Exercises — Surface integrals — scalar and vector (flux)

WHAT hum karte hain: map ko har parameter direction mein differentiate karo. ${\mathbf{r}}_u=(1,0,2),{\mathbf{r}}_v=(0,1,3).$ WHY: $u$ ko nudge karna hume $(1,0,2)$ ke along move karta hai; $v$ ko nudge karna $(0,1,3)$ ke along. Ye chhote parallelogram ke do edges hain — upar figure mein yellow aur pink arrows. Cross product ($\hat{\mathbf{i}},\hat{\mathbf{j}},\hat{\mathbf{k}}$ ka determinant do rows ke upar): ${\mathbf{r}}_u\times {\mathbf{r}}_v=(0\cdot 3-2\cdot 1,2\cdot 0-1\cdot 3,1\cdot 1-0\cdot 0)=(-2,-3,1).$ Uski length: $\Vert {\mathbf{r}}_u\times {\mathbf{r}}_v\Vert =\sqrt{(-2{)}^2+(-3{)}^2+1^2}=\sqrt{14}.$ Toh $dS=\sqrt{14}dudv$. Kyunki plane flat hai, yeh stretch har jagah same hai — constant.

$z$-component $+1>0$ hai, toh normal upward point karta hai. Plane ko downward orient karne ke liye, ${\mathbf{r}}_v\times {\mathbf{r}}_u=(2,3,-1)$ use karo, jo har sign flip kar deta hai.

Scalar engine mein $f=1$ set karo. L1.1 se, $dS=\sqrt{14}dudv$. $\text{Area}={\iint }_{D}1\cdot \sqrt{14}dudv=\sqrt{14}\cdot (\text{unit square ka area})=\sqrt{14}\approx \mathrm{3.742.}$ Sanity: flat base ka area $1$ hai; plane tilted hai, toh asli patch exactly stretch $\sqrt{14}$ se badi hai. ✔ (Yeh sirf ek constant ka Double integrals hai.)

WHAT: graph formula $dS=\sqrt{1+g_x^2+g_y^2}dxdy$ use karo. Yahan $g=x+y\Rightarrow g_x=1,g_y=1$, toh $\sqrt{1+1^2+1^2}=\sqrt{3}.$ Surface par $z=g=x+y$, toh ${\iint }_{S}zdS={\iint }_D(x+y)\sqrt{3}dxdy=\sqrt{3}{\iint }_D(x+y)dxdy.$ Base integral triangle $x+y\le 1$ par: ${\iint }_D(x+y)dxdy={\int }_0^1{\int }_0^{1-x}(x+y)dydx.$ Inner: ${\int }_0^{1-x}(x+y)dy=x(1-x)+\frac{1}{2}(1-x{)}^2$. Expand karo: $x-x^2+\frac{1}{2}(1-2x+x^2)=\frac{1}{2}-\frac{1}{2}{x}^2$. Outer: ${\int }_0^1(\frac{1}{2}-\frac{1}{2}{x}^2)dx=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}.$ $\boxed{{\iint }_{S}zdS=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}\approx \mathrm{0.5774.}}$

Flux-through-a-graph shortcut ${\iint }_D(-P{g}_x-Q{g}_y+R)dxdy$ use karo upward normal $(-g_x,-g_y,1)$ ke saath. (Yaad karo $\mathbf{F}=(P,Q,R)$; $(P,Q,R)$ ko $(-g_x,-g_y,1)$ mein dot karna exactly $-P{g}_x-Q{g}_y+R$ deta hai.) Yahan $P=0,Q=0,R=z=g=1-x-y$, aur $g_x=g_y=-1$ (lekin $P=Q=0$ toh woh terms vanish ho jaati hain). $\text{Flux}={\iint }_D(1-x-y)dxdy.$ L2.2 ke method se symmetry se: ${\int }_0^1{\int }_0^{1-x}(1-x-y)dydx$. Inner $=\frac{1}{2}(1-x{)}^2$. Outer $={\int }_0^1\frac{1}{2}(1-x{)}^{2}dx=\frac{1}{6}$. $\boxed{\text{Flux}=\frac{1}{6}\approx \mathrm{0.1667.}}$

Downward normal $(g_x,g_y,-1)$ hai — har sign flip. Toh flux sign flip kar leta hai: ${\text{Flux}}_{\text{down}}=-\frac{1}{6}\approx -\mathrm{0.1667.}$ Kyun: flux net flow through measure karta hai; reverse karo ki kaun si side ko "out" kehte ho aur wahi flow ab entering count hota hai, yaani negative. Scalar surface integrals unchanged rahenge (length ka koi sign nahi).

