Line integral: how can it be > 0?

[oneamp]

Using my understanding of calculus, I don't understand why line integrals in 3-d space
can give a result > 0. You are following a line and integrating under that line. The line
has some length. But according to my understanding of calculus, it does not have a width.
What is this arbitrary width, and where does it come from?

Thank you

 

[jbunniii]

Imagine a line segment $[a,b]$. Its length is $b-a$. If we embed this line segment in 3-d space, then its length is still $b-a$. Of course, its area and volume are zero. The line integral is measuring length (perhaps with a nonuniform weighting applied), not area or volume.

 

[oneamp]

Yes, that's where I get confused. It measures length with a non-uniform weight applied. Let's say that weight is the height. We have a length and a height. But the area is length*width*height, integrate to solve the integral. But width is zero, so the integral should be zero. What am I missing?

 

[jbunniii]

Area is a 2-dimensional measure: length*height. If you calculate length*width*height, you are measuring volume, not area.

Think about a very thin sheet of paper suspended in 3-dimensional space. To measure its area, I multiply length*height to get, say, 100 square inches. To measure its volume, I multiply length*height*width to get zero. Both calculations are legitimate, depending on what it is you are trying to measure.

 

[pasmith]

Yes, that's where I get confused. It measures length with a non-uniform weight applied. Let's say that weight is the height. We have a length and a height. But the area is length*width*height,

That would be a volume. Area is always a product of two lengths.

integrate to solve the integral. But width is zero, so the integral should be zero. What am I missing?

A line integral is best thought of as
\sum (\mbox{quantity per unit length})(\mbox{small length along a curve}).
It's true that we can find the signed area bounded by the curve y = f(x), the line y = 0 and the lines x = a and x = b by
<br /> \sum (\mbox{perpendicular distance of $(x,f(x))$ from $x = 0$})(\mbox{small length along $x$ axis})<br />
to get \int_a^b f(x)\,dx, and that's what motivates the formal definition of the Riemann integral^*, but from a physical point of view one has really
<br /> \sum (\mbox{small area on the $(x,y)$ plane})<br />
which is a special case of a surface integral, which can be thought of as
<br /> \sum (\mbox{quantity per unit area})(\mbox{small area on a surface}).<br />

Similarly one can have a volume integral, which may be thought of as
<br /> \sum (\mbox{quantity per unit volume})(\mbox{small volume}).<br />

^*One could equally well motivate the Riemann integral as
<br /> (\mbox{position}) = \sum (\mbox{instantaneous velocity})(\mbox{short time})<br />

 

[oneamp]

Thanks