4A (Vector Calculus)

Solution

Green's theorem states, for a positively oriented simple boundary

C = \partial R

\oint_{C} (P d x + Q d y) = \iint_{R} (Q_{x} - P_{y}) d A .

The defining polynomial becomes

r^{2} (r^{2} - 2 r cos θ - sin^{2} θ),

so the region is the cardioid

0 \leq r \leq 1 + cos θ, - π \leq θ \leq π .

The field is

e_{θ}

, so on the polar boundary

F \cdot d r = r d θ

. Hence

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

Here

Q_{x} - P_{y} = y - 2 y = - y

- y

over a region symmetric about the

x

axis, hence the answer is

0

On the boundary the integral is

\int_{- π}^{π} θ d y

. Integration by parts has zero endpoint term and leaves

- \int_{- π}^{π} y d θ = - \int_{- π}^{π} (1 + cos θ) sin θ d θ = 0.

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