MHB Calculate directly the curve integrals

[mathmari]

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)

 

[I like Serena]

Hey mathmari! (Smile)

We can't directly apply Green's theoreom on the rectangle because we do not have a continuous derivative in (0,0). (Worried)
However, we can apply it to a so called keyhole contour.

The circle parametrizations should have their center added shouln't they?
And for the proper orientation shouldn't they be clockwise?

And yes, we can parametrize the rectangle by joining those 4 line segments. (Nod)
But shouldn't they be oriented counter clockwise?

 

[mathmari]

We can't directly apply Green's theoreom on the rectangle because we do not have a continuous derivative in (0,0). (Worried)
However, we can apply it to a so called keyhole contour.

Does that means that we apply Green's Theorem on $D$ when we remove the poinr $(0,0)$ ? (Wondering)

The circle parametrizations should have their center added shouln't they?
And for the proper orientation shouldn't they be clockwise?

What do you mean? (Wondering)

And yes, we can parametrize the rectangle by joining those 4 line segments. (Nod)
But shouldn't they be oriented counter clockwise?

Isn't the orientation at the parametrization I wrote counter clockwise? We start from the point $(-2,-3)$, then we go to the point $(7,-3)$, then we go to the point $(7,6)$, then to the point $(-3,6)$ and then again to the point $(-2,-3)$. (Wondering)

 

[I like Serena]

Does that means that we apply Green's Theorem on $D$ when we remove the poinr $(0,0)$ ?

We can't remove only 1 point.
The area must be the area enclosed by a simple closed curve. (Nerd)

However, we can connect the rectangle contour to the region around (0,0) and treat them as a one curve (a keyhole contour).
See for example:
View attachment 7605
The connection is supposed to be zero width.
Note that the inner contour has to be clockwise instead of counter-clockwise for this to work. (Thinking)

What do you mean?

$\sigma_2$ and $\sigma_3$ are currently counter clockwise instead of clockwise aren't they?
And $\sigma_3$ currently has its center at (0,0) instead of (3,3) doesn't it?

Isn't the orientation at the parametrization I wrote counter clockwise? We start from the point $(-2,-3)$, then we go to the point $(7,-3)$, then we go to the point $(7,6)$, then to the point $(-3,6)$ and then again to the point $(-2,-3)$. (Wondering)

Oh yes.

 

[mathmari]

$\sigma_2$ and $\sigma_3$ are currently counter clockwise instead of clockwise aren't they?
And $\sigma_3$ currently has its center at (0,0) instead of (3,3) doesn't it?

So that the orientation is clockwise do we have to define $\sigma_2 = (\cos t, -\sin t), t\in [0,2\pi]$ ?
Is the parametrization of circle with center $(3,3)$ this one: $\sigma_3=(2\cos t+3,-2\sin t+3), t\in [0, 2\pi]$ ?

(Wondering)

 

[I like Serena]

So that the orientation is clockwise do we have to define $\sigma_2 = (\cos t, -\sin t), t\in [0,2\pi]$ ?
Is the parametrization of circle with center $(3,3)$ this one: $\sigma_3=(2\cos t+3,-2\sin t+3), t\in [0, 2\pi]$ ?

That will do it yes. (Nod)

 

[mathmari]

That will do it yes. (Nod)

For the curve integral over $\sigma_1$ I have found the following:
$$\oint_{\sigma_1}=\oint_{\rho_1}+\oint_{\rho_2}+\oint_{\rho_3}+\oint_{\rho_4}=\ldots =\int_{-2}^7-\frac{3}{t^2+9}dt+\int_{-3}^6\frac{7}{49+t^2}dt+\int_0^9\frac{6}{(7-t)^2+36}dt+\int_0^9\frac{2}{4+(6-t)^2}dt$$

The curve integral over $\sigma_2$ is equal to $$\oint_{\sigma_2}\left (-\frac{y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy\right )=\int_0^{2\pi}\left (-(-\sin t)\cdot (-\sin t)+\cos t (-\cos t)\right )dt=\int_0^{2\pi}dt=2\pi$$

The curve integral over $\sigma_3$ is equal to $$\oint_{\sigma_3}\left (-\frac{y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy\right )=\int_0^{2\pi}\left (-\frac{-2\sin t+3}{-12\sin t+12\cos t+22}\cdot (-2\sin t)+\frac{2\cos t+3}{-12\sin t+12\cos t+22}\cdot (-2\cos t)\right )dt=\int_0^{2\pi}\left (\frac{-4\sin^2 t+6\sin t-4\cos^2 t-6\cos t}{-12\sin t+12\cos t+22}\right )dt=\int_0^{2\pi}\left (\frac{-4+6\sin t-6\cos t}{-12\sin t+12\cos t+22}\right )dt=0$$

Can that be correct? (Wondering)

 

[I like Serena]

For the curve integral over $\sigma_1$ I have found the following:
$$\oint_{\sigma_1}=\oint_{\rho_1}+\oint_{\rho_2}+\oint_{\rho_3}+\oint_{\rho_4}=\ldots =\int_{-2}^7-\frac{3}{t^2+9}dt+\int_{-3}^6\frac{7}{49+t^2}dt+\int_0^9\frac{6}{(7-t)^2+36}dt+\int_0^9\frac{2}{4+(6-t)^2}dt$$

The curve integral over $\sigma_2$ is equal to $$\oint_{\sigma_2}\left (-\frac{y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy\right )=\int_0^{2\pi}\left (-(-\sin t)\cdot (-\sin t)+\cos t (-\cos t)\right )dt=\int_0^{2\pi}dt=2\pi$$

The curve integral over $\sigma_3$ is equal to $$\oint_{\sigma_3}\left (-\frac{y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy\right )=\int_0^{2\pi}\left (-\frac{-2\sin t+3}{-12\sin t+12\cos t+22}\cdot (-2\sin t)+\frac{2\cos t+3}{-12\sin t+12\cos t+22}\cdot (-2\cos t)\right )dt=\int_0^{2\pi}\left (\frac{-4\sin^2 t+6\sin t-4\cos^2 t-6\cos t}{-12\sin t+12\cos t+22}\right )dt=\int_0^{2\pi}\left (\frac{-4+6\sin t-6\cos t}{-12\sin t+12\cos t+22}\right )dt=0$$

Can that be correct? (Wondering)

Didn't we lose a minus sign for $\sigma_2$? (Wondering)

And yes, this is as expected.
With Green's theorem you've already found that any contour would result in 0, with only an exception when the contour contains the origin, since the function is not defined there.
It means that we need to handle the origin separately and calculate it by contour integral.
All other contours or combinations of contours (such as the keyhole contour that I mentioned) will result in 0.
And the contour you have for $\sigma_1$ should result in $2\pi$ since it includes the origin (the opposite of $\sigma_2$). (Thinking)