66. In the

*xy*-plane, if*C*is the circle $x^2 + y^2 = 9$, oriented counterclockwise, then $\oint \limits_C -2y \; dx + x^2 \; dy =$ (A) $0$

(B) $6\pi$

(C) $9\pi$

(D) $12\pi$

(E) $18\pi$.

**Solution:**The hypothesis for Green's Theorem is satisfied since $C$ is a positively oriented, piecewise smooth, simple closed curve. So

$I = \oint_C -2y dx + x^2 dy = \oint_C P dx + Q dy = \iint_D \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} dA$,

where $D$ is the region enclosed by the circle $C$ centered at the origin and radius 3.

$I = \iint_D 2x + 2 dA = \int_0^{2\pi} \int_0^3 2(r \cos \theta + 2) r dr d\theta$.

$I = \int_0^{2\pi} \int_0^3 2r^2 \cos \theta + 2r dr d\theta$

$I = \int_0^{2\pi} \frac{2r^3}{3} \cos \theta + r^2 |_0^3 d\theta$

$I = \int_0^{2\pi} 18 \cos \theta + 9 d\theta$

$I = 18 \sin \theta + 9 \theta \vert_0^{2\pi} = 18 \sin (2\pi) + 9(2\pi) - 18 \sin (0) - 9(0) = 18 \pi$.

The answer is E.