Why can one use Greens theorem in this problem?

Question

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 :)

Answer 1

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.

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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.

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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.

Answer 2

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$