Calculate the line integral of a half circle as a standing unit circle? Calculate the line integral
$$
\rm I=\int_{C}\mathbf{v}\cdot d\mathbf{r}
\tag{01}
$$
where
$$
\mathbf{v}\left(x,y\right)=y\mathbf{i}+\left(-x\right)\mathbf{j}
\tag{02}
$$
and $C$ is the semicircle of radius $2$ centred at the origin from $(0,2)$ to $(0,-2)$ to the negative x axis (left half-plane).

I have used the parametrisation of $\mathbf{r}\left(t\right) = (2\cos t, 2\sin t)$, $t \in [0,{\pi}]$. 

The answer I get is -4$\pi$


I have no idea if this is correct or not. Is my orientation correct, is my bound for $t$ correct, since this is a closed unit circle would it not be $[{\pi/2},{-\pi/2}]$, etc... 
 A: 
Since the problem is a special case, the answer is given in Figure without any integration.
If the curve (semicircle) lies to the positive $x$ as in above Figure, then
$$
\int_{C+} \mathbf{v}\circ d\mathbf{r}=\int_{C+}\| \mathbf{v}\|ds=\int_{C+}2\cdot ds=2\int_{C+}ds=2\cdot\left(\pi \rm R \right)=+4\pi
\tag{01}
$$
while if lies to the negative $x$, then 
$$
\int_{C-} \mathbf{v}\circ d\mathbf{r}=\int_{C-}\left[-\| \mathbf{v}\| \right]ds=-\int_{C+}2\cdot ds=-2\int_{C-}ds=2\cdot\left(\pi \rm R \right)=-4\pi
\tag{02}
$$

(01). Case $C_{\boldsymbol{+}}$ :  
A parametrization of the curve $C_{\boldsymbol{+}}$ in the right half-plane, as  in Figure, from positive $y$ to negative $y$ would be :
$$
\mathbf{r}=2\:\left(\sin t,\cos t\right),  \quad  t \in \left[0,\pi\right]
\tag{01-a}
$$
so
\begin{align}
\mathbf{v} & = 2\:\left(\cos t,-\sin t\right)
\tag{01-b}\\
d\mathbf{r} & =2\:\left(\cos t,-\sin t\right)dt
\tag{01-c}\\
\mathbf{v} \boldsymbol{\cdot}  d\mathbf{r} & = 4\:\left(\cos t,-\sin t\right) \boldsymbol{\cdot}\left(\cos t,-\sin t\right)dt =\:\boldsymbol{+}\:4\:dt
\tag{01-d}\\
{\rm I} &=\int_{C_{\boldsymbol{+}}}\mathbf{v} \boldsymbol{\cdot}  d\mathbf{r}  =\:\boldsymbol{+}\:4\:\int_{0}^{\pi}dt =\:\boldsymbol{+}\:4\:\pi
\tag{01-e} 
\end{align}
(02). Case $C_{\boldsymbol{-}}$ :  
A parametrization of the curve $C_{\boldsymbol{-}}$ in the left half-plane from positive $y$ to negative $y$ would be :
$$
\mathbf{r}=2\:\left(\cos t,\sin t\right),  \quad  t \in \left[\pi/2,3\pi/2\right]
\tag{02-a}
$$
so
\begin{align}
\mathbf{v} & = 2\:\left(\sin t,-\cos t\right)
\tag{02-b}\\
d\mathbf{r} & =2\:\left(-\sin t,\cos t\right)dt
\tag{02-c}\\
\mathbf{v} \boldsymbol{\cdot}  d\mathbf{r} & = 4\:\left(\sin t,-\cos t\right) \boldsymbol{\cdot}\left(-\sin t,\cos t\right)dt =\:\boldsymbol{-}\:4\:dt
\tag{02-d}\\
{\rm I} &=\int_{C_{\boldsymbol{-}}}\mathbf{v} \boldsymbol{\cdot}  d\mathbf{r}  =\:\boldsymbol{-}\:4\:\int_{\pi/2}^{3\pi/2}dt =\:\boldsymbol{-}\:4\:\pi
\tag{02-e} 
\end{align}
A: The integral becomes, with your parametrization:
$$\int_0^\pi(2\sin t,-2\cos t)\cdot(-2\sin t,2\cos t)dt=\int_0^\pi-4\;dt=-4\pi$$
so yes: I'd say it seems to be you got it right!
