I have this problem:
given the points $A(-1,-2)$ and $B(2,1)$ the path $T$ from $A$ to $B$ is on the circule $(x-0.5)^{2}+(y+0.5)^{2}=4.5$
caculate: $$\int _{T}\frac{ydx-xdy}{x^{2}+y^{2}}$$
after a short calculate I found out that: $P_{y}=Q_{x}$ and with this following simply connected space $D$, making $\vec{F}$ conservative vector field:
Therefore I can choose a different path between the two points.
Now my main error is that I calculate the integrals with two different paths and I've given different results.
- for the path $C$ on the circle $x^{2}+y^{2}=5$ I've got that:
the red path is part of the circule.
$$\left\{\begin{matrix}
x= \sqrt[]{5}\cos(t) & dx = (-1)\sqrt[]{5}\sin(t)dt \\
y= \sqrt[]{5}\sin(t) & dy = \sqrt[]{5}\cos(t)dt
\end{matrix}\right.$$
and by using caculator:$$\int _{T}\frac{ydx-xdy}{x^{2}+y^{2}} = \int _{C}\frac{((\sqrt[]{5}\sin(t)\cdot (-1)\sqrt[]{5}\sin(t))-(\sqrt[]{5}\cos(t)\cdot\sqrt[]{5}\cos(t)))dt}{5}$$$$=-1\cdot\int _{C}1\cdot dt = -1\cdot\int_{\arccos (\frac{-1}{\sqrt[]{5}})}^{\arccos (\frac{2}{\sqrt[]{5}})} 1 \cdot dt = \frac{\pi }{2}$$ - for the path $E$ wich is the sum of the following paths: $E1$, $E2$ I'm getting diffrent result.
$$E_{1}(t) = (t,-2)\rightarrow\left\{\begin{matrix} dx= dt \\ dy= 0 \end{matrix}\right., -1\leq t\leq 2$$ $$\int _{E_{1}}\frac{ydx-xdy}{x^{2}+y^{2}} = \int_{-1}^{2}\frac{-2}{t^{2}+4}dt$$
but
$$E_{2}(t) = (2,t)\rightarrow\left\{\begin{matrix}
dx= 0 \\
dy= dt
\end{matrix}\right., -2\leq t\leq 1$$
$$\int _{E_{2}}\frac{ydx-xdy}{x^{2}+y^{2}} = \int_{-2}^{1}\frac{-2}{t^{2}+4}dt$$
in this picture the paths $E_{1}$ and $E_{2}$ are drwan in pink: 
and I'm getting using caculator that: $$\int _{T}\frac{ydx-xdy}{x^{2}+y^{2}} = \int_{-1}^{2}\frac{-2}{t^{2}+4}dt + \int_{-2}^{1}\frac{-2}{t^{2}+4}dt \approx -2.49$$
so, where is my error?
The principal range for arccos(x) is from 0 to $\pi$, so when you calculate the angle for the (-1, -2) point, you're actually getting the angle for the (-1, +2) point. So your first approach's integral should be:
$$ -1\cdot\int_{-\arccos (\frac{-1}{\sqrt[]{5}})}^{\arccos (\frac{2}{\sqrt[]{5}})} 1 \cdot dt = -2.498 $$
The change is in the lower limit of integration. That doesn't account for all the error though.
The last part of your error is in the final calculation of the second approach. It should be -2.498, says wolframalpha