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  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. enter image description here

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    The contour of integration is a horizontal line segment from $(a,0)$ to $(b,0)$, so $y=0$. That's it. It's nothing more than the definition of the integral. – Randall Jan 03 '23 at 00:58
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    It's the ordinary integral of the function $g(x)=f(x,0)$. – Cheerful Parsnip Jan 03 '23 at 01:10
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    Stewart is being silly. Every line integral reduces to a normal single variable integral, after you parameterize the curve and do all the usual stuff. It’s just that some curves are easier to parameters than others. He’s encouraging you to take short cuts. Although that works here, you should be cautious in general about using short cuts. – Deane Jan 03 '23 at 03:05
  • @Deane "Every line integral reduces to a normal single variable integral, after you parameterize the curve and do all the usual stuff" Can you elaborate pls? –  Jan 03 '23 at 05:51
  • I'll answer in two parts. First, a line integral is $$ \int_C\vec{F}\cdot d\vec{r}\text{ or }\int_CP,dx+Q,dy+R,dz.$$ You can't compute the integral directly from these formulas. You need to parameterize the curve $C$ by $\vec{r}(t)=(x(t),y(t).z(t))$, where $a\le t\le b$. The line integral now becomes $$\int_C\vec{F}\cdot d\vec{r}=\int_{t=a}^{t=b}\vec{F}\cdot\vec{r}',dt$$ or $$\int_C P,dx+Q,dy+R,dz=\int_{t=a}^{t=b}(Px'+Qy'+Rz'),dt. $$ – Deane Jan 03 '23 at 17:14
  • Either way, after you've substituted all the formulas you have for $\vec{F},\vec{r},\vec{r}',P,Q,R,x,y,z,x',y',z'$, all of the variables have disappeared except $t$, so the line integral becomes $$\int_{t=a}^{t=b} (\text{formula with $t$ only}),dt.$$ – Deane Jan 03 '23 at 17:15

1 Answers1

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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.

Lemmon
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