Tangents: ${\mathbf{r}}_{\theta }=(-\sin \theta ,\cos \theta ,0),{\mathbf{r}}_z=(0,0,1).$ Cross product (yeh hamara vector area element hai): ${\mathbf{r}}_{\theta }\times {\mathbf{r}}_z=(\cos \theta \cdot 1-0,0-(-\sin \theta )\cdot 1,0)=(\cos \theta ,\sin \theta ,0).$ Orientation check karo: $\theta =0$ par yeh $(1,0,0)$ hai, axis se seedha bahar point karta hai — outward. ✔ Field se dot karo ($\mathbf{F}=(x,y,0)=(\cos \theta ,\sin \theta ,0)$ wall par): $\mathbf{F}\cdot ({\mathbf{r}}_{\theta }\times {\mathbf{r}}_z)={\cos }^2\theta +{\sin }^2\theta =1.$ $\text{Flux}={\int }_0^2{\int }_0^{2\pi }1d\theta dz=2\pi \cdot 2=4\pi \approx \mathrm{12.566.}$ Meaning: field wall ke har point par strength $1$ ke saath seedha bahar push karta hai; total $=$ (wall area) $=2\pi rh=4\pi$. ✔

Direct — teen pieces:

• Side wall: L3.2 se, flux $=4\pi$.
• Top disk ($z=2$, outward normal $=(0,0,1)$): $\mathbf{F}\cdot (0,0,1)=0$ kyunki $\mathbf{F}$ mein koi $z$-component nahi. Flux $=0$.
• Bottom disk ($z=0$, outward normal $=(0,0,-1)$): $\mathbf{F}\cdot (0,0,-1)=0$. Flux $=0$. Total $=4\pi +0+0=4\pi$.

Divergence Theorem ke through: ${\iint }_{\partial V}\mathbf{F}\cdot d\mathbf{S}={\iiint }_V(\nabla \cdot \mathbf{F})dV$. $\nabla \cdot \mathbf{F}=\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial 0}{\partial z}=1+1+0=2.$ Cylinder ka volume $=\pi r^{2}h=\pi \cdot 1\cdot 2=2\pi$. ${\iiint }_{V}2dV=2\cdot 2\pi =4\pi .$ Dono $4\pi$ dete hain. ✔ Caps kuch contribute nahi karte kyunki flow purely horizontal hai.

Graph formula: $g=\sqrt{x^2+y^2}$, toh $g_x=\frac{x}{\sqrt{x^2+y^2}},g_y=\frac{y}{\sqrt{x^2+y^2}},$ $1+g_x^2+g_y^2=1+\frac{x^2+y^2}{x^2+y^2}=1+1=2(\text{tip se door}).$ Toh $dS=\sqrt{2}dxdy$ aur surface par $z=\sqrt{x^2+y^2}$: $\text{Mass}={\iint }_D\sqrt{x^2+y^2}\cdot \sqrt{2}dxdy.$ Polar mein switch karo (Jacobian and change of variables: $dxdy=\rho d\rho d\theta$, aur $\sqrt{x^2+y^2}=\rho$): $=\sqrt{2}{\int }_0^{2\pi }{\int }_0^1\rho \cdot \rho d\rho d\theta =\sqrt{2}\cdot 2\pi \cdot \frac{1}{3}=\frac{2\sqrt{2}\pi }{3}\approx \mathrm{2.962.}$ Tip ka note: $g_x,g_y$ origin par blow up karte hain, lekin integrand $\rho \cdot \rho$ wahan itni tezi se vanish ho jaata hai — integral finite hai. ✔

Route chosen — sphere parametrization (toh outward normal clean hai): $\mathbf{r}(\varphi ,\theta )=(\sin \varphi \cos \theta ,\sin \varphi \sin \theta ,\cos \varphi ),\varphi \in [0,\frac{\pi }{2}],\theta \in [0,2\pi ].$ ${\mathbf{r}}_{\varphi }\times {\mathbf{r}}_{\theta }=\sin \varphi \mathbf{r}$ yahan kyun — local justification: map differentiate karo, ${\mathbf{r}}_{\varphi }=(\cos \varphi \cos \theta ,\cos \varphi \sin \theta ,-\sin \varphi ),{\mathbf{r}}_{\theta }=(-\sin \varphi \sin \theta ,\sin \varphi \cos \theta ,0).$ Cross product component by component lo:

• $x$-comp: $\cos \varphi \sin \theta \cdot 0-(-\sin \varphi )\cdot \sin \varphi \cos \theta ={\sin }^2\varphi \cos \theta$,
• $y$-comp: $(-\sin \varphi )(-\sin \varphi \sin \theta )-\cos \varphi \cos \theta \cdot 0={\sin }^2\varphi \sin \theta$,
• $z$-comp: $\cos \varphi \cos \theta \cdot \sin \varphi \cos \theta -\cos \varphi \sin \theta \cdot (-\sin \varphi \sin \theta )=\sin \varphi \cos \varphi$.

