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

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?

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}
• Hi thanks for the reply! So later in the lecture slides, they claimed: $\int_C f ds = \int_a^b f(\xi (u))||\mathbf{x}'(u)||du$ Could you explain why your integral is the same as what they have please? – user523384 Apr 15 at 12:37
• well the $x$ you just used in your formul is the same as my $\tilde x$ and we have $\tilde x = \Phi \circ \xi$ which results exactly in this equation. To get to this for you just have to expand the first term (that includes $\tilde f$ but not the second term (that includes the norm). – flawr Apr 15 at 12:39
• Just to clarify, is the line integral of the function on the LHS ($\tilde{f}$) different to the function those line integral is to be determined ($f$)? I see how they are related, but would the outputs of the two functions be the same in the range $u\in (a,b)$? Is that why the line integral remain unaffected? – user523384 Apr 15 at 12:43
• Yes $\tilde f$ is the same function as $f$ but with transformed input coordinates. The function $\tilde f$ "lives" on the rectilinear coordinates while the function $f$ "lives" on the curvilinear coordinates. The outputs are therefore the same, they just have a different parametrization. If we would only compute directly in the curvilinear coordinates, the velocity term $\xi'$ would have no meaning, however in the cartesian coordinates the length of $x'$ weighs the function values appropriately. – flawr Apr 15 at 12:48