Multivariable-Calculus · Unit 17 · Video 4 · Interactive Practice

Line Integrals by Geometry: Work Around a Circle with F · T ds

IKey Formulas

FormulaNameWhat it says
dr⃗=T^ ds,dr⃗dt=dsdt T^d\vec r = \hat T\,ds, \qquad \dfrac{d\vec r}{dt} = \dfrac{ds}{dt}\,\hat TDirection times lengthA small step points along the unit tangent T^\hat T and has length dsds
∫CF⃗⋅dr⃗=∫CM dx+N dy=∫CF⃗⋅T^ ds\displaystyle\int_C \vec F \cdot d\vec r = \int_C M\,dx + N\,dy = \int_C \vec F \cdot \hat T\,dsThree formsF⃗⋅T^\vec F \cdot \hat T is the tangential component of the force
∮CF⃗⋅dr⃗=0\displaystyle\oint_C \vec F \cdot d\vec r = 0Radial field x ı^+y ȷ^x\,\hat\imath + y\,\hat\jmath, circle about the originF⃗⊥T^\vec F \perp \hat T everywhere
∮CF⃗⋅dr⃗=2πa2\displaystyle\oint_C \vec F \cdot d\vec r = 2\pi a^2Rotation field −y ı^+x ȷ^-y\,\hat\imath + x\,\hat\jmath, radius aa, counterclockwiseF⃗∥T^\vec F \parallel \hat T with ∣F⃗∣=a|\vec F| = a, so the work is a⋅2πaa \cdot 2\pi a

Key Insight: when F⃗⋅T^\vec F \cdot \hat T is constant along CC, it comes out of the integral, and ∫Cds\int_C ds is just the length of CC.

IIThe Tangential Component

Only the part of F⃗\vec F along T^\hat T does work; on a circle about the origin a radial field has none.

IIIWork as Area Under F⃗⋅T^\vec F \cdot \hat T

Plotting F⃗⋅T^\vec F \cdot \hat T against arc length ss, the work around the circle is the area under the graph.

IVQuiz Questions

Problem 1 · Rotation Field, Radius 3

Given: F⃗=−y ı^+x ȷ^\vec F = -y\,\hat\imath + x\,\hat\jmath and CC the circle of radius 33 about the origin, counterclockwise — find ∮CF⃗⋅dr⃗\oint_C \vec F \cdot d\vec r.

✅ Correct! F⃗⋅T^=∣F⃗∣=3\vec F \cdot \hat T = |\vec F| = 3 at every point, times the length 6π6\pi.
❌ That is only the length of CC. F⃗⋅T^\vec F \cdot \hat T equals ∣F⃗∣=r=3|\vec F| = r = 3, not 11.
❌ That is the area of the disk. The work is (F⃗⋅T^)×length=3⋅6π(\vec F \cdot \hat T) \times \text{length} = 3 \cdot 6\pi.
❌ Too large. ∣F⃗∣=a=3|\vec F| = a = 3, not a2a^2: the work is a⋅2πaa \cdot 2\pi a.
Show solution

On the circle, F⃗\vec F is tangent and points counterclockwise, the same way as T^\hat T, so F⃗⋅T^=∣F⃗∣=x2+y2=3\vec F \cdot \hat T = |\vec F| = \sqrt{x^2 + y^2} = 3. Hence

∮CF⃗⋅T^ ds=3∮Cds=3⋅2π⋅3=18π.\oint_C \vec F \cdot \hat T\,ds = 3 \oint_C ds = 3 \cdot 2\pi \cdot 3 = 18\pi.

Problem 2 · Radial Field Around a Circle

Given: F⃗=x ı^+y ȷ^\vec F = x\,\hat\imath + y\,\hat\jmath and CC the circle of radius 33 about the origin, counterclockwise — find ∮CF⃗⋅dr⃗\oint_C \vec F \cdot d\vec r.

✅ Correct! The radial field is perpendicular to the tangent of a circle centred at the origin, so F⃗⋅T^=0\vec F \cdot \hat T = 0 everywhere.
❌ Use the tangential component, not the length. ∣F⃗∣=3|\vec F| = 3 here too, but F⃗\vec F points along the radius, perpendicular to the motion.
Show solution

F⃗\vec F points along the radius and T^\hat T along the circle; a circle's tangent is perpendicular to its radius, so F⃗⋅T^=0\vec F \cdot \hat T = 0 and the work is ∮C0 ds=0\oint_C 0\,ds = 0.

