $x y$ -plane mein measure kiya jaata hai. $0 \leq θ < 2 π$ .

YEH TEEN KYUN, $x, y, z$ NAHI. Kisi bhi centre-ke-around round cheez ke liye (ball, shell), boundary simply " $ρ =$ constant" hoti hai. Ek clean equation, Cartesian ke nested square roots $z = \pm a^{2} - x^{2} - y^{2}$ ki jagah le leti hai.

PICTURE. Arrows follow karo: lambi chalk-blue arrow $ρ$ hai, upar wali axis se neeche curve karti pink arc $ϕ$ hai, flat padi yellow arc $θ$ hai.

Figure — Triple integrals in Cartesian, cylindrical, spherical coordinates

Step 2 — Angles ko $x, y, z$ mein convert karo

KYA. Hum usi point ke ordinary Cartesian coordinates ko $ρ, ϕ, θ$ use karke likhte hain. Trick yeh hai ki point se $z$ -axis par ek perpendicular giraao, kaam ko do right triangles mein baanto.

KYUN. Hum Jacobian machine (honest tarika $d V$ nikalne ka) tab hi use kar sakte hain jab $x, y, z$ ko $ρ, ϕ, θ$ ke functions ke roop mein jaante hon. Aur right triangles hi wo ek tool hain jo angle ko sides ke ratio mein badalta hai — exactly yahi $sin$ aur $cos$ karte hain: right triangle par, $sin ϕ = \frac{opposite}{hypotenuse}$ aur $cos ϕ = \frac{adjacent}{hypotenuse}$ .

PICTURE. Vertical triangle mein hypotenuse $ρ$ hai, upar ka angle $ϕ$ hai. Axis ke saath waali side ( $ϕ$ ke adjacent) height $z$ hai; aakar waali side (opposite $ϕ$ ) flat plane mein shadow-length $r$ hai.

Us vertical right triangle se: $z = hyp ρ adjacent/hyp cos ϕ, r = hyp ρ opposite/hyp sin ϕ .$

Yahan $r = ρ sin ϕ$ flat shadow circle ka radius hai. Ab $θ$ se flat plane mein spin karo (doosra right triangle, is baar $x y$ -plane mein hypotenuse $r$ ke saath): $x = r cos θ = ρ sin ϕ cos θ, y = r sin θ = ρ sin ϕ sin θ .$

Step 3 — Har coordinate ko thoda sa nudge karo aur dekho kya hilta hai

KYA. Solid ko tiny cells mein kaato by $d ρ$ , $d ϕ$ , $d θ$ ke chhote steps lete hue alag alag. Hum poochhte hain: har nudge ke liye point physically kitna move karta hai? Distance, angle nahi, volume fill karta hai.

EK STEP KYUN, AUR DISTANCE KYUN MEASURE KARO. Volume = (length)×(length)×(length). Angle ek length nahi hai, isliye teen angles multiply karke volume nahi mil sakta. Har angular nudge ko pehle arc length mein convert karna hoga — point actually kitna slide karta hai — volume multiply karne se pehle. Yahi conversion $ρ^{2} sin ϕ$ factor ka poora origin hai.

PICTURE. Teen coloured arrows ek corner point se nikl rahi hain: radial nudge ke liye blue, $ϕ$ -nudge ke liye pink, $θ$ -nudge ke liye yellow. Dhyan do, teeno mutually perpendicular directions mein point karti hain.

Teen edge lengths, ek per nudge:

Nudge	Motion ki direction	Arc length	Kyun
$d ρ$	$ρ$ ke saath seedha bahar	$d ρ$	distance mein change hai hi ek distance
$d ϕ$	ek meridian ke saath (pole-to-pole circle)	$ρ d ϕ$	arc = radius × angle, aur us circle ka radius $ρ$ hai
$d θ$	ek latitude circle ke saath	$ρ sin ϕ d θ$	us circle ka radius $r = ρ sin ϕ$ hai, $ρ$ nahi

Step 4 — Latitude circle kyun chhota hai: $sin ϕ$ factor

KYA. Hum ek sabse aksar bhool jaane wali baat par zoom karte hain: $θ$ -nudge ek chhota arc travel karta hai jitna tum sochoge se, kyunki tum jis circle par spin karte ho woh $ρ$ wala bada circle nahi hai — woh $ρ sin ϕ$ radius wala chhota latitude circle hai.

