Worked examples — Surface integrals — scalar and vector (flux)

4.4.31 · D3 · Maths › Multivariable Calculus › Surface integrals — scalar and vector (flux)

Yeh page ek drill floor hai. Parent note ne do machines banaye the:

• scalar: ${\iint }_{S}fdS={\iint }_{D}f\Vert {\mathbf{r}}_u\times {\mathbf{r}}_v\Vert dudv$,
• flux: ${\iint }_S\mathbf{F}\cdot d\mathbf{S}={\iint }_D\mathbf{F}\cdot ({\mathbf{r}}_u\times {\mathbf{r}}_v)dudv$.

Ab hum inhe har tarah ki surface aur har trap se guzaarenge taaki tum koi bhi scenario pehli baar na dekho.

Shuru karne se pehle, do words jinhe hum baar baar use karenge:

• Component of a vector matlab uske teen numbers mein se ek, jaise $\mathbf{F}=(P,Q,R)$ ke components hain $P,Q,R$.
• Normal matlab "surface se seedha bahar nikalna, us par $90^{\circ }$ par" — jaise ek nail ko diwar mein perpendicular thoka ho.

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The scenario matrix

Har surface-integral problem in cells mein se kisi ek mein aata hai. Neeche ke examples [C#] se label hain taaki tum dekh sako ki hum sab cover kar rahe hain.

Cell	Kya cheez ise alag banati hai	Dhyan raho
C1 Graph $z=g(x,y)$, scalar	flat parameter square, square root chahiye	tilted patches flat se bade lagte hain
C2 Graph $z=g(x,y)$, flux	root cancel ho jaata hai, $(-P{g}_x-Q{g}_y+R)$ use karo	normal kis direction mein point kar raha hai?
C3 Explicit parametrization (cylinder / sphere)	tangents $\partial /\partial u,\partial /\partial v$ se	$\mathbf{F}$ mein parametrization plug karo
C4 Orientation sign flip	outward vs inward, upward vs downward	flux ka sign badal jaata hai
C5 Degenerate / zero input	flat plate ($g_x=g_y=0$), ya $\mathbf{F}$ purely sideways	flux bilkul zero ho sakta hai
C6 Closed surface + limiting check	faces jodo; Divergence Theorem se verify karo	kya tumne har face include kiya?
C7 Real-world word problem	density → mass, ya velocity → flow rate	units sahi aane chahiye
C8 Exam twist: constant field through slanted disk	$\mathbf{F}$ constant, tilted surface	sirf perpendicular part count hota hai

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C1 — Scalar integral over a slanted plane (sheet ki mass)

Forecast: $xy$-plane mein flat triangle ki area $\frac{1}{2}\cdot 2\cdot 2=2$ hai. Tilted sheet badi hogi. Guess karo: kya tumhara answer $>2$ hoga?

1. Graph ke roop mein parametrize karo $\mathbf{r}(x,y)=(x,y,2-x-y)$. Yeh step kyun? Surface $z=g(x,y)$ ke roop mein likhi hai, isliye hum parent note se graph formula reuse karte hain — koi nayi parametrization banane ki zaroorat nahi.
2. Slopes: $g_x=-1,g_y=-1$. Yeh step kyun? Stretch factor $\sqrt{1+g_x^2+g_y^2}$ ke liye ye slopes chahiye; ye measure karte hain ki plane kitni tezi se uthti hai.
3. Stretch factor $\sqrt{1+(-1{)}^2+(-1{)}^2}=\sqrt{3}$. Yeh step kyun? Yeh woh number hai jo kehta hai "neeche ka har flat patch slanted area ka $\sqrt{3}$ guna cover karta hai." Figure dekho — shadow patch aur uska slanted parent.
4. Integrate karo: kyunki factor constant hai, area $=\sqrt{3}\cdot (\text{area of }D)=\sqrt{3}\cdot 2=2\sqrt{3}$. Yeh step kyun? Ek constant double integral se bahar aa jaata hai; bacha hua ${\iint }_{D}dxdy$ bas base triangle ki area hai.

Verify karo: $2\sqrt{3}\approx 3.464$. Sanity: ek plane jis slope par woh horizontal run par $2$ uthti hai — tilt factor $\sqrt{3}\approx 1.73$ ek reasonable "shadow se kitna lamba" hai. ✔

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C2 — Usi graph se flux (koi square root nahi!)

Forecast: $\mathbf{F}$ seedha upar point karta hai strength $z$ ke saath. Surface upar ko tilt hai. Guess karo: kya flux positive hoga?