Check by parametrising x=3cos⁡θx = 3\cos\theta, y=3sin⁡θy = 3\sin\theta: x dx+y dy=(−9cos⁡θsin⁡θ+9sin⁡θcos⁡θ) dθ=0x\,dx + y\,dy = (-9\cos\theta\sin\theta + 9\sin\theta\cos\theta)\,d\theta = 0.

Problem 3 · Against the Flow

Given: F⃗=−2y ı^+2x ȷ^\vec F = -2y\,\hat\imath + 2x\,\hat\jmath and CC the unit circle about the origin traversed clockwise — find F⃗⋅T^\vec F \cdot \hat T on CC and the work ∮CF⃗⋅dr⃗\oint_C \vec F \cdot d\vec r.

What is F⃗⋅T^\vec F \cdot \hat T at each point of CC?

What is the work?

✅ Correct! F⃗\vec F points counterclockwise with length 2r=22r = 2 while T^\hat T points clockwise, so F⃗⋅T^=−2\vec F \cdot \hat T = -2 and the work is −2⋅2π-2 \cdot 2\pi.
❌ Check F⃗⋅T^\vec F \cdot \hat T. On the unit circle ∣F⃗∣=2r=2|\vec F| = 2r = 2, and F⃗\vec F points opposite to the clockwise T^\hat T.
❌ Check the work. Multiply the constant F⃗⋅T^\vec F \cdot \hat T by the length of the unit circle, 2π2\pi.
Show solution

F⃗=2⟨−y,x⟩\vec F = 2\langle -y, x\rangle is tangent to every circle about the origin, points counterclockwise and has length 2x2+y2=22\sqrt{x^2 + y^2} = 2 on the unit circle. Traversed clockwise, T^\hat T is opposite to F⃗\vec F, so F⃗⋅T^=−2\vec F \cdot \hat T = -2 and

∮CF⃗⋅T^ ds=−2⋅2π=−4π.\oint_C \vec F \cdot \hat T\,ds = -2 \cdot 2\pi = -4\pi.

Check with x=cos⁡θx = \cos\theta, y=−sin⁡θy = -\sin\theta, 0≤θ≤2π0 \le \theta \le 2\pi (clockwise): −2y dx+2x dy=(−2sin⁡2θ−2cos⁡2θ) dθ=−2 dθ-2y\,dx + 2x\,dy = (-2\sin^2\theta - 2\cos^2\theta)\,d\theta = -2\,d\theta.

Problem 4 · A Field That Weakens With Distance

Given: F⃗=−y ı^+x ȷ^x2+y2\vec F = \dfrac{-y\,\hat\imath + x\,\hat\jmath}{x^2 + y^2} and CC the circle of radius aa about the origin, counterclockwise — find ∮CF⃗⋅dr⃗\oint_C \vec F \cdot d\vec r.

✅ Correct! F⃗\vec F is tangent with ∣F⃗∣=r/r2=1/a|\vec F| = r/r^2 = 1/a, so the work is 1a⋅2πa=2π\tfrac1a \cdot 2\pi a = 2\pi.
❌ Check F⃗⋅T^\vec F \cdot \hat T. It is not 11: dividing by x2+y2=a2x^2 + y^2 = a^2 shrinks the length aa to 1/a1/a.
❌ That is the rotation field's answer. Here each arrow is divided by a2a^2.
❌ Not quite. F⃗\vec F is parallel to T^\hat T, not perpendicular, so F⃗⋅T^=∣F⃗∣>0\vec F \cdot \hat T = |\vec F| > 0.
Show solution

F⃗\vec F is the rotation field divided by r2r^2, so it is still tangent to the circle and points counterclockwise, with

∣F⃗∣=x2+y2x2+y2=1aon C.|\vec F| = \frac{\sqrt{x^2 + y^2}}{x^2 + y^2} = \frac{1}{a} \quad\text{on } C.

So F⃗⋅T^=1a\vec F \cdot \hat T = \tfrac1a is constant and ∮CF⃗⋅T^ ds=1a⋅2πa=2π\oint_C \vec F \cdot \hat T\,ds = \tfrac1a \cdot 2\pi a = 2\pi for every radius.

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