Toh ${\mathbf{r}}_{\varphi }\times {\mathbf{r}}_{\theta }=\sin \varphi (\sin \varphi \cos \theta ,\sin \varphi \sin \theta ,\cos \varphi )=\sin \varphi \mathbf{r}$. Yeh $\mathbf{r}$ ke along point karta hai (radially outward) length $\sin \varphi$ ke saath — exactly spherical area element. Surface par $z=\cos \varphi$, toh $\mathbf{F}=(0,0,\cos \varphi )$, aur $\mathbf{r}=(\dots ,\dots ,\cos \varphi )$: $\mathbf{F}\cdot (\sin \varphi \mathbf{r})=\sin \varphi (0\cdot r_x+0\cdot r_y+\cos \varphi \cdot \cos \varphi )=\sin \varphi {\cos }^2\varphi .$ $\text{Flux}={\int }_0^{2\pi }{\int }_0^{\pi /2}\sin \varphi {\cos }^2\varphi d\varphi d\theta .$ Inner: $c=\cos \varphi$ let karo, $dc=-\sin \varphi d\varphi$; ${\int }_0^{\pi /2}\sin \varphi {\cos }^2\varphi d\varphi ={\int }_0^1{c}^{2}dc=\frac{1}{3}$. $\text{Flux}=2\pi \cdot \frac{1}{3}=\frac{2\pi }{3}\approx \mathrm{2.094.}$ Closed surface + Divergence Theorem se confirm karo: hemisphere ko flat disk $z=0$ se close karo (wahan outward normal $(0,0,-1)$ hai, aur $\mathbf{F}=(0,0,0)$ $z=0$ par, toh woh cap $0$ contribute karta hai). Phir solid half-ball par $\nabla \cdot \mathbf{F}=1$, volume $=\frac{1}{2}\cdot \frac{4}{3}\pi =\frac{2\pi }{3}$. Total closed flux $=\frac{2\pi }{3}$; $0$ cap subtract karo → hemisphere flux $=\frac{2\pi }{3}$. ✔

Stokes' Theorem se, ${\iint }_S(\mathrm{curl}\mathbf{F})\cdot d\mathbf{S}={\oint }_{\partial S}\mathbf{F}\cdot d\mathbf{r}$, jo sirf boundary curve par depend karta hai, surface par nahi. Toh har $S$ us same boundary ke saath same number deta hai. Boundary line integral compute karo (Line integrals) $\mathbf{r}(t)=(\cos t,\sin t,0)$ par, $t\in [0,2\pi ]$, ${\mathbf{r}}^{\prime }(t)=(-\sin t,\cos t,0)$: $\mathbf{F}=(-\sin t,\cos t,0),\mathbf{F}\cdot {\mathbf{r}}^{\prime }={\sin }^{2}t+{\cos }^{2}t=1.$ ${\oint }_{\partial S}\mathbf{F}\cdot d\mathbf{r}={\int }_0^{2\pi }1dt=2\pi .$ Curl ka direct check (flat disk ke liye): $\mathrm{curl}\mathbf{F}=(0,0,{\partial }_{x}x-{\partial }_y(-y))=(0,0,2)$, aur flat disk ke through (normal $(0,0,1)$, area $\pi$): $2\cdot \pi =2\pi$. ✔ $\boxed{\text{Flux of }\mathrm{curl}\mathbf{F}=2\pi \text{for any such }S.}$

$z=2u+3v$ ka stretch factor? ::: $\sqrt{14}$ (constant, plane). $(0,0,z)$ ka flux triangle par $z=1-x-y$ ke upar upward? ::: $\frac{1}{6}$. Same flux, downward oriented? ::: $-\frac{1}{6}$. $(x,y,0)$ ka outward flux unit cylinder wall ke through, $0\le z\le 2$? ::: $4\pi$. Wahan caps kuch add kyun nahi karte? ::: $\mathbf{F}$ mein koi $z$-component nahi, toh $\mathbf{F}\cdot (0,0,\pm 1)=0$. Cone $z=\sqrt{x^2+y^2}$, $z\le 1$, density $z$ ka mass? ::: $\frac{2\sqrt{2}\pi }{3}$. $(0,0,z)$ ka flux upper unit hemisphere ke bahar? ::: $\frac{2\pi }{3}$. $\mathrm{curl}(-y,x,0)$ ka flux unit circle par kisi bhi cap ke through? ::: $2\pi$.