Geometry of integrating the $\hat{r}$ basis vector from $\theta = 0$ to $\pi/4$ Is there a geometrical interpretation of integrating the $\hat{r}$ unit basis vector over $\theta$ from 0 to $\pi/4$ in cylindrical coordinates?
$$\int_{0}^{\pi/4}\hat{r} d\theta$$
An old question (Integrating Basis Vectors of Cylindrical Coordinates) asked if this was the correct approach:
$$\hat{r} = \hat{x}\cos\theta + \hat{y}\sin \theta$$
\begin{align*}
\int_{0}^{\pi/4}\hat{r} d\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}
\end{align*}
Two answerers confirmed that this is correct and that it is a vector pointing in the direction of $\theta = \pi/8$ with length
$$\sqrt{\frac12+\frac32-\sqrt{2}}=\sqrt{2-\sqrt{2}}\approx 0.77$$
The $\hat{z}$ unit basis vector plays no role, so I assume that this problem might as well be in polar rather than cylindrical coordinates?  Is there a geometrical interpretation or, better yet, a figure associated with integrating the radial unit basis vector from  $\theta = 0$ to $\pi/4$ and getting a vector that makes an angle with the x-axis of $\theta = \pi/8$ and has length $\approx 0.77$?
Context:  This question came up at the beginning of an intermediate course on electricity and magnetism. I believe I understand what it means to integrate a vector field over a curve, surface, or volume, but what does it mean to integrate a vector field as the half-plane goes from $\theta =0$ to $\theta=\pi/4$?
 A: Technically, "the $\hat r$ unit basis vector" at at any one angle $\theta$ lives in a tangent space different from the $\hat r$ vector at any other angle, so it seems weird to integrate these vectors.
But you can certainly integrate a unit vector defined as
$\hat r(\theta) = \hat x\cos\theta + \hat y\sin\theta$
from $\theta = 0$ to $\theta = \pi/4.$
One way to think about the vector integration is to consider the vector as the velocity vector of a particle. Integrating a velocity vector over time, you end up with a vector representing the particle's displacement from its starting point.
Since the path direction changes at a uniform rate, the path that the particle follows is a circular arc. Since the speed of the particle is $1$ ($\hat r$ is a unit vector),
the particle travels $\pi/4$ distance along this arc while changing its direction by the angle $\pi/4$ radians. This tells us the radius of the arc is $1.$
If we put the center of the arc at the origin, the particle should start at
$(0,-1)$ so that its initial velocity is $\hat r(0) = \hat x$.
Then it travels $\pi/4$ of the distance around the unit circle starting at
$(0,-1),$ therefore ending up at $(1/\sqrt2, -1/\sqrt2).$
The displacement from $(0,-1),$ to $(1/\sqrt2, -1/\sqrt2)$ is
$$ (1/\sqrt2, 1-1/\sqrt2), $$
and that's the integral of  $\hat r(\theta)$ from $\theta = 0$ to $\theta = \pi/4.$
And since this vector is a chord of the unit circle subtending an angle $\pi/4$ at the center, it makes an angle $\pi/8$ with the tangent at $(0,-1),$
which tells you the angular direction of the displacement vector.
A: Maybe this is more computational than what you're looking for, but still...
If you rotate the whole situation by an angle of $\pi/8$ clockwise, you integrate over the interval $-\pi/8 \le \theta \le \pi/8$ instead. We may just as well consider the more general situation $-\beta \le \theta \le \beta$, and then we get
$$
\int_{-\beta}^{\beta} \hat{r}(\theta) \, d\theta
= \int_{-\beta}^{\beta} \begin{pmatrix} \cos\theta \\ \sin\theta \end{pmatrix} d\theta
= \begin{pmatrix} 2\sin\beta \\ 0 \end{pmatrix}
= 2 \sin\beta \, \hat{x}
.
$$
So the $y$-component is zero because of symmetry, and the amount of $x$-component that you get is obtained from knowing that the integral of cosine is sine, so to get a geometrical interpretation of the length of the resulting vector, you would need some way of visualizing that integral.
(I don't really have any good suggestions for that in this situation.)
If you then rotate the whole picture back by the angle $\beta$ counterclockwise, you get
$$
\int_{0}^{2\beta} \hat{r}(\theta) \, d\theta
= 2 \sin\beta \, \hat{u},
$$
where
$$
\hat{u} = \cos\beta \, \hat{x} + \sin\beta \, \hat{y}
$$
is the unit vector that you get when you rotate $\hat{x}$.
EDIT: This is the picture that I have in mind when I look at the integral $\int_{-\beta}^{\beta} \hat{r}(\theta) \, d\theta$. We approximate it with a finite sum instead, with angular step length $\Delta \theta$. For the purpose of integrating (= “adding up”) the vectors $\hat{r}(\theta)$, they can be moved so that their tails are at the origin, and then they form a “fan” as in the picture. When we add them all up, we get a rather long vector pointing in the $x$ direction (for symmetry reasons). Then we multiply the whole thing by $\Delta \theta$, which rescales the result so that its length is a little shorter than $1$ (since it's the average length of the $x$-components of all the vectors). And the integral is of course obtained in the limit as $\Delta \theta \to 0$.

