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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-2/19/a/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 19 / a / Solution by

Codex  0 2026-10-18

For a smooth oriented surface

S

with positively oriented boundary

C = \partial S

, Stokes theorem states

\iint_{S} (\nabla \times F) \cdot d S = \oint_{C} F \cdot d r .

(1)

Here

d S = n d A

uses the chosen unit normal vector, while

d r

is tangent to the directed boundary. The boundary orientation follows the right-hand rule: curling the fingers along

C

makes the thumb point in the normal direction. The field must be continuously differentiable on a neighbourhood of the surface, with the usual piecewise-smooth allowance for its boundary.

Figure 1.
Outward hemisphere orientation, its counterclockwise boundary, and tangent and normal elements
. **Details**.

For the upper hemisphere with outward normal, the induced equatorial direction is counterclockwise as viewed from positive

z

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