3B (Vector Calculus)

Parametrize the straight segment by

x = t

y = 1 + t

0 \leq t \leq 1

. The line integral becomes

\int_{0}^{1} [t^{2} - (1 + t) + (1 + t)^{2} + t] d t = \int_{0}^{1} (2 t^{2} + 2 t) d t = \frac{5}{3} .

ii

For the parabola,

x = t

y = 1 + t^{2}

d x = d t

, and

d y = 2 t d t

. Hence

\int_{0}^{1} [- 1 + 2 t ((1 + t^{2})^{2} + t)] d t = \int_{0}^{1} (- 1 + 2 t + 2 t^{2} + 4 t^{3} + 2 t^{5}) d t = 2 .

The difference from the straight-path value reflects that the one-form is not exact.

Green theorem states that for a positively oriented planar region

D

\oint_{\partial D} P d x + Q d y = \iint_{D} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d A .

Here

P = x y^{2} - λ y

and

Q = x^{2} y

, so

Q_{x} - P_{y} = 2 x y - (2 x y - λ) = λ .

The stated orientations make

C_{1} \cup C_{2}

the positive boundary of the region inside the radius-

a

circle and outside the radius-

b

circle. Its area is

π (a^{2} - b^{2})

regardless of the inner circle's center. Therefore

\int_{C_{1} \cup C_{2}} P d x + Q d y = πλ (a^{2} - b^{2}) .

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