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I know from vector calculus, that line integral is the area of the curtain under the curve. Then, i'm realize we can solve the line integral with respect to $x$ or $y$ .

Integration with respect to $y$ is just like ordinary definite integral. But what about with respect to $x$? Is it an arc length?

And i'm still confuse with the closed integral. Is this kind of integral calculate the area which is bounded by closed path?

Please give me the best explanation. Thanks. Anyway, my teacher just teached me about calculating, not a concept. That's why i don't understand a bit.

user516076
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  • Integral is nothing but an ideal version of 'summing up small quantities'. So the meaning of the integral entirely depends on what you are summing up. For instance, if $f(x)$ is the linear density so that the infinitesimal $f(x) , \mathrm{d}x$ represents the small mass deposited on the small interval $[x,x+\mathrm{d}x]$, then the integral $\int_{a}^{b}f(x),\mathrm{d}x$ is simply the sum of these small masses, and so, the integral itself also represents the mass. This is no different for the line integral or any other types of integrals that you might encounter elsewhere. – Sangchul Lee Nov 05 '19 at 23:59

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You have two kinds of line integrals.

One where $F(x,y) = z$ gives has a one dimensional output, and represents a surface in $\mathbb R^3$. You are integrating over some path $\mathbb r$ in the $xy$ plane that is parametrized by $t.$

$\int_r F(x,y)\ d\mathbb r = \int F(x,y) \|\mathbb r(t)\|\ dt$

One way to visualize this, is that you have some curtain that falls from the surface directly above your path down to the path, and what is the area of that curtain.

The other $F(x,y) = (P(x,y),Q(x,y))$ gives a multi-dimensional, or vector as an output. Now we traditionally think of $F$ as representing a force-field.

$\int_r F(x,y)\cdot d\mathbb r$ represents the amount of energy that will be required to move though this force-field.

user317176
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