 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1/13/b/solution

Put

L = 5 π /2

r^{'} = (- b sin θ, b cos θ, c)

(a \times r) \cdot r^{'} = a \cdot (r \times r^{'})

r \times r^{'} = (b c (sin θ - θ cos θ), - b c (θ sin θ + cos θ), b^{2}) .

\int_{0}^{L} (sin θ - θ cos θ) d θ = [- θ sin θ - 2 cos θ]_{0}^{L} = 2 - L,

and

\int_{0}^{L} (- θ sin θ - cos θ) d θ = [θ cos θ - 2 sin θ]_{0}^{L} = - 2

. Hence

I_{1} = b c (2 - \frac{5 π}{2}) a_{x} - 2 b c a_{y} + \frac{5 π b ^{2}}{2} a_{z} .

For the second line integral, substitute the curve into

v_{2} \cdot r^{'}

. The cubic terms combine to

b^{3} cos θ

, leaving

v_{2} \cdot r^{'} = b^{3} cos θ + b c^{2} θ^{2} cos θ + 2 b c^{2} θ sin θ = \frac{d}{d θ} [b sin θ (b^{2} + c^{2} θ^{2})] .

I_{2} = b (b^{2} + \frac{25 π ^{2} c ^{2}}{4}) .

Both calculations retain the full path from

θ = 0

5 π /2

, rather than inadvertently closing the helix after one turn.

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