Integrating Basis Vectors of Cylindrical Coordinates Hello I have a problem where $\hat{p}$  is a basevector in a cylindrical system
$$\int_{0}^{\pi/4}\hat{p} d\theta$$
I know that; $\hat{p} = \hat{x}\cos\theta + \hat{y}\sin \theta$
$$
\hat{x}\int_{0}^{\pi/4}\cos\theta d\theta + \hat{y}\int_{0}^{\pi/4}\sin \theta d\theta = (1/\sqrt{2})\hat{x} + (1-(1/\sqrt{2}))\hat{y}
$$
But then I get an answer in cartesian? Is this the right way to proceed?
 A: Yes. Your Process is right. If you want your final answer in cylindrical coordinates note that
$$\left\{ \matrix{
  \hat p\left( \theta  \right) = \cos \theta \hat x + \sin \theta \hat y \hfill \cr 
  \hat \theta \left( \theta  \right) =  - \sin \theta \hat x + \cos \theta \hat y \hfill \cr}  \right.$$
and then solving for ${\hat x}$ and ${\hat y}$ leads to
$$\left\{ \matrix{
  \hat x = \cos \theta \hat p - \sin \theta \hat \theta  \hfill \cr 
  \hat y = \sin \theta \hat p + \cos \theta \hat \theta  \hfill \cr}  \right.$$
and hence you can write your final answer in cylindrical coordinates.
Also, some other tricks may work. According to the first pair of equations above, you can simply conclude that
$$\left\{ \matrix{
  {{d\hat p} \over {d\theta }} = \hat \theta  \hfill \cr 
  {{d\hat \theta } \over {d\theta }} =  - \hat p \hfill \cr}  \right.$$
and hence your final answer should be
$$\int_0^{{\pi  \over 4}} {\hat pd\theta }  =  - \int_0^{{\pi  \over 4}} {{{d\hat \theta } \over {d\theta }}d\theta }  = \left. { - \hat \theta } \right|_0^{{\pi  \over 4}} =  - \hat \theta \left( {{\pi  \over 4}} \right) + \hat \theta \left( 0 \right)$$
A: Your answer is correct.  So, let's look at the resulting vector $\vec A$ given by
$$\vec A=\hat x\frac{\sqrt{2}}{2}+\hat y\left(1-\frac{\sqrt{2}}{2}\right)$$
What is the magnitude of the vector $\vec A$.  It is $|\vec A|=\sqrt{\frac12+\frac32-\sqrt{2}}=\sqrt{2-\sqrt{2}}$
What is the angle $\theta$ that this vector makes with the x-axis?  It is $\theta =\arctan(\sqrt{2}-1)=\pi/8$.  
Therefore, in terms of the radial unit vector $\hat \rho (\theta)$ is
$$\bbox[5px,border:2px solid #C0A000]{\vec A=\hat \rho(\pi/8)\left(\sqrt{2-\sqrt{2}}\right)}$$
where 
$$\begin{align}
\hat \rho(\pi/8)&=\hat x \cos \pi/8+\hat y \sin \pi/8\\\\
&=\hat x\frac{\sqrt{2+\sqrt{2}}}{2}+\hat y\frac{\sqrt{2-\sqrt{2}}}{2}
\end{align}$$