YEH KYUN MATTER KARTA HAI. Agar tumne naively $ρ d θ$ use kiya (jaise latitude circle sphere ke barabar bada ho), toh tum poles ke paas volume over-count kar lete. Haqeeqat: pole ke paas latitude circle ek dot mein simat jaata hai, toh $θ$ spin karna tumhe almost move nahi karta.

PICTURE. Sphere ko $z = ρ cos ϕ$ height par horizontally kaato. Horizontal cut ek circle hai. Uska radius Step 2 ka flat shadow $r = ρ sin ϕ$ hai — poles ke paas chhota, equator par sabse bada.

$latitude radius = r = ρ 0 at poles 1 at equator sin ϕ ⟹ θ -arc = ρ sin ϕ d θ .$

Step 5 — Teeno edges ko multiply karke volume banao

KYA. Chhota cell (almost exactly) ek rectangular box hai kyunki — Step 3 se — teeno nudges perpendicular directions mein point karti hain. Box ka volume = uske teeno edge lengths ka product.

PERPENDICULAR HONE SE SIRF MULTIPLY KYUN KAR SAKTE HAIN. Square corners wale box ke liye, volume = length × width × height, bina kisi cross-terms ke. (Step 6 ka honest Jacobian confirm karta hai ki yeh teeno directions sach mein orthogonal hain, isliye koi correction nahi chahiye.)

PICTURE. Teeno edges $d ρ$ , $ρ d ϕ$ , $ρ sin ϕ d θ$ right angles par milti hain, ek tiny curved brick box off karti hain.

$d V = radial edge (d ρ) \cdot meridian arc (ρ d ϕ) \cdot latitude arc (ρ sin ϕ d θ) = ρ^{2} sin ϕ d ρ d ϕ d θ .$

$ρ$ 's gino: ek meridian arc se, ek latitude arc se → $ρ^{2}$ . Akela $sin ϕ$ sirf latitude arc se aata hai.

Step 6 — Honest Jacobian se confirm karo

KYA. Geometric argument ne assume kiya tha ki teeno edges perpendicular hain. Ab algebra se prove karvaao. Jacobian $∣ J ∣$ coordinate box $d ρ d ϕ d θ$ aur real volume ke beech exact stretch factor hai — "almost" wali baat nahi.

YEH BHI KYUN KARO. Determinant automatically edges ke beech kisi bhi skew ko handle kar leta hai. Agar hamari picture galat hoti (non-perpendicular edges), toh determinant $ρ^{2} sin ϕ$ se alag hota. Yeh alag nahi hai — toh picture sahi thi.

PICTURE. Partial derivatives ki $3 \times 3$ table, har column Step 3 ka ek edge-direction vector hai.

