Line integrals — scalar and vector, work done: Worked examples

4.4.27 · D3 · Maths › Multivariable Calculus › Line integrals — scalar and vector, work done

Scenario matrix

Har line-integral problem jo tum kabhi bhi miloge, in cells mein se kisi ek mein aati hai. Neeche har worked example ko us cell ke saath tag kiya gaya hai jo woh cover karta hai, taaki tum dekh sako ki poora board fill ho raha hai.

#	Cell (scenario class)	Kya tricky banata hai	Example
A	Scalar $\int_{C} f d s$ , non-constant speed	$∣ r^{'} ∣$ rakhna zaroori, yeh 1 nahi hai	Ex 1
B	Scalar integral, orientation reversed	answer nahi badalna chahiye	Ex 2
C	Vector integral, non-conservative field	genuinely path-dependent	Ex 3
D	Vector integral, orientation reversed	answer sign flip karta hai	Ex 3 (part b)
E	Conservative field, sirf endpoints use karo	potential ke zariye shortcut	Ex 4
F	Closed loop, non-conservative	nonzero circulation	Ex 5
G	Closed loop, conservative	zero hona chahiye	Ex 6
H	Degenerate curve (ek single point)	zero length ⇒ zero integral	Ex 7
I	Piecewise path (corner)	pieces mein split karo, add karo	Ex 8
J	Real-world word problem + units	physics ko symbols mein translate karo	Ex 9

Across padhte hain: A–B scalar flavour exhaust karte hain (speed matters, orientation nahi). C–G vector flavour exhaust karte hain field type (conservative vs not) aur curve type (open vs closed) ke hisaab se. H–J degenerate, piecewise, aur applied twists pakdte hain jo exams ko bahut pasand hain.

Do tools baar baar aate hain, toh chaliye unhe ek ek line mein dobara earn karte hain:

Dot product $a \cdot b = a_{1} b_{1} + a_{2} b_{2}$ — measure karta hai "kitna $a$ , $b$ ke along point karta hai". Humein yeh chahiye kyunki work sirf force ka woh component count karta hai jo motion ke along ho.
Parametric curves — $C$ ko ek number $t$ se describe karna, taaki har integral ek ordinary $\int_{a}^{b} \dots d t$ ban jaye.

Ex 1 — Scalar integral, speed constant NAHI hai · Cell A

Worked example

$y = x^{2}$ ke along ek wire ka mass Wire $r (t) = (t, t^{2})$ follow karta hai $t \in [0, 1]$ ke liye, density $f (x, y) = x$ ke saath. Total mass $\int_{C} f d s$ nikalo.

Forecast: wire upar chadhte waqt steep hoti jaati hai, toh "per unit $t$ " mein woh upar ke paas "zyada lambi" hoti hai. Kya tumhe lagta hai speed factor $∣ r^{'} ∣$ $1$ hoga, ya $1$ se bada aur badhta hua? Padhne se pehle guess karo.

Step 1. Velocity: $r^{'} (t) = (1, 2 t)$ . Yeh step kyun? $d s$ ek chote se step ki length se aata hai, aur ek chota step $r^{'} (t) d t$ hai; hum har component differentiate karte hain ise paane ke liye.

Step 2. Speed: $∣ r^{'} (t)∣ = 1^{2} + (2 t)^{2} = 1 + 4 t^{2}$ . Yeh step kyun? $d s = ∣ r^{'} ∣ d t$ . Dhyan do yeh $1$ nahi hai — yeh $t$ ke saath badhta hai, bilkul wahi steepening jo humne forecast ki thi. (Figure dekho: step arrows lambe hote jaate hain.)

Step 3. Curve par density: $f (r (t)) = x = t$ . Yeh step kyun? Parametrisation ko $f$ mein substitute karo; yahan $x = t$ hai.

Step 4. Assemble karo aur integrate karo: $\int_{C} f d s = \int_{0}^{1} t 1 + 4 t^{2} d t .$ $u = 1 + 4 t^{2}$ lo, toh $d u = 8 t d t$ , yaani $t d t = \frac{1}{8} d u$ . Limits: $t = 0 \to u = 1$ , $t = 1 \to u = 5$ . $= \frac{1}{8} \int_{1}^{5} u d u = \frac{1}{8} \cdot \frac{2}{3} [u^{3/2}]_{1}^{5} = \frac{1}{12} (5^{3/2} - 1) .$ Yeh step kyun? Substitution messy $t 1 + 4 t^{2}$ ko $u$ ki ek clean power mein badal deti hai.

