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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/2011/paper-1/19/b/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 / 2011 / Paper 1 / 19 / b / Solution by

Codex  0 2026-10-18

Choose the plane normal to be

+ y

. Only the second field component contributes, and on

y = 0

it is

e^{- (x^{2} + z^{2})}

. Thus the surface integral factors into two Gaussian integrals:

\int_{S} F \cdot d S = \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} e^{- x^{2} - z^{2}} d x d z = (π)^{2} = π .

The original question does not specify an orientation; choosing

- y

instead gives $- π$ .

The printed first component

sin y / y

is undefined on the entire integration plane. Its removable continuous extension there is

1

, since

lim_{y \to 0} sin y / y = 1

. The stated flux uses this natural extension; in fact the normal component alone is already well-defined and integrable on the plane, so the first component does not affect either oriented value.

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