Integrating Basis Vectors of Cylindrical Coordinates

Question

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?

Answer 1

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

$$\int_0^{{\pi \over 4}} {\hat pd\theta } = - \int_0^{{\pi \over 4}} {{{d\hat \theta } \over {d\theta }}d\theta } = \left. { - \hat \theta } \right|_0^{{\pi \over 4}} = - \hat \theta \left( {{\pi \over 4}} \right) + \hat \theta \left( 0 \right)$$

Answer 2

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

$$\begin{align} \hat \rho(\pi/8)&=\hat x \cos \pi/8+\hat y \sin \pi/8\\\\ &=\hat x\frac{\sqrt{2+\sqrt{2}}}{2}+\hat y\frac{\sqrt{2-\sqrt{2}}}{2} \end{align}$$