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




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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 $36 \pi - 18 \pi + 0 = 18 \pi$.
The answer is E.
-sg-









2 comments:

  1. Late to the party but...

    What I did was to average the values of z=2*x+2 over the plane (since it is linear in z) then just multiply that by the area of the circle. It seems to me that this is valid, and it gave the correct answer.

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  2. Once you've got that double integral over D of (2x + 2) dA, you've got it in the bag without any computation.

    2x is odd in x, and D is a symmetric region, so the integral of 2x dA is 0.

    Then all you have left is integral over D of 2 dA, which is twice the area of the circle. Answer is 18*pi.

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