 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-v-resit/question-1/b/i/solution

(- y, x, 0) = r e_{θ}

and

x + r = r (1 + cos θ)

. Hence the vector field simplifies to

F = \frac{1 + c o s θ}{r} e_{θ} .

Take the orientation of

C

to be increasing

θ

, as in its given parametrization. On the circle,

d r = a e_{θ} d θ

\int_{C} F \cdot d r = \int_{- π}^{π} 1 + cos θ d θ = 2 \int_{- π}^{π} cos (θ /2) d θ = 42 .

We used

1 + cos θ = 2 ∣ cos (θ /2) ∣

; the cosine is nonnegative on this interval. In particular, the answer does not depend on

a

 New to topics? Read the docs here!