What did James Stewart mean by "the line integral reduces to an ordinary single integral in this case"?

Question

1. What did James Stewart mean by "the line integral reduces to an ordinary single integral in this case"? See the para. aside my two green question marks below.

2. How do you symbolize "the line integral reduces to an ordinary single integral in this case"? $\int^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a f(x) \, dx $?

3. From $\int^b_a f(x \color{goldenrod}{, 0)} \, dx $, how exactly do you deduce $= \int^b_a f(x) \, dx$? What warrants you to drop and disregard the $\color{goldenrod}{, 0)}$?

4. I disagree that $\int^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a f(x) \, dx $ for these reasons.

4.1. You're starting with different functions. The LHS is a BIvariate function, and the RHS is a UNIvariate function.

4.2. The left side requires you to evaluate $f(x, y)$ at $y = 0$. $f(x)$ requires no evaluation!

I scanned James Stewart, Calculus Early Transcendentals, 9 edn 2021, pp. 1132-3.

Answer 1

Let us define a bivariate function $f:\mathbb{R}^2 \to \mathbb{R}$. We wish to find the line integral along the straight-line path $C$ from $(a,0)$ to $(b,0)$, and will assume that this integral is defined.

We can parametrize this path as $(x(t),y(t))$ where $x(t) := t$, and $y(t) := 0$ for $t \in [a,b]$. Thus, our line integral is

$$\begin{aligned}\int_C{f(x,y)\,\mathrm{d}s}&=\int_a^b{f(x(t),y(t)) \sqrt{\left(x'(t)\right)^2 + \left(y'(t)\right)^2}\,\mathrm{d}t} \\&=\int_a^b{f(t,0) \sqrt{\left(\frac{\mathrm{d}}{\mathrm{d}t}t\right)^2+\left(\frac{\mathrm{d}}{\mathrm{d}t}0\right)^2}\,\mathrm{d}t} \\&=\int_a^b{f(t,0) \sqrt{1^2+0^2}\,\mathrm{d}t} \\&=\int_a^b{f(t,0)\,\mathrm{d}t}\end{aligned}$$

Now, by change-of-variable, rename $t$ to $x$,

$$=\int_a^b{f(x,0)\,\mathrm{d}x}$$ And defining a new univariate function $g:\mathbb{R}\to\mathbb{R}$ as a restriction of $f(x,y)$ to $y=0$, such that $g: x \mapsto f(x,0)$, we can see that

$$\int_a^b{f(x,0)\,\mathrm{d}x}=\int_a^bg(x)\,\mathrm{d}x$$

And so our line integral simplifies to an ordinary single integral.

You can see that the RHS univariate function is a restriction of the LHS bivariate function, and that we are still making implicitly the same evaluation.

I hope that this elucidates Stewart's reasoning here.