1. Graph-flux formula yaad karo: upward normal deta hai ${\iint }_D(-P{g}_x-Q{g}_y+R)dxdy$ jahan $\mathbf{F}=(P,Q,R)$. Yeh step kyun? Flux seedha ${\mathbf{r}}_x\times {\mathbf{r}}_y=(-g_x,-g_y,1)$ use karta hai — norm cancel ho jaata hai, isliye koi square root nahi (C1 se contrast karo).
2. Read off karo $P=0,Q=0,R=z=2-x-y$. Toh integrand $=-0-0+(2-x-y)=2-x-y$. Yeh step kyun? Hum surface ka $z=2-x-y$, $R$ mein substitute karte hain; integrate karne se pehle sab kuch $x,y$ ka function banna chahiye.
3. Triangle par integrate karo: ${\int }_0^2{\int }_0^{2-x}(2-x-y)dydx.$ Yeh step kyun? Fixed $x$ ke liye, $y$ $0$ se line $x+y=2$ tak jaata hai, yaani $y=2-x$.
4. Inner: ${\left[(2-x)y-\frac{y^2}{2}\right]}_0^{2-x}=(2-x{)}^2-\frac{(2-x{)}^2}{2}=\frac{1}{2}(2-x{)}^2$. Outer: ${\int }_0^2\frac{1}{2}(2-x{)}^{2}dx=\frac{1}{2}\cdot \frac{(2-x{)}^3}{-3}{|}_0^2\cdot (-1)=\frac{1}{2}\cdot \frac{8}{3}=\frac{4}{3}$.

Verify karo: flux $=\frac{4}{3}>0$, forecast se match karta hai (up-field up-tilted surface se through). Dhyan do ki humne kabhi $\sqrt{3}$ nahi likha — yahi flux ki khoobsoorti hai. ✔

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C3 — Explicit parametrization: cylinder ka side

Forecast: $\mathbf{F}$ radially outward point karta hai (iska $xy$-part position hai). Wall ka normal bhi radially bahar point karta hai. Toh field wall ko seedha pierce karta hai har jagah. Guess karo: bada positive flux.

1. Wall ko parametrize karo angle $\theta$ aur height $z$ ke saath: $\mathbf{r}(\theta ,z)=(2\cos \theta ,2\sin \theta ,z),\theta \in [0,2\pi ],z\in [0,5].$ Yeh step kyun? Cylinder graph $z=g(x,y)$ nahi hai (woh vertical hai, $z$ ko $x,y$ ka function nahi bana sakte). Isliye hum do natural knobs chunte hain: ghoomna ($\theta$) aur upar jaana ($z$).
2. Tangent vectors ${\mathbf{r}}_{\theta }=(-2\sin \theta ,2\cos \theta ,0)$, ${\mathbf{r}}_z=(0,0,1)$. Yeh step kyun? ${\mathbf{r}}_{\theta }$ circular direction mein chalta hai, ${\mathbf{r}}_z$ seedha upar; unka cross product dono ke normal hai, yaani wall se bahar nikal raha hai.
3. Cross product Cross product use karke: ${\mathbf{r}}_{\theta }\times {\mathbf{r}}_z=(2\cos \theta ,2\sin \theta ,0).$ Yeh step kyun? Iska $z$-component $0$ hai aur iska $xy$-part outward point karta hai (position ke same direction mein) — exactly woh outward normal jo problem chahta hai. Figure dekho.
4. $\mathbf{F}$ mein substitute karo: wall par $\mathbf{F}=(2\cos \theta ,2\sin \theta ,0)$. Normal se dot karo: $\mathbf{F}\cdot ({\mathbf{r}}_{\theta }\times {\mathbf{r}}_z)=4{\cos }^2\theta +4{\sin }^2\theta =4.$ Yeh step kyun? ${\cos }^2+{\sin }^2=1$ ise constant mein collapse kar deta hai — field wall ko har jagah same strength ke saath perpendicularly milta hai.
5. Integrate karo: ${\int }_0^5{\int }_0^{2\pi }4d\theta dz=4\cdot 2\pi \cdot 5=40\pi .$

Verify karo: wall area $2\pi rh=2\pi \cdot 2\cdot 5=20\pi$ hai, aur $\mathbf{F}\cdot \hat{\mathbf{n}}=2$ (radial strength $=r=2$) constant, toh flux $=2\cdot 20\pi =40\pi$. ✔ Match karta hai.

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C4 — Orientation flip: same cylinder, inward normal

Forecast: humne normal ko cylinder ke andar point karne ke liye flip kiya. Instantly answer guess karo.

