VC5.6

From Exampleproblems

If R is a plane region bounded by a simple closed curve C,then by Green's theorm in plane,we have $\oint_C (\psi dx+\phi dy)=\iint_R [\frac{\partial\phi}{\partial x}-\frac{\partial\psi}{\partial y}]dx dy\,$ --(1)

putting $\psi=-y,\phi=x\,$ in (1),we have

$\oint_C (xdy-ydx)=\iint_R [\frac{\partial}{\partial x}(x)-\frac{\partial}{\partial y}(-y)]dx dy=2\iint_R dx dy=2A\,$ where A is the area of plane region R bounded by C.

Therefore,A= $\frac{1}{2}\oint_C(xdy-ydx)\,$

The parametric equations of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\,$ are $x=a\cos\theta,y=b\sin\theta,0\le\theta\le 2\pi\,$ -(2)

The area of the ellipse= $\frac{1}{2}\oint_C(xdy-ydx)=\frac{1}{2}\int_{\theta=0}^{2\pi}[x\frac{\partial y}{\partial\theta}-y\frac{\partial x}{\partial\theta}]d\theta\,$

= $\frac{1}{2}\int_{0}^{2\pi}[(a\cos\theta)(b\cos\theta)-(b\sin\theta)(-a\sin\theta)]d\theta\,$

= $\frac{1}{2}\int_{0}^{2\pi} ab(\cos^2\theta+\sin^2\theta)d\theta\,$

= $\frac{1}{2}ab\int_{0}^{2\pi}d\theta=\pi ab\,$

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VC5.6

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