Multivariable-Calculus · Unit 18 · Video 1 · Interactive Practice

Work Around a Closed Curve: Integrating Piece by Piece

IKey Formulas

Formula	Name	What it says
$\displaystyle\oint_C \vec F \cdot d\vec r = \int_{C_1} + \int_{C_2} + \int_{C_3}$	Piece by piece	Split $C$ at its corners; each piece gets its own parametrisation
$\displaystyle\int_{-C} \vec F \cdot d\vec r = -\int_C \vec F \cdot d\vec r$	Reversing a path	Every step $d\vec r$ flips, so the work changes sign
$x = \cos\theta,\ y = \sin\theta:\quad y\,dx + x\,dy = \cos 2\theta\,d\theta$	The arc $C_2$	$\int_0^{\pi/4} \cos 2\theta\,d\theta = \tfrac12$
$0 + \tfrac12 - \tfrac12 = 0$	Total for $\vec F = y\,\hat\imath + x\,\hat\jmath$	Around the sector from $0$ to $\pi/4$

Key Insight: on every piece the integrand is the same expression $y\,dx + x\,dy$ ; only the substitution changes.

IIWalking the Loop

The work of $\vec F = \langle y, x\rangle$ around the sector builds up piece by piece: $0$ , then $\tfrac12$ , then $-\tfrac12$ .

position along C C₂

IIIReversing the Diagonal

Along $-C_3$ the steps $d\vec r$ point the other way, so every $\vec F \cdot d\vec r$ changes sign.

Direction position 0.5

IVQuiz Questions

Problem 1 · A Single Segment

Given: $\vec F = y\,\hat\imath + x\,\hat\jmath$ and $C$ the segment from $(0, 0)$ to $(1, 2)$ — find $\int_C y\,dx + x\,dy$ .

✅ Correct! With

x = t

y = 2t

the integrand is

4t\,dt

, and

\int_0^1 4t\,dt = 2

❌ Only one of the two terms.

y\,dx = 2t\,dt

and

x\,dy = t \cdot 2\,dt

each give

1

; the integral is their sum.

❌ That is the integrand, not the integral.

y\,dx + x\,dy = 4t\,dt

, and integrating from

0

1

gives

2

❌ The parameter range is right, the integrand is not. Both

dx = dt

and

dy = 2\,dt

must be substituted:

2t\,dt + 2t\,dt

Show solution

Parametrise: $x = t$ , $y = 2t$ , $0 \le t \le 1$ , so $dx = dt$ and $dy = 2\,dt$ . Then

\int_C y\,dx + x\,dy = \int_0^1 2t\,dt + t \cdot 2\,dt = \int_0^1 4t\,dt = \big[2t^2\big]_0^1 = 2.

Problem 2 · Backwards Along the Same Curve

Given: for some field $\vec F$ and curve $C$ , $\int_C \vec F \cdot d\vec r = \tfrac34$ — find $\int_{-C} \vec F \cdot d\vec r$ .

✅ Correct! Reversing the path reverses every step

d\vec r

, so the work changes sign.

❌ The direction matters.

-C

has the same points but every

d\vec r

points the other way, so each

\vec F \cdot d\vec r

flips sign.

❌ That would be $C$ followed by $-C$ . Going out and coming back gives

\tfrac34 - \tfrac34 = 0

; the question asks for

-C

alone.

❌ The parametrisation does not matter. Any parametrisation of

-C

in its direction gives the same value,

-\int_C \vec F \cdot d\vec r

Show solution

$-C$ is the same set of points traversed in the opposite direction. Each small step $d\vec r$ is replaced by $-d\vec r$ , so

\int_{-C} \vec F \cdot d\vec r = -\int_C \vec F \cdot d\vec r = -\tfrac34.

Problem 3 · A Longer Arc

Given: $\vec F = y\,\hat\imath + x\,\hat\jmath$ and $C$ the arc of the unit circle from $\theta = 0$ to $\theta = \pi/3$ , counterclockwise — find $\int_C y\,dx + x\,dy$ .

✅ Correct!

\int_0^{\pi/3} \cos 2\theta\,d\theta = \tfrac12 \sin\tfrac{2\pi}{3} = \tfrac{\sqrt3}{4}

❌ Missing the factor $\tfrac12$ . An antiderivative of

\cos 2\theta

\tfrac12 \sin 2\theta

, not

\sin 2\theta

❌ Check the sign of the integrand.

y\,dx = -\sin^2\theta\,d\theta

and

x\,dy = \cos^2\theta\,d\theta

, so the integrand is

\cos^2\theta - \sin^2\theta = \cos 2\theta

, not its negative.

❌ That is the value for the arc up to $\pi/4$ . Here the upper limit is

\pi/3

, so

2\theta

ends at

2\pi/3

Show solution

On the unit circle $x = \cos\theta$ , $y = \sin\theta$ , so $dx = -\sin\theta\,d\theta$ and $dy = \cos\theta\,d\theta$ :

\int_C y\,dx + x\,dy = \int_0^{\pi/3} (\cos^2\theta - \sin^2\theta)\,d\theta = \int_0^{\pi/3} \cos 2\theta\,d\theta.

= \Big[\tfrac12 \sin 2\theta\Big]_0^{\pi/3} = \tfrac12 \sin\tfrac{2\pi}{3} = \tfrac{\sqrt3}{4}.

Problem 4 · The Rotation Field Around the Sector

Given: $\vec G = -y\,\hat\imath + x\,\hat\jmath$ and the same closed curve as the video: $C_1$ along the $x$ -axis from $(0,0)$ to $(1,0)$ , $C_2$ along the unit circle to $(\tfrac{1}{\sqrt2}, \tfrac{1}{\sqrt2})$ , $C_3$ back along the diagonal — find $\oint_C \vec G \cdot d\vec r$ .

✅ Correct! Only the arc contributes:

\vec G \cdot \hat T = 1

there, times the arc length

\pi/4

❌ The zero total was special to $\langle y, x\rangle$ . For

\vec G

the arc gives

\pi/4

and nothing cancels it.

❌ That is the work around the whole unit circle. The curve here contains only an eighth of the circle.

❌ The diagonal contributes nothing. On

y = x = t

-y\,dx + x\,dy = -t\,dt + t\,dt = 0

\vec G

is perpendicular to the diagonal.

Show solution

The integrand is $-y\,dx + x\,dy$ .

$C_1$ : $y = 0$ , $dy = 0$ , so the integrand is $0$ .

$C_2$ : $x = \cos\theta$ , $y = \sin\theta$ : $-\sin\theta(-\sin\theta)\,d\theta + \cos\theta\cos\theta\,d\theta = d\theta$ , so the work is $\int_0^{\pi/4} d\theta = \tfrac{\pi}{4}$ .

$C_3$ : on $y = x$ , $dy = dx$ , so $-y\,dx + x\,dy = -x\,dx + x\,dx = 0$ .

\oint_C \vec G \cdot d\vec r = 0 + \tfrac{\pi}{4} + 0 = \tfrac{\pi}{4}.

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