1. Cross-product order swap karo: inward normal $={\mathbf{r}}_z\times {\mathbf{r}}_{\theta }=-({\mathbf{r}}_{\theta }\times {\mathbf{r}}_z)$. Yeh step kyun? Parent note ka mistake box: ${\mathbf{r}}_u\times {\mathbf{r}}_v\to {\mathbf{r}}_v\times {\mathbf{r}}_u$ reverse karna sign flip karta hai. Orientation ek choice hai aur yeh answer ka sign badal deta hai.
2. Har integrand apna negative ban jaata hai: $-4$.
3. Integrate karo: $-40\pi$.

Verify karo: flux ka sign flip hua, $|value|$ unchanged raha: $-40\pi$. Negative flux matlab field bahar se ja rahi hai, yaani inward normal ke against flow kar rahi hai. ✔ Yeh "flux has a sign" lesson concretely dikhayi di.

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C5 — Degenerate / zero cases

Forecast (a): plate bilkul tilted nahi hai. Iska area apne shadow ke barabar hona chahiye, $1$. Forecast (b): field horizontal hai, surface flat hai aur iska normal seedha upar point karta hai — kya koi fluid through jaata hai?

Part (a):

1. $g(x,y)=3\Rightarrow g_x=g_y=0$. Kyun? Ek constant ka slope zero hota hai — koi tilt nahi, toh koi stretching nahi.
2. Stretch factor $\sqrt{1+0+0}=1$. Area $={\iint }_{D}1dxdy=1$. Yeh kyun important hai: yeh C1 ka degenerate end hai — square root $1$ ban jaata hai aur sheet apne shadow ke barabar ho jaati hai. Pure $\sqrt{1+g_x^2+g_y^2}$ machinery ka sanity check.

Part (b):

1. Disk plane $z=0$ mein hai; upward normal $\hat{\mathbf{n}}=(0,0,1)$. Kyun? Disk horizontal hai, toh "seedha bahar" matlab seedha upar.
2. $\mathbf{F}\cdot \hat{\mathbf{n}}=(1,0,0)\cdot (0,0,1)=0$. Yeh step kyun? Dot product $\mathbf{F}$ ka woh part pick karta hai jo normal ke saath hai. Ek horizontal field ka vertical direction mein kuch nahi hota — woh disk ke across slide karta hai, kabhi through nahi.
3. Flux $={\iint }_{D}0dS=0$.

Verify karo: (a) $=1$, (b) $=0$. Dono forecasts se match karte hain. (b) pure "sideways flow slides along, not through" picture hai — flux genuinely zero hai, bas chhota nahi. ✔

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C6 — Closed surface + Divergence-Theorem limiting check

Forecast: closed surface, divergence $\nabla \cdot \mathbf{F}=1+1+1=3$. Divergence Theorem se total flux hona chahiye $3\times (\text{volume})$. Volume $=\pi r^{2}h=\pi \cdot 4\cdot 5=20\pi$. Guess: $60\pi$.

Hum teen faces ka sum karte hain — tumhe har face include karna hoga warna undercount hoga.

1. Side wall. C3-style kaam se $\mathbf{F}=(x,y,z)$ ke saath: $\mathbf{F}\cdot ({\mathbf{r}}_{\theta }\times {\mathbf{r}}_z)=(2\cos \theta ,2\sin \theta ,z)\cdot (2\cos \theta ,2\sin \theta ,0)=4$. Flux${}_{\text{side}}={\int }_0^5{\int }_0^{2\pi }4d\theta dz=40\pi$. Kyun: $\mathbf{F}$ ka $z$-part wall ke zero $z$-normal se dot karta hai → drop ho jaata hai, wahi $4$ pehle jaisa.
2. Top disk $z=5$, outward normal $=(0,0,1)$. $\mathbf{F}\cdot \hat{\mathbf{n}}=z=5$. Area $=\pi \cdot 2^2=4\pi$. Flux${}_{\text{top}}=5\cdot 4\pi =20\pi$. Normal $(0,0,1)$ kyun hai: top par "outward" matlab upward.
3. Bottom disk $z=0$, outward normal $=(0,0,-1)$ (outward yahan downward matlab hai!). $\mathbf{F}\cdot \hat{\mathbf{n}}=(x,y,0)\cdot (0,0,-1)=0$. Flux${}_{\text{bottom}}=0$. Normal $(0,0,-1)$ kyun hai: bottom face ka bahar neeche ki taraf hai — ek classic sign trap (C4 flavour).
4. Total $=40\pi +20\pi +0=60\pi$.

Verify karo (limiting / theorem check): Divergence Theorem deta hai $\iiint 3dV=3\cdot 20\pi =60\pi$. ✔ Exactly match karta hai, aur usne humein bottom face ka normal flip karna yaad dilaya.

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C7 — Real-world word problem: roof par rainfall

Forecast: baarish neeche aati hai, roof upar face karta hai. Hum dekhna chahte hain ki kitna through jaata hai har second — m³/s mein flow rate. Sign convention guess karo: upward normal use karo, phir physically-caught rate $-$(flux) hoga kyunki baarish neeche jaati hai.

