 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/16/a/iii/solution

x^{2} + y^{2} = r^{2}

and

r > 0

, so

ψ = a ln r .

The potential vector has components

(A_{r}, A_{θ}, A_{z}) = (0, 0, a ln r)

. Expanding the prescribed cylindrical curl formula gives

u_{r} = \frac{1}{r} \partial_{θ} A_{z} - \partial_{z} A_{θ} = 0, u_{θ} = \partial_{z} A_{r} - \partial_{r} A_{z} = - \frac{a}{r}, u_{z} = \frac{1}{r} [\partial_{r} (r A_{θ}) - \partial_{θ} A_{r}] = 0.

Therefore

(u_{r}, u_{θ}, u_{z}) = (0, - a / r, 0), r > 0.

Since

e_{θ} = (- sin θ, cos θ, 0)

, this agrees with the Cartesian result and its clockwise orientation.

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