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…
Best Answer
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}