Define the path integral when scalar $f$ and curve $\mathbf{c}$ is in *curvilinear* coordinates

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I'm doing a multivariable calculus course at the moment. I've seen path integrals in the cartesian coordinate system as the following definition:

Definition. The path integral of $f(x,y,z)$ along the curve $C$ is

$$ \int_C f ds = \int_a^bf(\mathbf{x}(u))||\mathbf{x}'(u)||du$$

where $\mathbf{x}:[a,b]\to\mathbb R^3$ is the parametric representation of $C$. The definition didn't make any reference to the coordinate system.

Question. Suppose that $ (x,y,z)=\mathbf \Phi(\xi_1,\xi_2,\xi_3) $ where the transformation $\mathbf \Phi:U\to V$ is sufficiently differentiable and has a inverse $\mathbf{\Phi^{-1}}:V\to U$ where $U,V$ are open subsets of $\mathbb R^3$.

What would be the definition of a path integral over $C$ when $f$ is a function of curvilinear coordinates ($\xi_1,\xi_2, \xi_3$), and the curve was parameterized in curvilinear coordinates as $\mathbf{\xi}=\mathbf{\xi}(u)$ for $a\le > u\le b$?

Is the definition the same as a regular "cartesian" path integral? If so, can you please derive how you would get a path integral for a function in curvilinear coordinates from the definition above for cartesian?

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Note that the curve $f$ in Cartesian coordinates is given by $\tilde f = f \circ \Phi^{-1}$ and similarly the curve parametrization is given by $\tilde x = \Phi \circ \xi$. So in the Cartesian coordinates the path integral would be

$$ \begin{align} \int_a^b \tilde f (\tilde x(u)) \| \tilde x'(u) \| du &= \int_a^b (f \circ \Phi^{-1} \circ \Phi \circ \xi)(u) \| (\Phi \circ \xi)'(u)\|du \\ &=\int_a^b f(\xi(u)) \|(D\Phi)(\xi(u)) \xi' (u)\|du \end{align}$$

So yes there is no new definition, you just have to write it in terms of a cartesian path integral. You see that the difference is in the Jacobian of the coordinate transformation that we have to respect.