I want to determine the following curve integral
$$ \int_{\gamma} \vec{F} \cdot d\vec{r} \qquad,\qquad \vec{F}= \bigl( \sin(y-x) , 2xy+\sin(x-y) \bigr) \quad,\quad \gamma \quad : \quad y=\sqrt{x} \quad 0 \leq x \leq 1. $$
So I thought that Green's formula would be useful, and therefore we need a closed curve to use it, i choose the following
So we get
$$ \int_\gamma + \int_u + \int_v \quad= \quad \iint_K \qquad , \quad \text{K: The inside of the closed curve}. $$
For the first line segment we get the following parameterization
$$ \int_u \quad : x=1 \quad \rightarrow \quad dx=0 \quad : \quad \int_1^0 \sin(t-1) \cdot 0 + 2t+\sin(1-t) \, dt = [t^2+ \cos(1-t)]_1^0 \quad=\quad \cos(1)-2 $$
And the second
$$\int_v \quad : y=0 \quad \rightarrow \quad dy=0 \quad : \quad \int_1^0 \sin(0-t) \cdot dt + (0 + \sin(t-0) \cdot 0 dt = [\cos(t)]_1^0 \quad=\quad 1 - \cos(1) $$
Applying Green's formula:
$$ \iint_K \bigl( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \bigr) dxdy = \int_0^1 \int_0^{\sqrt{x}} 2y- \cos(x-y) + \cos(x-y) dxdy =$$ $$ \int_0^1 \int_0^{\sqrt{x}} 2y dxdy = \int_0^1 [y^2]_0^{\sqrt{x}} dx \quad=\quad \int_0^1 x dx \quad=\quad \frac{1}{2} $$
Finally we get:
$$ \int_\gamma = \iint_K - \int_u - \int_v \quad=\quad \frac{1}{2} - \cos(1) + 2 - 1 + \cos (1) = \frac{3}{2} $$
But if try another approach with the following closed curve
I get for the line integral
$$\int_W \quad : y=t \quad \rightarrow \quad x=t \quad : \quad \int_1^0 sin(t-t) \cdot dt + (2t^2 + sin(t-t) \cdot dt = [\frac{2t^3}{3}]_1^0 = \frac{-2}{3} $$
Green's formula
$$ \iint_K \bigl( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \bigr) dxdy = \int_0^1 \int_x^{\sqrt{x}} 2y dxdy \quad=\quad \int_0^1 [y^2]_x^{\sqrt{x}} dx = \int_0^1 x-x^2 dx = \frac{1}{6} $$
So we get
$$ \int_\gamma = \frac{1}{6} + \frac{2}{3} = \frac{5}{6} $$
Which is not the same answer! But I should get the same answer regardless of the curve I choose right? What did I go wrong here?
Recall that in the Green's formula the linear integral around the domain should be positively oriented i.e. counterclockwise. On the other hand, your closed curves $\gamma\cup u \cup v$ and $\gamma\cup w$ are both clockwise. Then, with the given orientations, you should change the sign to the double integral.
For the first one, $$\int_\gamma= -\iint_K - \int_u - \int_v =-\frac{1}{2}- \cos(1) + 2 - 1 + \cos (1)=\frac{1}{2}$$ and for the second one, $$\int_\gamma= -\iint_K- \int_w =-\frac{1}{6}+\frac{2}{3}=\frac{1}{2}.$$ As expected, in both cases, we get the same result.