Exercises — Fundamental theorem for line integrals

Jo humein dikha: integrand pehle se hi $\nabla f$ ke roop mein likha hua hai. FTLI se answer hai $f(B)-f(A)$ — path irrelevant hai. Kyun: FTLI kehta hai ${\int }_C\nabla f\cdot d\mathbf{r}=f(B)-f(A)$.

• $f(3,2)=3^2\cdot 2=18$.
• $f(1,0)=1^2\cdot 0=0$. ${\int }_C\nabla f\cdot d\mathbf{r}=18-0=\boxed{18}.$

Jo hum karte hain: sirf endpoints. $f(2,1,3)=2\cdot 1\cdot 3=6$, aur $f(1,1,1)=1$. ${\int }_C\nabla f\cdot d\mathbf{r}=6-1=\boxed{5}.$

Jo humein dikha: integral par chhota circle $\oint$ matlab curve closed hai — tum wahin khatam hote ho jahan shuru kiya tha, toh $A=B$. Kyun yeh collapse ho jaata hai: $f(B)-f(A)=f(A)-f(A)=0$, chahe $f$ kuch bhi ho. ${\oint }_C\mathbf{F}\cdot d\mathbf{r}=\boxed{0}.$ Specific ellipse aur specific $f$ kabhi matter hi nahi kiya.

Step 1 — Conservativeness test karo. Likho $\mathbf{F}=(P,Q)$ jahan $P=2xy$, $Q=x^2$. Poore plane par (simply connected) humein chahiye $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$. $P_y=2x,Q_x=2x.✓$ Kyun: agar potential $f$ exist karta hai toh $P=f_x,Q=f_y$, aur mixed partials commute karte hain, jo force karta hai $P_y=Q_x$. Step 2 — $P$ ko $x$ mein integrate karo. $f=\int 2xydx=x^{2}y+h(y)$. Integration ka "constant" $y$ par depend kar sakta hai kyunki hum $y$ ko freeze karke integrate kar rahe the. Step 3 — $Q$ se match karo. $f_y=x^2+h^{\prime }(y)$ equal hona chahiye $Q=x^2$ ke, toh $h^{\prime }(y)=0\Rightarrow h(y)=$ const. Lo $h=0$: $f=x^{2}y$. Step 4 — FTLI. $f(2,3)=4\cdot 3=12$, $f(0,0)=0$. ${\int }_C\mathbf{F}\cdot d\mathbf{r}=12-0=\boxed{12}.$

Step 1 — Test. $P=2xy,Q=x^2+2y$; $P_y=2x$, $Q_x=2x$. Equal ✓. Step 2 — $P$ integrate karo: $f=x^{2}y+h(y)$. Step 3 — $Q$ se match karo: $f_y=x^2+h^{\prime }(y)=x^2+2y\Rightarrow h^{\prime }(y)=2y\Rightarrow h(y)=y^2$. Toh $f=x^{2}y+y^2$. Step 4 — FTLI: $f(1,2)=2+4=6$, $f(0,0)=0$. ${\int }_C\mathbf{F}\cdot d\mathbf{r}=6-0=\boxed{6}.$

Step 1 — Test. $P=y\cos x,Q=\sin x+2y$. $P_y=\cos x$, $Q_x=\cos x$. Equal ✓. Step 2 — $P$ ko $x$ mein integrate karo: $\int y\cos xdx=y\sin x+h(y)$. Step 3 — $Q$ se match karo: $f_y=\sin x+h^{\prime }(y)=\sin x+2y\Rightarrow h^{\prime }(y)=2y\Rightarrow h(y)=y^2$. Toh $f=y\sin x+y^2$. Step 4 — FTLI: $f\left(\frac{\pi }{2},1\right)=1\cdot \sin \frac{\pi }{2}+1=1+1=2$; $f(0,0)=0$. ${\int }_C\mathbf{F}\cdot d\mathbf{r}=2-0=\boxed{2}.$

Step 1 — Test. $P=-y,Q=x$; $P_y=-1$, $Q_x=1$. Equal nahi ⇒ koi potential exist nahi karta ⇒ FTLI apply nahi hota. Integral genuinely path par depend karta hai. Step 2 — Seedha compute karo kyunki shortcut nahi le sakte. Parametrise karo $x=\cos t,y=\sin t,t\in [0,2\pi ]$, toh $dx=-\sin tdt,dy=\cos tdt$. ${\oint }_C\mathbf{F}\cdot d\mathbf{r}={\int }_0^{2\pi }(Pdx+Qdy)={\int }_0^{2\pi }((-\sin t)(-\sin t)+(\cos t)(\cos t))dt.$ Bracket hai ${\sin }^{2}t+{\cos }^{2}t=1$, toh $\oint ={\int }_0^{2\pi }1dt=\boxed{2\pi }\ne 0.$ Nonzero loop integral ek non-conservative field ka fingerprint hai.

