Why can one use Greens theorem in this problem? Find the lineintegral $$\oint F\bullet dr$$ given the vector function
$$F(x,y)=(x^2-y^2-3x)i+e^{x/ \sqrt{y}}j$$
and the curve C being the boundary of the area in the first quadrant where $x,y\geq 0$ and the curves $x=0, y=x^2$ and $y=1$ define the area.
So the solution tells me to use Greens theorem. But I don't understand why I can use this theorem, because is it not a hole in the domain? The vector function has $\sqrt{y}$ in the denominator, so the point $(0,0)$ is a hole? Then Greens theorem is not valid in this case? Really appreciate some help understanding this :)
 A: From wiki:

Let $C$ be a positively oriented, piecewise smooth, simple closed curve in a plane, and let $D$ be the region bounded by $C$. If $L$ and $M$ are functions of $(x, y)$ defined on an open region containing $D$ and having continuous partial derivatives there, then



where the path of integration along $C$ is anticlockwise.


Update: While the function is not defined at the origin in your setup, the line integral is well-defined as an improper integral, and Green's theorem thus works. It's  similar to how the integral $\int_0^1 \frac{1}{\sqrt x}dx$ is well-defined even though the integrand is not defined at $x=0$.
If it helps, you can explicitly compute the line integral and convince yourself it is well-defined along your contour. Let
$$L(x,y):=x^2-y^2-3x\\
M(x,y):=\exp\{x/\sqrt y\}.$$
Moving counterclockwise, parametrize the parabolic, horizontal, and vertical segments respectively by $\mathcal{C}_1,-\mathcal C_2, -\mathcal C_3,$ where
$$\mathcal C_1:(x,y)=(t,t^2),t\in [0,1]\\
\mathcal C_2: (x,y)=(t,1),t\in[0,1]\\
\mathcal C_3:(x,y)=(0,t),t\in[0,1].$$
Then the line integral is given by
$$\oint_{\mathcal C_1} F\cdot dr+\oint_{-\mathcal C_2} F\cdot dr+\oint_{-\mathcal C_3} F\cdot dr\\
=\int_0^1 t^2-(t^2)^2-3t dt+\lim_{\epsilon\downarrow 0}\int_\epsilon^1 \exp\{t/\sqrt {t^2}\} dt\quad (\mathcal C_1)\\
-\int_0^1 t^2-(1)^2-3t dt\quad (\mathcal -C_2)\\
-\lim_{\epsilon\downarrow 0}\int_\epsilon^1 \exp\{0/\sqrt {t}\} dt\quad (\mathcal -C_3)\\
=\left(1/3-1/5-3/2+e\right)-\left(1/3-1-3/2\right)-\left(1\right)\\
=e-1/5.$$
This should agree with the answer obtained via Green's theorem. Note the contour is important here; the integral need not converge along another contour, say, one where the parabolic segment is replaced by $y=x^4$; Green's theorem would likewise give a divergent result in this case.

So in summary, singularities on the boundary do not stop you from using Green's theorem. At such points, one may use an improper integral and proceed.
A: Yes you can apply Green's theorem as the point $(0, 0)$ is on the boundary. Also, you can consider a region in the first quadrant with boundary curves given by $x = 0, y = x^2, y = \delta, y = 1$. This excludes point $(0, 0)$.
Given Vector field, $F(x,y) = \left((x^2-y^2-3x), e^{x/ \sqrt y}\right)$
Applying Green's theorem, the desired line integral in counter-clockwise direction is -
$ \lim_{\delta \to 0} \displaystyle \int_{\delta}^{1} \left[\int_0^{\sqrt y} \left(\frac 1 {\sqrt y} e^{x/ \sqrt y} + 2y\right) dx\right] ~ dy$
