Line integral of vector field using polar coordinates I am solving 
$$\int_C \vec F(x,y) \cdot \,d\vec r$$
where
$$ \vec F(x,y)= \begin{bmatrix}5y \\10x\end{bmatrix}$$
and $C$ is the quarter-circle arc centered at $(0,2)$ and going from the origin to $(2,2)$.
I am able to get the correct answer using Cartesian coordinates as follows:
$$\begin{align}
\vec r(t) &= \begin{bmatrix}2\cos t \\2-2\sin t\end{bmatrix}, \quad t \in (\pi/2, 0) \\
\vec F(t) &= \begin{bmatrix}10 - 10\sin t \\20 \cos t\end{bmatrix}\\
d\vec r &= \begin{bmatrix}-2\sin t \\-2 \cos t\end{bmatrix}\, dt\\
\int_C \vec F(x,y) \cdot \,d\vec r & = \int_{\pi/2}^0 -20 \sin t + 20 \sin^2 t - 40 \cos^2 t\,dt = 35.71
\end{align}$$
(Notice that the integral is done "backwards" because of how I parameterized the curve.)
I am trying to do the same using polar coordinates and getting stuck. Since the curve as stated is not centered around the origin, I first define
$$\vec G(x, y) = \vec F(x, y+2)$$
Then $$\int_C \vec F(x,y) \cdot \,d\vec r = \int_{C'} \vec G(x,y) \cdot \,d\vec s $$
where $C'$ is the quarter-circle arc centered at the origin and going, in Cartesian coordinates, from $(0,-2)$ to $(2,0)$.
Switching to polar coordinates $(r,\theta)$, we have 
$$\begin{align}
\vec s(t) &= \begin{bmatrix}2 \\ t \end{bmatrix}, \quad t \in (-\pi/2, 0) \\
\vec G_\mathrm{polar}(x,y) &= \begin{bmatrix} \sqrt{\left(5y+10\right)^2 + \left(10x\right)^2} \\ \arctan\left(\frac{10x}{5y+10}\right)\end{bmatrix}
&\text{convert $(5y, 10x)$ to polar}\\
\vec G_\mathrm{polar}(r,\theta) & = \begin{bmatrix} \sqrt{\left(5r\sin\theta+10\right)^2 + \left(10r\cos\theta\right)^2} \\ \arctan\left(\frac{10 r\cos\theta}{5r\sin\theta+10}\right)\end{bmatrix}
&\text{replace x and y with their polar equivalents}\\
\vec G_\mathrm{polar}(t) & = \begin{bmatrix} \sqrt{\left(10\sin t+10\right)^2 + \left(20 \cos t\right)^2} \\ \arctan\left(\frac{20 \cos t}{10\sin t+10}\right)\end{bmatrix}
&\text{sub in parameterized $r, \theta$}\\
d\vec s &= \begin{bmatrix}0 \\ 1 \end{bmatrix}\, dt \\
\int_{C'} \vec G(x,y) \cdot \,d\vec s & = \int_{-\pi/2}^0 \arctan\left(\frac{20 \cos t}{10\sin t+10}\right) \,dt = 2.13 \neq 35.71
\end{align}$$
What am I doing wrong? I think I have failed to apply the chain rule somewhere.
 A: To convert $G$ properly we need the definitions of $\hat{r}$ and $\hat{\theta}$:
$$\begin{cases}\hat{r} = \cos\theta\:\hat{x} + \sin\theta\:\hat{y} \\ \hat{\theta} = -\sin\theta\:\hat{x} + \cos\theta\:\hat{y} \\ \end{cases} \implies \begin{cases}\hat{x} = \cos\theta\:\hat{r} - \sin\theta\:\hat{\theta} \\ \hat{y} = \sin\theta\:\hat{r} + \cos\theta\:\hat{\theta} \\ \end{cases}$$
Then we get that
$$\vec{G}(x,y) = (5y+10)\:\hat{x} + 10x\:\hat{y}$$
$$ = (5r\sin\theta+10)\cdot(\cos\theta\:\hat{r} - \sin\theta\:\hat{\theta}) + (10r\cos\theta)\cdot(\sin\theta\:\hat{r} + \cos\theta\:\hat{\theta})$$
$$= (15r\sin\theta\cos\theta+10\cos\theta)\:\hat{r}+(10r\cos^2\theta - 5r\sin^2\theta - 10\sin\theta)\cdot\hat{\theta} \equiv \vec{G}(r,\theta)$$
In this case, we parametrize $r(t) = 2$ and $\theta(t) = t$, giving us that
$$d\vec{s} = d\vec{r} + rd\vec{\theta} = (0\:\hat{r} + 2 \: \hat{\theta})dt \equiv \begin{pmatrix}0 \\ 2\\ \end{pmatrix}dt$$
giving us a line integral of
$$\int \vec{G}(r,\theta)\cdot d\vec{s} = \int_{-\frac{\pi}{2}}^0 40\cos^2t-20\sin^2t-20\sin t\:dt = 5\pi + 20$$