Hum FTLI (2.1 ka potential $f=x^{2}y$) se pehle se jaante hain ki answer hona chahiye $f(1,1)-f(0,0)=1-0=1$. Chalte hain hard way se verify karte hain.

Path $C_1$ (line): $x=t,y=t,t\in [0,1]$; $dx=dt,dy=dt$. ${\int }_0^1(2t\cdot tdt+t^{2}dt)={\int }_0^{1}3t^{2}dt=[t^3{]}_0^1=1.$ Path $C_2$ (parabola): $x=t,y=t^2,t\in [0,1]$; $dx=dt,dy=2tdt$. ${\int }_0^1(2t\cdot t^{2}dt+t^2\cdot 2tdt)={\int }_0^1(2t^3+2t^3)dt={\int }_0^{1}4t^{3}dt=[t^4{]}_0^1=1.$ Dono $\boxed{1}$ ke equal hain — path independence, confirmed. Kyunki $\mathbf{F}=\nabla (x^{2}y)$, FTLI ne kisi bhi integration se pehle hi yeh guarantee kar diya tha.

Check karo $P_y=Q_x$ (toh yeh lagta hai conservative): $P=\frac{-y}{x^2+y^2}$ ke saath, $P_y=\frac{-(x^2+y^2)-(-y)(2y)}{(x^2+y^2{)}^2}=\frac{y^2-x^2}{(x^2+y^2{)}^2},$ aur $Q=\frac{x}{x^2+y^2}$ ke saath, $Q_x=\frac{(x^2+y^2)-x(2x)}{(x^2+y^2{)}^2}=\frac{y^2-x^2}{(x^2+y^2{)}^2}.$ Yeh har jagah match karte hain jahan field defined hai ✓. Ab loop compute karo. $x=\cos t,y=\sin t$ toh $x^2+y^2=1$. Phir $P=-\sin t,Q=\cos t$, $dx=-\sin tdt,dy=\cos tdt$: $\oint ={\int }_0^{2\pi }({\sin }^{2}t+{\cos }^{2}t)dt={\int }_0^{2\pi }1dt=\boxed{2\pi }.$ Paradox resolved: $P_y=Q_x$ potential guarantee karta hai sirf ek simply connected region par — jisme koi holes nahi hote. Yeh field origin par undefined hai, toh jis disk ko yeh enclose karta hai usme ek puncture hai. FTLI ki hypothesis fail ho jaati hai, aur loop vanish karna zaroori nahi. (Origin se door, kisi bhi aisi region par jo hole se bache, ek local potential $f=\arctan (y/x)$ exist karta hai — lekin usse consistently poore chakkar mein define nahi kiya ja sakta.)

Step 1 — $P=yz$ ko $x$ mein integrate karo: $f=xyz+h(y,z)$ (constant baaki dono variables par depend kar sakta hai). Step 2 — $Q$ se match karo: $f_y=xz+h_y=Q=xz\Rightarrow h_y=0$, toh $h=h(z)$ hi. Step 3 — $R$ se match karo: $f_z=xy+h^{\prime }(z)=R=xy\Rightarrow h^{\prime }(z)=0$, toh $h=$ const. Lo $f=xyz$. Step 4 — FTLI: $f(2,3,4)=24$, $f(1,1,1)=1$. ${\int }_C\mathbf{F}\cdot d\mathbf{r}=24-1=\boxed{23}.$