\begin{vmatrix} \sin\phi\cos\theta & \rho\cos\phi\cos\theta & -\rho\sin\phi\sin\theta\\ \sin\phi\sin\theta & \rho\cos\phi\sin\theta & \rho\sin\phi\cos\theta\\ \cos\phi & -\rho\sin\phi & 0 \end{vmatrix}=\rho^2\sin\phi.$$ Har column ek nudge-direction hai: column 1 hai "$\rho$ badhao", column 2 hai "$\phi$ badhao", column 3 hai "$\theta$ badhao". Edge-vectors ka determinant *hi* box volume hai — Step 5 se exactly match karta hai. Hum $|J|=\rho^2\sin\phi$ lete hain ($0\le\phi\le\pi$ ke saath, $\sin\phi\ge0$, toh absolute-value ka koi drama nahi). --- ## Step 7 — Degenerate cases: formula ko edges par check karo **KYA.** Ek formula tabhi trustworthy hai jab woh apne extremes par sensibly behave kare. Hum teen jagah test karte hain jahan kuch vanish hota hai. **KYUN.** Yahi woh jagahein hain jahan galat formula blow up ya double-count karta. Agar $dV$ finite rehta hai aur wahan shrink karta hai jahan karna chahiye, toh hum beech mein bhi usse believe karte hain. **PICTURE.** North pole ($\phi=0$) aur south pole ($\phi=\pi$): latitude circle ek single point hai, toh yellow $\theta$-arc ki length zero hai. Origin ($\rho=0$): saari edges collapse ho jaati hain. ![[deepdives/dd-maths-4.4.20-d2-s07.png]] - **Poles, $\phi=0$ ya $\phi=\pi$:** $\sin\phi=0\Rightarrow dV=0$. Correct — latitude circle ek dot mein simat gaya, toh ek $\theta$-step *koi* volume sweep nahi karta. $\sin\phi$ ke bina hum yahan kuch nahi se volume fabricate kar lete. - **Origin, $\rho=0$:** $\rho^2=0\Rightarrow dV=0$. Correct — sab kuch ek point mein pile ho jaata hai; koi volume nahi. - **Equator, $\phi=\pi/2$:** $\sin\phi=1$, toh $dV=\rho^2\,d\rho\,d\phi\,d\theta$ — latitude circle full size $\rho$ par hai, sabse mote cells yahan rehte hain. > [!recall]- Poles par $dV=0$ ek integral ko kyun nahi bigaadta? > Ek line (polar axis) ka measure zero hota hai — woh total mein kuch contribute nahi karti. $\sin\phi\to0$ sahi tarike se un cells ko bhookha rakhta hai; yeh kabhi real volume ko vanish nahi karta. ::: Poles measure zero ka ek set hain, isliye unke cells ko kuch nahi tak shrink karna bilkul sahi hai — real solids axis ke *aas paas* ki space fill karte hain. --- ## Ek-picture summary **KYA.** Ek diagram jisme poora derivation hai: tiny spherical brick apne teeno labelled edges ke saath, har ek ka origin trace kiya hua. ![[deepdives/dd-maths-4.4.20-d2-s08.png]] $$\underbrace{d\rho}_{\text{radial}}\times\underbrace{\rho\,d\phi}_{\text{meridian}}\times\underbrace{\rho\sin\phi\,d\theta}_{\text{latitude}}=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta.$$ > [!mnemonic] Do rho aur ek sine > "**Ek $\rho$ neeche swing karne se ($\phi$), ek $\rho$ around spin karne se ($\theta$), aur ek $\sin\phi$ jo poles ko kill karta hai.**" > [!recall]- Feynman retelling — poora walkthrough ek dost ko explain karo > Ek globe imagine karo. Iske andar ek tiny brick ka volume nikalne ke liye, main uske teeno sides measure karta hoon. Pehla side: centre se thoda aur door jao — woh side bas $d\rho$ hai, ek plain distance. Doosra side: north pole se equator ki taraf thoda tilt karo — main $\rho$ radius ke bade pole-to-pole circle par chal raha hoon, toh woh arc $\rho\,d\phi$ hai. Teesra side: axis ke around thoda spin karo — lekin yahan ek catch hai, main bade circle par nahi chal raha, main apne point se guzarne wale *horizontal ring* par chal raha hoon, jo chhota hai. Uska radius $\rho\sin\phi$ hai: equator par mota, poles par zero. Toh woh arc $\rho\sin\phi\,d\theta$ hai. Teeno perpendicular sides multiply karo aur milta hai $\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$. Do $\rho$ do curved sides se aaye; $\sin\phi$ nature ki yaad dilaayi hai ki pole ke paas spin karna tumhe mushkil se move karta hai. Jacobian determinant phir arithmetic double-check karta hai aur agree karta hai. --- ## Connections - [[Triple integrals in Cartesian, cylindrical, spherical coordinates]] — parent note jo yeh result *state* karta hai; yahan hum ise *earn* karte hain. - [[Jacobian and change of variables]] — Step 6 ek worked Jacobian hai; yeh general stretch-factor machine hai. - [[Spherical coordinates geometry]] — Steps 1–2 ki naming aur right triangles. - [[Double integrals & polar coordinates]] — cylindrical/polar ka single $r$ is $\rho^2\sin\phi$ ka flat cousin hai. - [[Center of mass and moments of inertia]] — ball ke upar har aisa integral is $dV$ ko use karta hai.