 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/d/solution

Integrating

ϕ_{x} = 2 x y

gives

ϕ = x^{2} y + h (y, z)

. Matching

ϕ_{y} = x^{2} + 3 y^{2} + z^{2}

then gives

h = y^{3} + y z^{2} + k (z)

. Finally

ϕ_{z} = 2 yz

k

to be constant. A scalar potential is

ϕ (x, y, z) = x^{2} y + y^{3} + y z^{2} + C .

The initial and final points of the helix are

(b, 0, 0)

and

(0, b, c L)

, where

L = 5 π /2

ϕ (0, b, c L) - ϕ (b, 0, 0) = b^{3} + b c^{2} L^{2} = b (b^{2} + \frac{25 π ^{2} c ^{2}}{4}),

agreeing with the directly evaluated

I_{2}

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