 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-1/1/e/solution

The ellipse bounding the surface is

u = 1

. With the upward normal, its positively oriented boundary is traversed by increasing

v

from zero to

2 π

r (v) = (a cosh 1 cos v, a sinh 1 sin v, c), d r = (- a cosh 1 sin v, a sinh 1 cos v, 0) d v .

The given components are Cartesian components, so their ordinary dot product with this tangent is

F \cdot d r = - 2 a sinh 1 cosh 1 (sin^{2} v + cos^{2} v) d v .

Therefore Stokes theorem gives

\int_{S} (\nabla \times F) \cdot d S = - 4 πa sinh 1 cosh 1 = - 2 πa sinh 2.

A direct check uses

F_{x} = 2 y / a

and

F_{y} = - 2 x / a

, hence

(\nabla \times F)_{z} = - 4/ a

, and multiplies by the ellipse area

π a^{2} sinh 1 cosh 1

. The third component cannot affect either calculation because the surface and its boundary lie at fixed

z

. Equivalently one may apply Stokes to the smooth planar field

(2 y / a, - 2 x / a, 0)

, avoiding any need to differentiate the elliptic-coordinate expression for the third component at its foci.

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