Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 10

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- Solution 10

Solution

On the disc

z = 0

, the upward normal selects only the third component, which equals

e^{0} / (1 + 0) = 1

. Hence the surface integral is the area of the unit disk:

\int_{D} F \cdot n d S = \int_{D} 1 d S = π .

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