Answer: $\frac{5 5 - 1}{12} \approx 0.8844.$

Verify: $5^{3/2} = 55 \approx 11.18$ , toh $(11.18 - 1) /12 \approx 0.848$ … exactly ho: $(11.1803 - 1) /12 = 0.8484$ . Positive, sensible (density aur length dono positive hain), aur dimensionally ek mass hai. ✓

Worked example Wahi wire, ulti taraf chalte hue

Ex 1 ki wire lo lekin use $(1, 1)$ se $(0, 0)$ tak traverse karo: $r (t) = (1 - t, (1 - t)^{2})$ , $t \in [0, 1]$ . Dikhao ki mass unchanged rehta hai.

Forecast: density unsigned hai, arc-length unsigned hai. Kya tum same $0.8484$ expect karte ho ya uska negative?

Step 1. $r^{'} (t) = (- 1, - 2 (1 - t))$ , toh $∣ r^{'} ∣ = 1 + 4 (1 - t)^{2}$ . Yeh step kyun? Same machine; minus signs ek square ke andar hain, toh woh disappear ho jaate hain — length negative nahi ho sakti.

Step 2. $f = x = 1 - t$ . Integral: $\int_{0}^{1} (1 - t) 1 + 4 (1 - t)^{2} d t .$ $s = 1 - t$ substitute karo (toh $d s = - d t$ , aur limits swap karne se minus khatam ho jaata hai) to milta hai $\int_{0}^{1} s 1 + 4 s^{2} d s$ — Ex 1 se bilkul same. Yeh step kyun? Change of variable expose karta hai ki dono integrals literally ek hi object hain.

Answer: $\frac{5 5 - 1}{12} \approx 0.8484$ , Ex 1 jaisa hi.

Verify: Ex 1 se exactly match karta hai ⇒ rule confirm hota hai: scalar line integrals orientation-independent hote hain. ✓

Ex 3 — Vector integral, non-conservative, dono directions · Cells C & D

Worked example Ek swirling field ka work, aage phir peeche

Field $F = (- y, x)$ unit circle $r (t) = (cos t, sin t)$ ke along. (a) Forward quarter, $t \in [0, π /2]$ . (b) Wahi quarter reversed.

Forecast: yeh field counter-clockwise circulate karta hai (arrows motion ki direction mein point karte hain). Forward ko positive work dena chahiye; kya (b) bilkul uska negative hoga?

(a) Step 1. $r^{'} (t) = (- sin t, cos t)$ . Kyun? Displacement direction — har component differentiate karo.

Step 2. $F (r (t)) = (- sin t, cos t)$ (kyunki $x = cos t, y = sin t$ ). Kyun? Point ko field mein substitute karo.

Step 3. Dot product: $F \cdot r^{'} = (- sin t) (- sin t) + (cos t) (cos t) = sin^{2} t + cos^{2} t = 1$ . Kyun? Work sirf tangential part count karta hai; yahan field motion ke saath perfectly aligned hai, toh dot product constant $1$ hai.

Step 4. $W = \int_{0}^{π /2} 1 d t = \frac{π}{2} \approx 1.5708.$

(b) Reverse: orientation rule se, $W \to - \frac{π}{2} \approx - 1.5708.$ Kyun? $d r$ sign flip karta hai, dot product flip karta hai, integral flip karta hai — vector asymmetry yahi hai.

Verify: forward $+ π /2$ , backward $- π /2$ ; dono ka sum $0$ hai jaisa hona chahiye (aage aur peeche cancel ho jaata hai). Ex 2 se contrast karo, jahan reversing ne kuch nahi kiya. ✓

Ex 4 — Conservative field: sirf endpoints use karo · Cell E

Worked example Potential se integral skip karo

$F = (2 x y, x^{2})$ kisi bhi path ke along $(0, 0)$ se $(2, 3)$ tak. Work nikalo.

Forecast: agar hum ek scalar $ϕ$ dhundh sako jisme $\nabla ϕ = F$ ho, toh Fundamental Theorem for Line Integrals kehta hai $W = ϕ (end) - ϕ (start)$ path se independent. Guess: kya answer route par depend karta hai?

