Exercises — Divergence theorem (Gauss's theorem) — statement, flux-divergence connection

WHAT: Teeno "matching" partial derivatives ko add karo. $\nabla \cdot \mathbf{F}=\frac{\partial (3x)}{\partial x}+\frac{\partial (-2y)}{\partial y}+\frac{\partial (5z)}{\partial z}=3+(-2)+5=6.$ WHY har term: ${\partial }_x$ sirf pehle component $P=3x$ ko dekhta hai ($x$ mein uska slope $3$ hai); beech wala ${\partial }_y(-2y)=-2$; aakhri ${\partial }_z(5z)=5$. Cross-terms (jaise $-2y$ ka ${\partial }_x$) zero hain kyunki $-2y$ mein koi $x$ nahi hai. Answer: $\nabla \cdot \mathbf{F}=6$, ek constant — field har jagah same rate par pump out karta hai.

Answer: (a). WHY: Theorem ko ek closed surface $S=\partial V$ chahiye — ek complete skin jo ek solid $V$ ko enclose kare. (b) ek curve ke saath flow ka line integral hai — woh circulation hai, Stokes' theorem se handle hota hai, is theorem se nahi. (c) ek open surface hai (bina lid ke ek bowl): yeh akele kisi solid ko bound nahi karta, isliye theorem tab tak apply nahi hoga jab tak tum ek closing disk add nahi karte.

WHAT: Aasaan (volume) side use karo. $\nabla \cdot \mathbf{F}=1+1+1=3$. $\Phi ={\iiint }_{V}3dV=3\cdot \text{Vol}=3\cdot \frac{4}{3}\pi R^3=4\pi R^3.$ WHY volume side: divergence ek constant $3$ hai, isliye triple integral bas $3\times$(ball ka volume) hai — bilkul bhi surface parameterisation nahi. Answer: $\Phi =4\pi R^3$. (Check: $R=1$ se $4\pi$ milta hai, parent ke Example 1 se match karta hai.)

WHAT: $\nabla \cdot \mathbf{F}=3x^2+3y^2+3z^2$. Cube par integrate karo. ${\iiint }_{[0,1{]}^3}(3x^2+3y^2+3z^2)dV.$ Symmetry se teeno terms equal hain, isliye ek compute karo aur teen se multiply karo: $3{\int }_0^1{\int }_0^1{\int }_0^1{x}^{2}dxdydz\times 3=3\cdot \left(\frac{1}{3}\cdot 1\cdot 1\right)\times 3.$ WHY: ${\int }_0^1{x}^{2}dx=\frac{1}{3}$; $y$ aur $z$ integrals dono $1$ dete hain ($[0,1]$ par $1$ integrate karna). Ek term $=3\cdot \frac{1}{3}=1$; teen identical terms $=3$. Answer: $\Phi =3$.

$\nabla \cdot \mathbf{F}=2+3-1=4$ (constant). $\Phi =4\cdot \frac{4}{3}\pi (1{)}^3=\frac{16\pi }{3}.$ Answer: $\frac{16\pi }{3}$.

WHAT: $\nabla \cdot \mathbf{F}=1+2+3=6$. Cylinder ka volume $\pi r^{2}h=\pi (2{)}^2(5)=20\pi$ hai. $\Phi =6\cdot 20\pi =120\pi .$ WHY surface way se better hai: cylinder ki skin ke teen pieces hain — ek curved wall, ek bottom disk, ek top disk — har ek ko apna normal aur integral chahiye. Divergence theorem teeno ko ek multiplication mein collapse kar deta hai. Answer: $\Phi =120\pi$.

Volume side: $\nabla \cdot \mathbf{F}=1$, isliye ${\iiint }_{V}1dV=\text{Vol}=1$.

Surface side (figure dekho): Sirf woh faces contribute karte hain jinke nonzero $x$-normal hai, kyunki $\mathbf{F}=(x,0,0)$ purely $x$ mein point karta hai.

• $x=1$ par face (outward normal $+\hat{x}$): $\mathbf{F}\cdot \hat{\mathbf{n}}=x=1$, area $=1$ ⟹ flux $+1$.
• $x=0$ par face (outward normal $-\hat{x}$): $\mathbf{F}\cdot \hat{\mathbf{n}}=-x=0$ ⟹ flux $0$.
• Chaar side faces ke normals $\pm \hat{y},\pm \hat{z}$ hain, aur $\mathbf{F}$ mein koi $y$ ya $z$ part nahi hai, isliye $\mathbf{F}\cdot \hat{\mathbf{n}}=0$.

Surface total $=1+0=1$. Dono sides $1$ ke equal hain. ✓ WHY kaam karta hai: field $x$-direction mein badhta hai; sirf $x=1$ par front face actually bahar jaata hua flow feel karta hai, aur woh unit area par rate $1$ se nikalta hai.

$\nabla \cdot \mathbf{F}={\partial }_x(y)+{\partial }_y(z)+{\partial }_z(x)=0+0+0=0.$ Isliye $\Phi ={\iiint }_{V}0dV=0$ har closed surface ke liye (koi singularity nahi — yeh field har jagah smooth hai). WHY: har component ek alag variable par depend karta hai, us derivative se alag jo use catch karta. Yeh ek solenoidal (divergence-free) field hai: yeh cheezein idhar-udhar shuffle karta hai bina kuch create ya destroy kiye. Answer: $\Phi =0$.

