Communities

Writing
Writing
Codidact Meta
Codidact Meta
The Great Outdoors
The Great Outdoors
Photography & Video
Photography & Video
Scientific Speculation
Scientific Speculation
Cooking
Cooking
Electrical Engineering
Electrical Engineering
Judaism
Judaism
Languages & Linguistics
Languages & Linguistics
Software Development
Software Development
Mathematics
Mathematics
Christianity
Christianity
Code Golf
Code Golf
Music
Music
Physics
Physics
Linux Systems
Linux Systems
Power Users
Power Users
Tabletop RPGs
Tabletop RPGs
Community Proposals
Community Proposals
tag:snake search within a tag
answers:0 unanswered questions
user:xxxx search by author id
score:0.5 posts with 0.5+ score
"snake oil" exact phrase
votes:4 posts with 4+ votes
created:<1w created < 1 week ago
post_type:xxxx type of post
Search help
Notifications
Mark all as read See all your notifications »
Q&A

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

+1
−5
  1. Please see the question in the title, in reference to the paragraph beside my two green question marks in the image 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 the following reasons.
    1. You're starting with different functions. The LHS is a BIvariate function, and the RHS is a UNIvariate function.
    2. 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.

Pages 1132 and 1133 of Calculus Early Transcendentals

History
Why does this post require moderator attention?
You might want to add some details to your flag.
Why should this post be closed?

3 comment threads

wrong interpretation of the phrase in 2 (1 comment)
Title equation doesn't appear in excerpt (2 comments)
Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be... (1 comment)

2 answers

You are accessing this answer with a direct link, so it's being shown above all other answers regardless of its activity. You can return to the normal view.

+2
−0

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

$$ \int_0^1f(x,0)\;dx = \int_0^1 2x\;dx=1 $$
History
Why does this post require moderator attention?
You might want to add some details to your flag.

1 comment thread

I don't understand the context of the "no" (3 comments)
+0
−0

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.

History
Why does this post require moderator attention?
You might want to add some details to your flag.

0 comment threads

Sign up to answer this question »