Sunday, May 23, 2010

GR9768.66: Green's Theorem

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.
$I = \iint_D 2x + 2 dA = \int_0^{2\pi} \int_0^3 2(r \cos \theta + 2) r dr d\theta$.
$I = \int_0^{2\pi} \int_0^3 2r^2 \cos \theta + 2r dr d\theta$
$I = \int_0^{2\pi} \frac{2r^3}{3} \cos \theta + r^2 |_0^3 d\theta$
$I = \int_0^{2\pi} 18 \cos \theta + 9 d\theta$
$I = 18 \sin \theta + 9 \theta \vert_0^{2\pi} = 18 \sin (2\pi) + 9(2\pi) - 18 \sin (0) - 9(0) = 18 \pi$.

::

Alternative Solution:
Parametrize using the circle centered at the origin with radius 3:
$x = r \cos \theta = 3 \cos \theta$ and $y = r \sin \theta = 3 \sin \theta$,
where $\theta \in [0, 2 \pi]$. Thus,
$dx = -3 \sin \theta d\theta$ and $dy = 3 \cos \theta d\theta$. And so
$I = \oint_C -2y dx + x^2 dy = \int_0^{2\pi} -2(3\sin \theta) (-3 \sin \theta d\theta) + (3 \cos \theta)^2 (3 \cos \theta d\theta)$
$I = \int_0^{2\pi} 18 \sin^2 \theta d\theta + 27 \cos^3 \theta d\theta$.
$I = \int_0^{2\pi} 18 - 18 \cos^2 \theta + 27 \cos^3 \theta d\theta$.
Then:
$\int_0^{2\pi} 18 d\theta = 18(2\pi) = 36 \pi$.
$\int_0^{2pi} 18 \cos^2 \theta d\theta = 18 \int_0^{2\pi}\frac{1}{2} \left(\cos (2\theta) + 1\right) d\theta = 9 \left(\frac{1}{2}\sin (2 \theta) + \theta \right) \big \vert_0^{2\pi}$
$= 9 \left(\frac{1}{2}\sin (2 \cdot 2\pi) + 2\pi \right) - \frac{1}{2} \sin 0 - 2 (0)) = 9(2\pi) = 18\pi$.
$\int_0^{2\pi} 27 \cos^3 \theta d\theta = 27 \int_0^{2\pi} \cos^3 \theta= 0$ by observing that the area under the $x$-axis cancels the area above the $x$-axis from $0$ to $2\pi$.
The result is $36 \pi - 18 \pi + 0 = 18 \pi$.
-sg-