Here is the definition of line integral as it appears in Apostol's Calculus, Volume II
Let $\pmb{\alpha}$ be a piecewise smooth path in n-space defined on an interval $[a,b]$, and let $\pmb{f}$ be a vector field defined and bounded on the graph of $\pmb{\alpha}$. The line integral of $\pmb{f}$ along $\pmb{\alpha}$ is denoted by the symbol $\int\pmb{f}\cdot d\pmb{\alpha}$ and is defined by the equation
$$\int_a^b\pmb{f}[\pmb{\alpha}(t)]\cdot\pmb{\alpha}'(t)dt$$
whenever the integral on the right exists, either as a proper or improper integral.
Consider the following path $\pmb{\alpha}$ in 3-space
$$\pmb{\alpha}(\theta)=R(\sin{\theta}\hat{i}+\cos{\theta}\hat{k}), \theta\in [0,\pi]\tag{1}$$
and a vector field
$$\pmb{f}(r,\theta,\phi)=r^2(\cos{\theta}\hat{r}+\cos{\phi}\hat{\theta}-\cos{\theta}\sin{\phi}\hat{\phi})\tag{2}$$
We see that $\pmb{\alpha}$ is a function of one parameter $\theta$, and the righthand side expression in (1) gives a linear combination of the standard basis vectors. This coincides with Cartesian coordinates.
$\pmb{f}$ is a function of the three spherical coordinates, and (2) gives a linear combination of the spherical coordinate vectors.
Now, we can convert between standard basis vectors and spherical coordinate vectors, so we can have both $\pmb{\alpha}$ and $\pmb{f}$ in terms of either the standard basis vectors or the spherical coordinate vectors.
My question is how do $\pmb{\alpha}$ and $\pmb{f}$ fit into the definition of line integral. That is what is the line integral of $\pmb{f}$ along $\pmb{\alpha}$?
More specifically, what is $\pmb{f}[\pmb{\alpha}(\theta)]$ in this example?
After all, $\pmb{\alpha}$ is in Cartesian in (1) and $\pmb{f}$ takes spherical in (2).
If $f$ is defined in terms of the spherical coordinates it is best to translate the expression of $\alpha(\theta)$ into spherical coordinates so that you can easily compute $f$ at $\alpha(\theta).$ Alternatively you could express $f$ into cartesian coordinates and leave $\alpha$ as is but this would be a bad idea. Indeed the curve $\alpha$ is a circular arc. The choice of using spherical coordinates is more appropriate for this problem.
The path $\alpha$ is a circular arc $$ (R \sin \theta,0,R \cos \theta) : \theta \in [0,\pi]$$ Recall that the spherical coordinates are given by
\begin{aligned}x&=r\sin \theta \,\cos \phi ,\\y&=r\sin \theta \,\sin \phi ,\\z&=r\cos \theta .\end{aligned}
So the path $\alpha$ can be described in spherical coordinates as the place where the radius $r$ and the angle $\phi$ are constant with $r = R$ and $\phi = 0$.
From this we deduce two facts :
$\alpha'(\theta) = \frac{d \alpha}{d \theta} = R \hat \theta.$
The restriction of $f$ to the curve $\alpha$ is $$f(R,\theta,0) = R^2 \cos \theta \hat r + R^2 \hat \theta$$
It follows that
$$ \int_0^\pi f(\alpha (\theta)) \cdot\alpha'(\theta) d\theta = \int_0^\pi R^2 \cos \theta \;(\hat r \cdot R\hat \theta) + R^2 \; (\hat \theta \cdot R\hat \theta) \; d\theta$$ Now use the fact the the vectors $\hat r, \hat \phi , \hat \theta$ are orthonormal this last integral can be reduced to
$$ \int_0^\pi R^3 d\theta = \pi R^3.$$