What did James Stewart mean by "the line integral reduces to an ordinary single integral in this case" ?
- Please see the question in the title, in reference to the paragraph beside my two green question marks in the image below.
- 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 $?
- 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 $\color{goldenrod}{, 0)}$?
- I disagree that $\int^b_a f(x {\color{goldenrod}{, 0)}} \, dx = \int^b_a f(x) \, dx $ for the following reasons.
- You're starting with different functions. The LHS is a BIvariate function, and the RHS is a UNIvariate function.
- The left side requires you to evaluate $f(x, y)$ at $y = 0$. $f(x)$ requires no evaluation!
I scanned James Stewart, Daniel Clegg, Saleem Watson's Calculus Early Transcendentals, 9 edn 2021, pp. 1132-3.
2 answers
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A line integral is integrated with respect to arc parameter. If the path you're integrating along is the $x$-axis, then the arc parameter can be taken to be just $x$, and so the integral is identical to an ordinary integral with respect to $x$.
You are right that it's a bit improper to write $f(x,0) = f(x)$, since the left-hand side is a two variable function with one variable evaluated at $y=0$, while the right-hand side is just a single variable function (though it may still be useful if you can tolerate some sloppiness).
However I don't see anywhere that Stewart wrote such an equation.
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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 $?
NO.
For any function with two variables, $f(x,y)$, if you fix the value of $y$ at $0$, then $$x\mapsto f(x,0)$$ gives you a function in only one variable. Call this function $g(x)$. Then
$$ \int_a^b f(x,0)\;dx = \int_a^b g(x)\;dx $$is an "ordinary single integral" (of the function $g$).
Consider for instance, $f(x,y)=2x+x^2y$. Then $f(x,0)=2x$ and
$$ \int_0^1f(x,0)\;dx = \int_0^1 2x\;dx=1 $$
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