So I have the following question where $$F=(x,y)/\sqrt{1-x^2-y^2}$$ is a vector field. I have to define the domain for $F$, I have rewritten the function as $1/\sqrt{1-x^2-y^2} $ multiplied with $(x,y),$ where $r^2= x^2+y^2$ and thus $x^2+y^2$ is not equal to $1$. But is that it? If not, how do I get the domain for such a function please.
I have not started with the second part of the question yet, where i have to determine the area where $$F$$ is conservative.
Thereafter, calculate the line integral for $$\int_{\mathrm{C1}}F\, dr\, $$ where $$C1$$ is a circle with the center in origin and with radius $1/2$.
i have calculated param $x= 1/2cost$ and $y=1/2sint$ is what I got, where $$0\le t\le 2\pi$$.
It became really complicated when I tried to solve the problem, which got me wondering if it could be related to me skipping the second part of the question?
$\vec{F} = \big(\frac{x}{\sqrt{1-x^2-y^2}}, \frac{y}{\sqrt{1-x^2-y^2}}\big)$.
The domain for the vector field is all real values of $x,y$ where $x^2 + y^2 \lt 1$ which is all points inside a circle of radius $1$ centered at the origin (area of $\pi$).
Now we take derivative and check whether $P_y = Q_x$ where $P$ is the $x-$component of the vector field and $Q$ is the $y-$ component.
$P_y = \displaystyle \frac{\partial}{\partial y} \big(\frac{x}{\sqrt{1-x^2-y^2}}\big) = \frac{xy}{(1-x^2-y^2)^{3/2}}$ (for $x^2+y^2 \lt 1)$
Similarly find $Q_x$ and we see $P_y = Q_x$. That tells us the vector field is conservative where $x^2+y^2 \lt 1$.
As the vector field is conservative, its line integral over any closed curve (whose all points lie in the domain defined above) will simply be zero and that is also true for the circle of radius $\frac{1}{2}$ centered at the origin.