For the integral $\int_{0}^{2\beta} \hat{r}(\theta) \, d\theta$, just rotate the picture.
A: $
\newcommand\dd{\mathrm d}
\newcommand\INT{\int_0^{\pi/4}}
\renewcommand\vec\mathbf
\newcommand\bvec\boldsymbol
$It is the displacement between the points $(r,\theta)=(1,0)$ and $(1,\pi/4)$ rotated clockwise by $\pi/2$. You get the same interpretation replacing $0$ and $\pi/4$ by any two angles.
Let's use some geometric calculus to evalutate this integral in a coordinate-free manner:
$$
  \INT\hat{\vec r}\,\dd\bvec\theta
    = \INT\hat{\vec r}\hat{\bvec\theta}\hat{\bvec\theta}\,\dd\theta
    = I\int_C\dd\vec r
    = I\int_C\dd\vec r\,\partial\vec r
    = I\Bigl[\vec r(1, \pi/4) - \vec r(1, 0)\Bigr].
$$
Multiplying a vector by $I$ on the left rotates it by $\pi/2$ clockwise, giving exactly the interpretation I gave at the start. I will discuss each equality in turn:

*

*$\hat{\bvec\theta}$ is a unit vector so $\hat{\bvec\theta}\hat{\bvec\theta} = \hat{\bvec\theta}^2 = 1$.

*$I = \hat{\vec r}\hat{\bvec\theta}$ is the right-handed unit pseudoscalar, which is a constant and comes out of the integral. $\dd\theta$ is the line element of the unit circle and $\hat{\bvec\theta}$ is a unit vector tangential to the unit circle, so under the conventions I am using $\hat{\bvec\theta}\,\dd\theta = \dd\vec r$ is the oriented line element. We decide to think of the integral as being over a circular arc $C$ of unit radius starting at $(r,\theta)=(1,0)$ and ending at $(r,\theta)=(1,\pi/4)$. $\int_C\dd\vec r$ could be interpreted as the sum of all infinitesimal tangents along the path $C$, which intuitively is the displacement between the endpoints of $C$. The next two steps confirm this intuition; they are unnecessary since $\int_C\dd\vec r$ is well-known, but it is a fun exercise.

*$\partial$ is the tangential derivative; $\nabla = \partial + \partial_\perp$ where $\partial$ differentiates along the curve and $\partial_\perp$ differentiates orthogonal to it. $\partial_\perp\vec r = \hat{\vec r}(\partial\vec r/\partial r) = \hat{\vec r}\hat{\vec r} = 1$ and $\nabla\vec r = \nabla\cdot\vec r = 2$ (the dimension of the space), so $\partial\vec r = 2 - 1 = 1$.

*By the Fundamental Theorem of Geometric Calculus (an incarnation of the generalized Stokes' Theorem)
$$
  \int_S\dd^{k+1}\vec r\,\partial F(\vec r) = \int_{\partial S}\dd^k\vec r\,F(\vec r)
$$
where $S$ is a $(k+1)$-dimensional surface, $\partial S$ is its boundary, and e.g. $\dd^k\vec r$ is the $k$-vector valued oriented $k$-surface measure. In our case, $k = 0$, $S = C$, and $F(\vec r) = \vec r$. The $\dd^0\vec r$ integral is to be interpreted as a sum over the oriented endpoints of $C$, the starting point getting $-1$ and the finishing point $+1$. By $\vec r(r, \theta)$ we mean the position vector with polar coordinates $(r, \theta)$.

To show that this is indeed correct and not gibberish, let's calculate:
$$
  \vec r(1, \pi/4) - \vec r(1, 0)
= \hat x\cos\pi/4 + \hat y\sin\pi/4 - \hat x\cos0 - \hat y\sin0
= \left(\frac1{\sqrt2} - 1\right)\hat x + \frac1{\sqrt2}\hat y.
$$
$I = \hat x\hat y$ and
$$
  \hat x\hat y\hat x = -\hat y,\quad \hat x\hat y\hat y = \hat x,
$$
so our final expression is
$$
   \frac1{\sqrt2}\hat x + \left(1 - \frac1{\sqrt2}\right)\hat y
$$
as it should be.
