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mathmari

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Hey!

We consider the space $D$ that we get if we remove from the square $[-2,7]\times [-3,6]$ the open discs with center the point $(0,0)$ and radius $1$ and with center $(3,3)$ and radius $2$.

I want to calculate $$\sum_{j=1}^3\oint_{\sigma_j}\left (-\frac{y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy\right )$$ where $\sigma_1, \sigma_2, \sigma_3$ are the boundary curves of $D$ with the positive orientation in relation to $D$.

Can we calculate directly the curve integrals or do we have to apply Green's Theorem? (Wondering)

If we apply Green's Theorem we get the following: $$\sum_{j=1}^3\oint_{\sigma_j}\left (-\frac{y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy\right )=\iint_D\left (\frac{\partial{\left (\frac{x}{x^2+y^2}\right )}}{\partial{x}}-\frac{\partial{\left (-\frac{y}{x^2+y^2}\right )}}{\partial{y}}\right )dxdy=\iint_D\left (\frac{-x^2+y^2}{(x^2+y^2)^2}+\frac{x^2-y^2}{(x^2+y^2)^2}\right )dxdy=\iint_D\left (\frac{0}{(x^2+y^2)^2}\right )dxdy=0$$ right?

To calculate the curve integral we have to find a parametrization of $D$, right? The space $D$ is:

View attachment 7602

So, $\sigma_1$ is the parametrization of the rectangle, $\sigma_2$ the parametrization of the small circle and $\sigma_3$ of the big circle, right?

We have that $\sigma_2=(\cos t,\sin t), t\in [0, 2\pi]$ and $\sigma_3=(2\cos t,2\sin t), t\in [0, 2\pi]$. Is everything correct?

How could we parametrize the rectangle? Do we consider for that the parametraization $\sigma_1$ as the union of the following? $$\rho_1=(t, -3), -2\leq t\leq 7 \\ \rho_2=(7, t), -3\leq t\leq 6 \\ \rho_3=(7-t, 6), 0\leq t\leq 9 \\ \rho_4=(-2, 6-t), 0\leq t\leq 9$$

(Wondering)

We consider the space $D$ that we get if we remove from the square $[-2,7]\times [-3,6]$ the open discs with center the point $(0,0)$ and radius $1$ and with center $(3,3)$ and radius $2$.

I want to calculate $$\sum_{j=1}^3\oint_{\sigma_j}\left (-\frac{y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy\right )$$ where $\sigma_1, \sigma_2, \sigma_3$ are the boundary curves of $D$ with the positive orientation in relation to $D$.

Can we calculate directly the curve integrals or do we have to apply Green's Theorem? (Wondering)

If we apply Green's Theorem we get the following: $$\sum_{j=1}^3\oint_{\sigma_j}\left (-\frac{y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy\right )=\iint_D\left (\frac{\partial{\left (\frac{x}{x^2+y^2}\right )}}{\partial{x}}-\frac{\partial{\left (-\frac{y}{x^2+y^2}\right )}}{\partial{y}}\right )dxdy=\iint_D\left (\frac{-x^2+y^2}{(x^2+y^2)^2}+\frac{x^2-y^2}{(x^2+y^2)^2}\right )dxdy=\iint_D\left (\frac{0}{(x^2+y^2)^2}\right )dxdy=0$$ right?

To calculate the curve integral we have to find a parametrization of $D$, right? The space $D$ is:

View attachment 7602

So, $\sigma_1$ is the parametrization of the rectangle, $\sigma_2$ the parametrization of the small circle and $\sigma_3$ of the big circle, right?

We have that $\sigma_2=(\cos t,\sin t), t\in [0, 2\pi]$ and $\sigma_3=(2\cos t,2\sin t), t\in [0, 2\pi]$. Is everything correct?

How could we parametrize the rectangle? Do we consider for that the parametraization $\sigma_1$ as the union of the following? $$\rho_1=(t, -3), -2\leq t\leq 7 \\ \rho_2=(7, t), -3\leq t\leq 6 \\ \rho_3=(7-t, 6), 0\leq t\leq 9 \\ \rho_4=(-2, 6-t), 0\leq t\leq 9$$

(Wondering)

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