# Given a vector field, $\vec{F}$, and let $D$ be the domain of $\vec{F}$, how can we tell if $D$ is connected and/or simply connected?

by 0000005   Last Updated August 14, 2019 15:20 PM

For example: How can we tell that for $$\vec{F} = \bigg(\dfrac{x}{x^2+y^2},\dfrac{y}{x^2+y^2},z^2\bigg),$$ $$D$$ is connected but not simply connected? Where $$D$$ is the domain of $$\vec{F}$$.

Is this true? We can tell it is connected because $$\vec{F}$$ is conservative since we can test that the $$(\nabla\times\vec{F})=0$$. If this is true, is there an easier way to see that this is connected?

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#### Answers 1

In essence, $$D$$ is connected but not simply connected because there are a series of points which are undefined in the domain.

Consider the origin, $$(0,0,0)$$; well this point is undefined because we have $$0$$ in the denominator of the $$x$$ and $$y$$ coordinates.

Additionally, there are singularities $$\forall (0,0,z)$$ points, i.e. all points along the $$z$$-axis.

As you have asserted, we know that $$\vec{F}$$ is conservative because $$\nabla\times\vec{F}=0$$. These aforementioned singularities are removable discontinuities. If you put these ingredients together we have that $$D$$ is connected.