WHAT: $\nabla \cdot \mathbf{F}=3x^2+3y^2+3z^2=3(x^2+y^2+z^2)=3r^2$ spherical coordinates mein, jahan $r$ origin se distance hai. WHY spherical: divergence sirf $r^2$ par depend karta hai, isliye natural volume element radius $r$, thickness $dr$, aur area $4\pi r^2$ ki ek thin spherical shell hai: $dV=4\pi r^{2}dr.$

WHAT: $\nabla \cdot \mathbf{F}={\partial }_x(xz)+{\partial }_y(yz)+{\partial }_z(z^2)=z+z+2z=4z.$ WHY volume side: solid hemisphere par $4z$ integrate karna ek single triple integral hai; surface mein ek curved dome aur ek flat disk lid hai — do pieces jinse hum bachna chahte hain. Spherical shells use karo lekin height se weighted: hemisphere ke liye, $\iiint 4zdV$. Likho $z=r\cos \varphi$, $dV=r^2\sin \varphi drd\varphi d\theta$ jahan $r\in [0,1]$, $\varphi \in [0,\pi /2]$, $\theta \in [0,2\pi ]$: $\Phi ={\int }_0^{2\pi }{\int }_0^{\pi /2}{\int }_0^{1}4(r\cos \varphi )r^2\sin \varphi drd\varphi d\theta .$ Pieces alag karo: $=4\left({\int }_0^{2\pi }d\theta \right)\left({\int }_0^{\pi /2}\cos \varphi \sin \varphi d\varphi \right)\left({\int }_0^1{r}^{3}dr\right).$

• ${\int }_0^{2\pi }d\theta =2\pi$.
• ${\int }_0^{\pi /2}\cos \varphi \sin \varphi d\varphi =\frac{1}{2}$ (kyunki $\int \sin \varphi d(\sin \varphi )=\frac{1}{2}{\sin }^2\varphi {|}_0^{\pi /2}=\frac{1}{2}$).
• ${\int }_0^1{r}^{3}dr=\frac{1}{4}$.

$\Phi =4\cdot 2\pi \cdot \frac{1}{2}\cdot \frac{1}{4}=\pi .$ Answer: $\Phi =\pi$.

Naive (galat) path: $r\ne 0$ ke liye yeh dikhaya ja sakta hai ki $\nabla \cdot \mathbf{F}=0$, jo hume $\Phi =0$ bolne ka lalach deta hai. Yeh kyun galat hai: field origin par singular hai — $|\mathbf{r}{|}^3\to 0$ wahan, isliye $\mathbf{F}$ aur uske partials us point par undefined hain. Theorem ko chahiye ki partials poore $V$ mein continuous hon, aur origin unit ball ke andar hai. Hypothesis violate hoti hai, isliye hum theorem ko poori ball par apply nahi kar sakte. Sahi path — surface directly compute karo. Unit sphere par $|\mathbf{r}|=1$, isliye $\mathbf{F}=\mathbf{r}$ aur $\hat{\mathbf{n}}=\mathbf{r}$ (outward radial). Isliye $\mathbf{F}\cdot \hat{\mathbf{n}}=\frac{\mathbf{r}\cdot \mathbf{r}}{|\mathbf{r}{|}^3}=\frac{|\mathbf{r}{|}^2}{|\mathbf{r}{|}^3}=\frac{1}{|\mathbf{r}|}=1.$ $\Phi ={\iint }_{S}1dS=\text{area}=4\pi .$ Answer: $\Phi =4\pi$, $0$ nahi. Yeh exactly Gauss's law wala behaviour hai — origin par ek point charge sphere size chahe jo bhi ho, $4\pi$ flux produce karta hai.

WHAT: Maano $V_{\epsilon }$ = shell $\epsilon \le |\mathbf{r}|\le 1$. Ab $\mathbf{F}$ $V_{\epsilon }$ mein har jagah smooth hai (koi origin nahi), isliye theorem apply hota hai: ${\iint }_{\partial V_{\epsilon }}\mathbf{F}\cdot \hat{\mathbf{n}}dS={\iiint }_{V_{\epsilon }}\underset{=0}{\underbrace{\nabla \cdot \mathbf{F}}}dV=0.$ WHY boundary ke do pieces hain: $\partial V_{\epsilon }$ mein outer unit sphere (outward normal $+\hat{r}$) aur radius $\epsilon$ ki inner sphere hai jiska normal hole ki taraf point karta hai, yaani $-\hat{r}$ (shell se outward matlab origin ki taraf). $\underset{4\pi }{\underbrace{{\Phi }_{\text{outer}}}}+{\Phi }_{\text{inner}}=0\Rightarrow {\Phi }_{\text{inner}}=-4\pi .$ Inner sphere par outward-from-origin normal $+\hat{r}$ ke saath, flux $+4\pi$ hai (Exercise 5.1 jaisi hi computation radius $\epsilon$ ke saath: $\mathbf{F}\cdot \hat{r}=1/{\epsilon }^2$ times area $4\pi {\epsilon }^2=4\pi$). Shell ke outward normal $-\hat{r}$ ke saath sign flip hoke $-4\pi$ ho jaata hai. Sab consistent hai. Answer: shell flux $=0$; "missing" $4\pi$ singularity ke aas-paas inner sphere par chhupa hua hai. Yeh continuity / Gauss's-law derivations ka standard trick hai.

WHAT: Divergence seedha diya gaya hai, isliye volume side par directly jao. Spherical shells mein $\nabla \cdot \mathbf{F}=r^2$, $dV=4\pi r^{2}dr$: $\Phi ={\int }_0^2{r}^2\cdot 4\pi r^{2}dr=4\pi {\int }_0^2{r}^{4}dr=4\pi \cdot \frac{2^5}{5}=4\pi \cdot \frac{32}{5}=\frac{128\pi }{5}.$ WHY shells: divergence sirf centre $r$ se distance par depend karta hai, isliye constant-$r$ shells mein slice karne se integrand har shell par constant ho jaata hai — sabse clean possible decomposition. Answer: $\Phi =\frac{128\pi }{5}$.