Question

Can anyone explain the gist of line integrals?

Answer 1

So when you are taking regular integral, you are finding area under the curve right

Answer 2

Correct, at least for single integrals.

Answer 3

In double integral, you have a region, you are trying to find area over it. In line integral you are going along the line instead of region

Answer 4

Alright, I see how it is different now. But what is this parmeterization about?

Answer 5

You are adding up value of function along the line would be better statement

Answer 6

So, if I have \[\int\limits_{?}^{?}y^3\] over c (c= x=t^3, y=t, 0<t<2), that means I am integrating the function y^3 over the line x=t^3 ?

Answer 7

Say that you are trying to find line integral of 1 along a circle of radius 1. We know from geometry that we should get circumference which is 2pi.Let's try using line integral.
\[\int 1\text{ }\text{ds}\]
r = {Cos[t], Sin[t]} 0<t<2pi

\[\text{ds}=\sqrt{\left(r'\right)^2}\text{dt} \]
\[\int_0^{2 \pi } 1 \, dt\]
=2pi

Answer 8

In a typical integral, you are summing the area of infinitely many small rectangles each one having a width of dx and a height of f(x).

In a line integral you are summing up the length of infinitely many line segments, where each line segment is \(\sqrt{dx^2 + dy^2}\) long. We can rewrite this:

\[\sqrt{dx^2 + dy^2} \]\[= \sqrt{\frac{dx^2}{dx^2}(dx^2 + dy^2)}\]\[=\sqrt{1 + (\frac{dy}{dx})^2}\ dx\]

Answer 9

So, an integral can be thought of as an area, a line integral can be thought of as a perimeter. So in my example, I am trying to find the perimeter of y^3 over that region. Is that correct?

Answer 10

well, it may be that you don't want the perimeter. You may want to know how much energy you use moving along a line, or something similar. In that case you'd then multiply the length by some other function that would give you the energy per distance, etc.

Answer 11

but in the most basic example yes, you're finding the perimeter.

Answer 12

In this picture, you have black curve and function value in red. you are adding up function value over the black curve

Answer 13

thanks guys!

Answer 14

It is a crude drawing but that's what got me intution