Step 1. $ϕ$ dhundho jisme $ϕ_{x} = 2 x y$ aur $ϕ_{y} = x^{2}$ ho. Kyun? Gradient and conservative fields: gradient field ka work ek potential difference hota hai. $ϕ_{x}$ ko $x$ mein integrate karo: $ϕ = x^{2} y + g (y)$ . Phir $ϕ_{y} = x^{2} + g^{'} (y)$ ko $x^{2}$ ke barabar hona chahiye, toh $g^{'} (y) = 0$ , $g =$ const. $ϕ = x^{2} y$ lo.

Step 2. Conservativeness ka endpoint check: $\partial_{y} (2 x y) = 2 x = \partial_{x} (x^{2})$ — match karta hai, toh $F$ sach mein conservative hai (shortcut legal hone ke liye zaroori). Kyun? Agar "cross-derivatives" disagree karte, toh koi $ϕ$ exist nahi karta aur hume lambe tarike se integrate karna padta.

Step 3. $W = ϕ (2, 3) - ϕ (0, 0) = (2^{2} \cdot 3) - 0 = 12.$

Verify (ek straight line ke along integrate karo pakka karne ke liye): $r (t) = (2 t, 3 t)$ , $t \in [0, 1]$ : $r^{'} = (2, 3)$ , $F = (2 (2 t) (3 t), (2 t)^{2}) = (12 t^{2}, 4 t^{2})$ , dot $= 24 t^{2} + 12 t^{2} = 36 t^{2}$ , $\int_{0}^{1} 36 t^{2} d t = 12$ . ✓ Endpoint shortcut se same.

Ex 5 — Closed loop, non-conservative: nonzero circulation · Cell F

Worked example Swirl ka poora ek chakkar

$F = (- y, x)$ poore unit circle ke around, $t \in [0, 2 π]$ , counter-clockwise.

Forecast: Ex 3 ne $1$ per unit $t$ of tangential push diya tha. Ek poora chakkar $2 π$ of $t$ hai. Saath hi — ek conservative field ka closed loop $0$ deta; kya yeh field conservative hai?

Step 1. Ex 3 se, $F \cdot r^{'} = 1$ unit circle par. Kyun? Same field, same curve; dot product ab bhi constant $1$ hai.

Step 2. $\oint_{C} F \cdot d r = \int_{0}^{2 π} 1 d t = 2 π \approx 6.2832.$

Step 3. Conservativeness cross-check: $\partial_{y} (- y) = - 1$ , $\partial_{x} (x) = 1$ . Yeh disagree karte hain ( $- 1 \neq = 1$ ), toh $F$ conservative nahi hai — yahi reason hai ki ek closed loop nonzero answer deta hai. Kyun? Ek conservative field har loop par $0$ force karta (Ex 6). Nonzero $2 π$ genuine swirl ki signature hai.

Green's Theorem se verify karo: $\oint P d x + Q d y = \iint_{D} (Q_{x} - P_{y}) d A = \iint_{D} (1 - (- 1)) d A = 2 \cdot (area of disc) = 2 π$ . ✓ Match karta hai.

Ex 6 — Closed loop, conservative: zero hona zaroori · Cell G

Ex 7 — Degenerate curve: ek single point · Cell H

Worked example Woh curve jo kahin nahi jaati

"Curve" $r (t) = (1, 2)$ $t \in [0, 3]$ ke liye — parameter chalta rehta hai lekin point kabhi move nahi karta. Dono $\int_{C} f d s$ (koi bhi $f$ ) aur $\int_{C} F \cdot d r$ (koi bhi $F$ ) compute karo.

Forecast: koi motion nahi, koi length nahi. Padhne se pehle dono answers guess karo.

Step 1. $r^{'} (t) = (0, 0)$ , toh speed $∣ r^{'} ∣ = 0$ . Kyun? Ek constant position ki zero velocity hoti hai — kuch trace hi nahi ho raha.

Step 2. Scalar: $\int_{0}^{3} f (1, 2) \cdot 0 d t = 0$ . Vector: $\int_{0}^{3} F \cdot (0, 0) d t = 0$ . Kyun? Dono machines ek step ( $d s$ ya $d r$ ) se weight karte hain jo identically zero hai.

Answer: dono integrals $0$ hain.

