# The line integral

I want to check my understanding of the line integral:

For a scalar line integral, what we have geometrically is
the area between a curve a given function, yes? Hence,
it can be thought of as a kind of thin wall, correct? And
where our function is f(x,y)=1, we have the length of the
curve we are integrating over.

For a vector line integral, we actually sum of the unit tangent
vectors along some curve, right?

Char. Limit
Gold Member
Well, for a scalar line integral, at least, I managed to gain some intuition of what it is.

Imagine a surface over the x y plane where the height is determined by f1(x,y). Now draw a path on the surface, where the x and y coordinates of the path are determined by x(t) and y(t). You should have a line on a surface. Now extend the line down (or up) to the plane determined by f2(x,y)=0. Now you should be imagining something like a curtain, I suppose. The line integral is the area of that curtain.

Vector line integrals I have no idea.

LCKurtz
Homework Helper
Gold Member
I want to check my understanding of the line integral:

For a scalar line integral, what we have geometrically is
the area between a curve a given function, yes? Hence,
it can be thought of as a kind of thin wall, correct? And
where our function is f(x,y)=1, we have the length of the
curve we are integrating over.

Yes, the "thin wall" interpretation is correct, but in my opinion hardly ever a good way to think of it. A better model is to think of the curve representing a wire with density per unit length f(x,y). The the line integral ∫c f(x,y) ds represents the mass of the wire. If f(x,y) ≡ 1 you get the length of the wire.

For a vector line integral, we actually sum of the unit tangent
vectors along some curve, right?

Yes. For this think of doing work by moving in a force field F. Then the line integral

$$\int_C \vec F \cdot d \vec R = \int_C \vec F \cdot \hat T\ ds$$

represents the work done by the force in moving along the curve. Notice that the dot product gives the component of the force tangent to the path.

The same ideas hold in 3-D.

LCKurts: thanks for that density interpretation, that's a good way to think about it!