 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-2/16/a/iv/solution

Put

R = 8

and parametrize the circle counterclockwise, as viewed from positive

z

, by

r (ϕ) = (R cos ϕ, R sin ϕ, 3)

for

0 \leq ϕ \leq 2 π

. The restricted field is

V (ϕ) = 2 R^{3} cos^{3} ϕ + R^{2} sin^{2} ϕ + 27 - R sin ϕ .

The PDF does not specify the differential in this line integral. An oriented vector line integral weighted by a scalar field uses

d r = (- R sin ϕ, R cos ϕ, 0) d ϕ

. Its two nonzero components are

\oint_{Γ} V d x \oint_{Γ} V d y = - R \int_{0}^{2 π} V (ϕ) sin ϕ d ϕ = π R^{2}, = R \int_{0}^{2 π} V (ϕ) cos ϕ d ϕ = 2 R^{4} \int_{0}^{2 π} cos^{4} ϕ d ϕ = \frac{3}{2} π R^{4} .

All other terms vanish by sine-cosine symmetry. Thus this interpretation gives

\oint_{Γ} V d r = (8 π, 96 π, 0) .

If the intended convention is the scalar arc length integral, then

d s = R d ϕ

and

\oint_{Γ} V d s = R (R^{2} \int_{0}^{2 π} sin^{2} ϕ d ϕ + 54 π) = 1242 π .

That scalar integral is independent of traversal direction. Giving both values retains the actual distinction left unspecified by the wording.

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