Verify: Arc length $= \int_{C} 1 d s = 0$ , "ek point ki koi length nahi hoti" se match karta hai. ✓ Yeh woh boundary case hai jo hume reassure karta hai ki definitions gracefully degrade hoti hain.

Ex 8 — Corner wala piecewise path · Cell I

Worked example L-shaped route par work

$F = (y, x)$ $(0, 0)$ se $(1, 0)$ tak (segment $C_{1}$ ) phir upar $(1, 1)$ tak (segment $C_{2}$ ). Total work nikalo.

Forecast: $F = (y, x) = \nabla (x y)$ conservative hai (parent ka Ex), toh answer $ϕ (1, 1) - ϕ (0, 0) = 1$ hona chahiye. Chaliye har piece grind karke confirm karte hain.

Step 1 — $C_{1}$ : $r (t) = (t, 0)$ , $t \in [0, 1]$ , $r^{'} = (1, 0)$ , $F = (y, x) = (0, t)$ . Dot $= 0 \cdot 1 + t \cdot 0 = 0$ . Toh $\int_{C_{1}} = 0$ . Kyun? Bottom edge ke along force ka " $y$ -part" $0$ hai aur motion purely horizontal hai.

Step 2 — $C_{2}$ : $r (t) = (1, t)$ , $t \in [0, 1]$ , $r^{'} = (0, 1)$ , $F = (t, 1)$ . Dot $= t \cdot 0 + 1 \cdot 1 = 1$ . Toh $\int_{C_{2}} = \int_{0}^{1} 1 d t = 1$ . Kyun? Ab motion vertical hai aur is edge par har jagah force ka " $x$ -part" $1$ hai.

Step 3. Total $= 0 + 1 = 1$ . Kyun? Line integrals concatenated pieces par add hote hain: $\int_{C_{1} + C_{2}} = \int_{C_{1}} + \int_{C_{2}}$ .

Verify: potential shortcut $ϕ (1, 1) - ϕ (0, 0) = 1 - 0 = 1$ . ✓ Corner sahi se handle kiya.

Ex 9 — Units wala real-world word problem · Cell J

Worked example Ek headwind mein crate khiinchna

Ek crate $r (t) = (t, t)$ metres ke along $t \in [0, 4]$ ke liye move karta hai. Wind force (newtons) $F = (- 1, 0)$ N hai — yeh poore time $- x$ direction mein blow karta hai. Wind crate par kitna work karta hai?

Forecast: crate $+ x, + y$ direction mein move karta hai; wind $+ x$ part ko oppose karta hai. Negative work expect karo (wind motion se lad raha hai).

Step 1. $r^{'} (t) = (1, 1)$ m per unit $t$ . Kyun? Displacement direction; yahan motion ek $4 5^{\circ}$ line hai.

Step 2. Dot product: $F \cdot r^{'} = (- 1) (1) + (0) (1) = - 1$ . Kyun? Sirf wind ka woh component jo motion ke along ho, work karta hai; wind ka $x$ -component crate ki $x$ -motion ko oppose karta hai, $- 1$ deta hai.

Step 3. $W = \int_{0}^{4} (- 1) d t = - 4$ joules (N·m). Kyun? Constant integrand times parameter range.

Answer: $W = - 4 J$ — wind $4$ joules remove karta hai.

Verify (units + sign): N × m $=$ J. ✓ Sign negative hai, "wind motion oppose karta hai" se match karta hai, jaisa forecast kiya tha. ✓

Recall

Recall Kaun se cells reversal par sign flip karte hain, kaun se nahi?

Scalar (Cells A–B, H) ::: orientation-independent — koi sign flip nahi. Vector (Cells C–J) ::: curve reverse hone par sign flip karta hai.

Recall Fast test: kya ek closed-loop integral automatically zero hota hai?

Sirf tab jab field conservative ho ::: $\nabla ϕ$ ka closed loop $0$ deta hai (Ex 6); ek swirl nonzero circulation deta hai (Ex 5).

Connections

Parametric curves — har example yahan se shuru hua.
Dot product — har vector example ka engine.
Arc length — $f = 1$ scalar case (Ex 7 ka check).
Gradient and conservative fields · Fundamental Theorem for Line Integrals — Ex 4, 6, 8 shortcuts.
Green's Theorem — Ex 5 cross-check.

Worked examples — Line integrals — scalar and vector, work done