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Let

$G = \{ (x,y) \in \mathbb{R}^2 : x^2+4y^2 >1, x^2+y^2 < 4 \} $

$ \int_G x^2+y^2 d(x,y) $

I want to verify Green's Theorem : $ \oint_{ \partial G } f n ds = \int_G \operatorname{div}f\, dx $

I solved for he righthand side:

$ \int_G \operatorname{div}f dx = \frac{25}{32} \pi $,

just by calculating the Integral for the circle and the ellipse separately and subtract.

For the lefthand side, I need to find $ f: G \rightarrow \mathbb{R^2} $ so that $ \operatorname{div}f = x^2+y^2 $ that can be $ f=( \frac{1}{3} x^3 , \frac{1}{3} y^3 ) $

now my question is, how to proceed? :0 Do I have to split the region? Is $ n= \frac{1}{ \sqrt{ (\frac{1}{3} x^3 )^2+ ( \frac{1}{3}y^3 )^2}} \begin{pmatrix} \frac{1}{3} x^3 \\ \frac{1}{3} y^3 \end{pmatrix} $ ?

Looking forward to any help! :-)

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  • $\begingroup$ Is $x^2+y^2$ already the divergence, or are you getting the divergence from that some how? $\endgroup$ – TurlocTheRed Nov 28 '18 at 19:55
  • $\begingroup$ yes, it is $ div f = x^2+y^2 $ $\endgroup$ – wondering1123 Nov 28 '18 at 20:04
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$\mathbf r$ is not normal to the ellipse. If the boundary is parametrized as $\mathbf r(t) = (x(t), y(t))$, then $\mathbf n(t) = (y'(t), -x'(t))$ is normal to the boundary and $$\int_{\partial D} \mathbf F \cdot \frac {\mathbf n} {|\mathbf n|} \,ds = \int_a^b \mathbf F \cdot \frac {\mathbf n} {|\mathbf n|} \,|\mathbf r'| \,dt = \int_a^b \mathbf F \cdot \mathbf n \,dt.$$ We have $\mathbf F = (x^3/3, y^3/3), \,\mathbf r_1(t) = (\cos t, 1/2 \sin t), \,\mathbf r_2(t) = (2 \cos t, 2 \sin t)$, thus $$\begin{aligned} &\begin{aligned} \int_{\partial D} \mathbf F \cdot \mathbf n \,dt = &\int_0^{2 \pi} \left( \frac {\cos^3 t} 3, \frac {\sin^3 t} {24} \right) \cdot \left( -\frac {\cos t} 2, -\sin t \right) dt + {} \\ &\int_0^{2 \pi} \left( \frac {8 \cos^3 t} 3, \frac {8 \sin^3 t} {3} \right) \cdot (2 \cos t, 2 \sin t) \,dt, \end{aligned} \\ &\int_D \nabla \cdot \mathbf F \,dS = \int_0^{2 \pi} \int_{1/\sqrt{\cos^2 t + 4 \sin^2 t}}^2 \,r^3 dr dt. \end{aligned}$$ Both are equal to $251 \pi/32$.

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$f=x^2+y^2=r^2=f=\nabla \cdot (r^3/4 \hat{r})$

So $\vec{F}=\frac{r^3}{4}\hat{r}$

The normal to a "surface" in 2d is the unit vector perpendicular to the tangent line.

The curves here can be parameterized as

$x=a\cos{\theta}$

$y=b\sin{\theta}$

Where for the ellipse $a=1$, $b=1/2$. For the circle $a=b=2$

The tangent line vector is $-a\sin{\theta}\hat{i}+b\cos{\theta}\hat{j}$

So the normal line vector is $b\cos{\theta}\hat{i}+a\sin{\theta}\vec{j}$

So the unit normal

$\hat{n}= \frac{<b\cos{\theta}, a\sin{\theta}>}{\sqrt{b^2\cos^2{\theta}+a^2\sin^2{\theta}}}$

$\vec{r}=<a\cos{\theta}, b\sin{\theta}>$

So $\vec{F}\cdot \hat{n}dA =\frac{r^2}{4}\frac{ab}{\sqrt{b^2\cos^2{\theta}+a^2\sin^2{\theta}}} d\theta $

You probably want to run that integral clockwise around the ellipse and counter clockwise around the circle. Integrate from 0 to $2\pi$

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  • $\begingroup$ The norm of $n$ doesn't contain an $a b$ term. $\endgroup$ – Maxim Dec 2 '18 at 9:14
  • $\begingroup$ Correct. I took r from r^3 and multiplied it with $\hat{r}$, then dotted that with $\hat{n}$ producing the result. $\endgroup$ – TurlocTheRed Dec 3 '18 at 14:44
  • $\begingroup$ I'm saying that if $\mathbf n = (b \cos \theta, a \sin \theta)$, then $|\mathbf n| = \sqrt{b^2 \cos^2 \theta + a^2 \sin^2 \theta}$. Also, the divergence of $r^3/3 \,\hat {\mathbf r}$ is not $r^2$, it's $4 r^2/3$. And what is $dA$? Area element? $\endgroup$ – Maxim Dec 3 '18 at 16:07
  • $\begingroup$ $\nabla \cdot \vec{A}$ in cylindrical coordinates = $\frac{\partial A_r}{\partial r}+\frac{\partial A_\theta}{r\partial \theta}+\frac{\partial A_z}{\partial z}$ $\vec{F}=r^3/3\hat{r}$ So: $\nabla \cdot \vec{F}=\frac{\partial(r^3/3)}{\partial r}$ $\nabla \cdot \vec{F}=r^2$ $\endgroup$ – TurlocTheRed Dec 4 '18 at 15:58
  • $\begingroup$ I asee what you mean about the norm. Corrected! Thanks! $\endgroup$ – TurlocTheRed Dec 4 '18 at 16:00

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