1. Graph flux formula, $g=4-\frac{1}{2}x$, toh $g_x=-\frac{1}{2},g_y=0$, upward normal $(-g_x,-g_y,1)=(\frac{1}{2},0,1)$. Kyun: roof $z=g(x,y)$ hai, seedha graph case.
2. $\mathbf{F}=(0,0,-3)=(P,Q,R)$, integrand $=-P{g}_x-Q{g}_y+R=0-0+(-3)=-3$. Yeh step kyun: sirf $R=-3$ bachta hai kyunki $P=Q=0$ (baarish ka horizontal motion nahi). Dot product sirf vertical part rakhta hai.
3. Base rectangle par integrate karo (area $2\times 3=6$): flux$=-3\cdot 6=-18$ m³/s. Base area kyun, roof area nahi: graph-flux formula pehle se flat $xy$-region $D$ par integrate karta hai; tilt $(-g_x,-g_y,1)$ normal mein baked in hai (koi alag  ​ nahi).
4. Har second paani pakda gaya $=-(\text{upward flux})=18$ m³/s. Negate kyun: upward flux negative hai kyunki baarish upward normal ke opposite flow karti hai; actually collect ki gayi amount woh magnitude hai.

Verify karo (units & geometry): rain speed $3$ m/s $\times$ horizontal footprint $6$ m² $=18$ m³/s. Horizontal footprint vertical baarish ko pakadta hai — roof ki tilt vertical baarish ke liye catch nahi badhati, isliye $\frac{1}{2}$ slope cancel ho gaya. Units: (m/s)(m²) $=$ m³/s. ✔

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C8 — Exam twist: constant field through a tilted disk

Forecast: constant field ka flat surface se flux $=(\mathbf{F}\cdot \hat{\mathbf{n}})\times (\text{area})$ — koi integral nahi kyunki kuch vary nahi karta. Guess karo: positive lekin $6\pi$ se kam kyunki plate wind se duri par tilt hai.

1. Dot product $\mathbf{F}\cdot \hat{\mathbf{n}}=(0,0,6)\cdot \frac{1}{\sqrt{3}}(1,1,1)=\frac{6}{\sqrt{3}}=2\sqrt{3}$. Kyun: sirf $\mathbf{F}$ ka component $\hat{\mathbf{n}}$ ke along through jaata hai; dot product use uthata hai.
2. Plate ki area $=\pi r^2=\pi$. Kyun: field aur normal dono constant hain, toh $\iint (\mathbf{F}\cdot \hat{\mathbf{n}})dS=(\mathbf{F}\cdot \hat{\mathbf{n}})\iint dS=(\mathbf{F}\cdot \hat{\mathbf{n}})\cdot \text{Area}$.
3. Flux $=2\sqrt{3}\cdot \pi =2\sqrt{3}\pi$.

Verify karo: $2\sqrt{3}\pi \approx 10.88$. Compare karo seedha wind face karne wali plate se ($\hat{\mathbf{n}}=(0,0,1)$): flux hota $6\pi \approx 18.85$. Hamari tilted plate us ka fraction $\frac{2\sqrt{3}}{6}=\frac{1}{\sqrt{3}}\approx 0.577$ pakadti hai — woh tilt "cosine" — jo match karta hai: sirf $57.7\%$ wind perpendicular hai. ✔ Positive aur $6\pi$ se kam, jaise forecast tha.

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Recall Har cell ke liye one-line recall

C1 slanted-plane area apne shadow se badi hoti hai ::: $\sqrt{1+g_x^2+g_y^2}$ ke factor se C2 graph flux integrand (upward) hai ::: $-P{g}_x-Q{g}_y+R$, koi square root nahi C3 cylinder-wall outward normal hai ::: ${\mathbf{r}}_{\theta }\times {\mathbf{r}}_z=(2\cos \theta ,2\sin \theta ,0)$ C4 normal reverse karna flux ke saath kya karta hai? ::: iska sign flip karta hai C5 horizontal disk se horizontal field ka flux hai ::: exactly $0$ C6 closed-surface flux check karne ke liye use karo ::: the Divergence Theorem, $\iiint \nabla \cdot \mathbf{F}dV$ C7 vertical baarish pakadna depend karta hai ::: horizontal footprint area par, tilt par nahi C8 constant field, flat plate: flux $=$ ::: $(\mathbf{F}\cdot \hat{\mathbf{n}})\times \text{Area}$

Related machinery: Double integrals · Jacobian and change of variables · Cross product · Divergence Theorem · Stokes' Theorem · Line integrals.