Step 1 — Test force karo. $P=a{y}^2,Q=2xy+3y$. Chahiye $P_y=Q_x$: $P_y=2ay$, $Q_x=2y$. Sab $y$ ke liye equal ⇒ $2a=2\Rightarrow \boxed{a=1}$. Step 2 — $a=1$ ke saath $f$ banao. $P=y^2$; $x$ mein integrate karo: $f=x{y}^2+h(y)$. Step 3 — $Q$ se match karo: $f_y=2xy+h^{\prime }(y)=2xy+3y\Rightarrow h^{\prime }(y)=3y\Rightarrow h(y)=\frac{3}{2}{y}^2$. Toh $f=x{y}^2+\frac{3}{2}{y}^2$. Step 4 — FTLI: $f(0,2)=0\cdot 4+\frac{3}{2}\cdot 4=6$; $f(1,0)=0+0=0$. ${\int }_C\mathbf{F}\cdot d\mathbf{r}=6-0=\boxed{6}.$

Key insight: round trip $A$ par start aur end hoti hai. Toh yeh ek closed loop hai, aur conservative field ke liye $\oint \nabla f\cdot d\mathbf{r}=f(A)-f(A)=0$ — chahe dono legs alag curves hoon. $\oint \mathbf{F}\cdot d\mathbf{r}=\boxed{0}.$ (Sirf outbound leg ka sanity check: $f(1,\pi )=e^1\sin \pi =0$, $f(0,\frac{\pi }{2})=e^0\sin \frac{\pi }{2}=1$, toh out = $0-1=-1$; return leg dena chahiye $+1$, aur $-1+1=0$.)

Forward direction (conservative ⇒ path independence). Kyunki $\mathbf{F}=\nabla f$, FTLI $A$ se $B$ tak kisi bhi curve $C$ par apply hota hai: ${\int }_C\mathbf{F}\cdot d\mathbf{r}={\int }_C\nabla f\cdot d\mathbf{r}=f(B)-f(A).$ Right-hand side mein $C$ ka koi reference nahi — sirf endpoints hain. Toh do curves $C_1,C_2$ dono $f(B)-f(A)$ dete hain; isliye ${\int }_{C_1}\mathbf{F}\cdot d\mathbf{r}={\int }_{C_2}\mathbf{F}\cdot d\mathbf{r}.■$ Corollary (path independence ⇒ loops vanish). Koi bhi closed loop $C$ lo jiska base point $A=B$ ho. Usse do arcs mein split karo: $C_1$ ($A$ se kisi point $M$ tak) aur $C_2$ ($M$ se wapas $A$ tak). $C_2$ ko reverse karne se curve $-C_2$ milti hai $A$ se $M$ tak. Path independence kehta hai ${\int }_{C_1}\mathbf{F}\cdot d\mathbf{r}={\int }_{-C_2}\mathbf{F}\cdot d\mathbf{r}=-{\int }_{C_2}\mathbf{F}\cdot d\mathbf{r}.$ Isliye ${\int }_{C_1}+{\int }_{C_2}=0$, yaani ${\oint }_C\mathbf{F}\cdot d\mathbf{r}=0$. $■$

Step 1 — FTLI. Work $={\int }_C\nabla f\cdot d\mathbf{r}=f(4,3)-f(1,1)$, path ignored.

• $f(4,3)=100-16-2\cdot 9=100-16-18=66$.
• $f(1,1)=100-1-2=97$. $\text{Work}=66-97=\boxed{-31\text{J (per unit)}}.$ Interpretation: hiker neechi altitude par gaya ($97\to 66$), yaani downhill. Field $\nabla f$ uphill point karta hai (increasing $f$ ki taraf), toh uske against move karne par field negative work karta hai. Switchbacks irrelevant hain — sirf $31$-metre net descent matter karta hai.

Kyun curl vanish hona chahiye. Agar $\mathbf{F}=\nabla f$ toh $P=f_x,Q=f_y,R=f_z$. Curl ka pehla component hai $(\nabla \times \mathbf{F}{)}_x=R_y-Q_z=f_{zy}-f_{yz}=0,$ kyunki mixed second partials commute karte hain (Clairaut). Wahi cancellation baaki do components deta hai: $Q_x-P_y=f_{yx}-f_{xy}=0,P_z-R_x=f_{xz}-f_{zx}=0.$ Toh $\nabla \times \mathbf{F}=0$. Dekho Curl — vanishing curl 3D conservativeness test hai (simply connected domains par). $\mathbf{F}=(yz,xz,xy)$ par verify karo:

• $R_y-Q_z=x-x=0$.
• $P_z-R_x=y-y=0$.
• $Q_x-P_y=z-z=0$. $\nabla \times \mathbf{F}=(0,0,0).✓$ 4.1 ke consistent, jahan potential $f=xyz$ mila tha.