Line integral over Archimedes's spiral Compute $\int \limits_{C} F.dr$ for $F(x,y)=(x,y)$ and $C$ is the Archimedes's spiral given in polar coordinates by $r=1+\theta$ for $\theta \in [0,2\pi]$
How do I parametrize this? And how do find I the limits?
 A: Switching the vector field $\vec{F}$ to polar coordinates, we get $\vec{F}(r,\theta) = (r\cos\theta, r \sin\theta)$. Now, since we are on Archimedes's spiral, we can substitute $r = 1 + \theta$ to find 
$$
\vec{F}(\theta) = \bigg( (1+\theta)\cos\theta, (1+\theta)\sin\theta \bigg)
$$
Similarly, 
\begin{align*}
d\vec{r} = \bigg( dx, dy \bigg) &= \bigg( \cos\theta dr -r\sin\theta d\theta, \sin\theta dr + r\cos\theta d\theta \bigg)\\
&= \bigg( \cos\theta d\theta -(1+\theta)\sin\theta d\theta, \sin\theta d\theta +(1+\theta)\cos\theta d\theta \bigg)\\
&= \bigg( \left( \cos\theta -(1+\theta)\sin\theta \right)d\theta, \left(\sin\theta+(1+\theta)\cos\theta \right)d\theta \bigg)
\end{align*}
As we are only interested in the portion of the spiral cut out by the angles $\theta \in [0,2\pi]$, the line integral becomes
\begin{align*}
\int_C \vec{F} \cdot d\vec{r} 
&= \int_{0}^{2\pi} (1+\theta)\cos^2\theta -(1+\theta)^2 \cos\theta\sin\theta  + (1+\theta)\sin^2\theta + (1+\theta)^2 \sin\theta\cos\theta d\theta\\
&= \int_0^{2\pi} (1+\theta)d\theta\\
&=\theta + \frac{\theta^2}{2} \bigg|_0^{2\pi}\\
&= 2\pi + 2\pi^2.
\end{align*}
It's important in this problem to distinguish between the two terms $r$ and $\vec{r}$, which play very different roles: $r$ is the variable for the polar coordinates in the plane, and $\vec{r}$ is the radial vector in the line integral.
