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Use Green's theorem to find the area enclosed by the ellipse: $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$


We parameterize the ellipse by the equations: $$x = a\cos{t}\\y = b\sin{t}$$ ... for $t \in [0, 2\pi]$. Green's theorem tell sus that the area of $R$, the region enclosed by this simple closed curve, is: $$\text{area of R} = \oint_c x~ dy = \int _0 ^{2\pi} (a\cos{t})(b\cos{t}dt)$$

Green's theorem is: $$\oint_c Mdx + Ndy = \int \int_R \left (\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right ) dA$$ I don't understand the usage of Green's theorem here. Where does the integral $\oint_c x~dy$ come from? I assume we're letting $N = x$, right?

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the area of the ellipse

$\iint \ dy\ dx$

Lets find a function $f(x,y) = M(x,y)\mathbf i+ N(x,y)\mathbf j$ such that

$\frac {\partial N}{\partial x} - \frac {\partial M}{\partial y} = 1$

how about $f(x) = 0 \mathbf i + x \mathbf j$

then

$\iint \ dy\ dx = \iint \left(\frac {\partial N}{\partial x} - \frac {\partial M}{\partial y}\right) \ dy\ dx = \oint f(x,y)\cdot dr$

$r = (a\cos t, b\sin t)\\ dr = (-a\sin t, b\cos t)\ dt$

$\oint f(x,y)\cdot dr = \int_0^{2\pi} (a\cos t)(b\cos t) \ dt$

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Let $C$ be a closed and smooth curve and $R$ the region enclosed by $C$, then the area defined by $R$ is $$Area_R = \oint_R xdy\,\,\,\,\,\,\,\,\,\,(1)$$ provided $C$ is defined as x = x(t) and y = y(t). To see this, we have that the area of $R$ is $\iint_R dA$, if we apply Green's Theorem to (1), we get $M=0,\,N=x,\,\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}=1$, so $Area_R =\oint_R xdy=\iint_R 1dA.$

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  • $\begingroup$ The line integral is a formula for the area of the closed region. $\endgroup$ – id500 Jan 30 '18 at 18:05

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