 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2002/paper-1/9/i/solution

Parametrize the segment by

r (t) = (1 - t, t, 0)

0 \leq t \leq 1

. Its tangent is

(- 1, 1, 0)

and

F = (- t, 1 - t, t - 1)

, so

F \cdot r^{'} = 1

. Hence its line integral is

\int F \cdot d r = 1.

For the other line integral,

∣ r ∣^{2} = (1 - t)^{2} + t^{2}

and

r \cdot r^{'} = 2 t - 1

. Thus

\int G \cdot d r = \int_{0}^{1} \frac{2 t - 1}{[ 1 + ( 1 - t ) ^{2} + t ^{2} ] ^{3/2}} d t = [- \frac{1}{1 + ( 1 - t ) ^{2} + t ^{2}}]_{0}^{1} = 0 .

This also equals the change in

Φ

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