Line Integral Interpretations: Physical and Geometric Uses

[gnome]

I understand that an example of a physical interpretation of the line integral of a scalar function with respect to arc length
\int_C f(x,y,z)ds
might be the total mass of a wire where f describes the linear density of the wire.

But can anybody give an example of a physical or geometric interpretation of the line integral with respect to x (or y, or z)?

i.e. given some specific function f(x,y,z) and points A and B on that curve I can calculate
\int_C f(x,y,z)dx
from A to B to answer an exercise in the textbook.

But what would it mean?

 

[HallsofIvy]

One important application of the line (or path) integral

∫_a^bf(x,y,z)ds is calculating work done by a force.

You know that if you apply a constant force F over a distance x, the work done is Fx. If the force is not a constant, then it is the integral ∫F(x)dx.

If the path over which the work is applied is a curve, then you think of "dx" as the vector "ds" tangent to the curve at each point and "F(x)dx" becomes the dot product of the force vector f(x,y,z) and the ds: ∫_a^bf(x,y,z)ds

 

[gnome]

Thanks, HallsofIvy, but I wasn't referring to integrals of vector functions. Here are examples of the types of integrals I meant.

Say we have a function
f(x,y)=2x (so it's actually independent of y)
and the curve y=x² and we want the integral of that function along the curve from (0,0) to (1,1).

The first type of integral is defined as
\int_C f(x,y)ds = \lim_{n\rightarrow\infty} \sum_{i=1}^n f(x_i^*,y_i^*)\Delta s_i<br />

We write
\int_C 2x \, ds \,\,\,\,and\,\,\,\, C=\{(x,y)|0\leq x \leq 1, 0 \leq y \leq x^2 \}
so dy/dx = 2x, therefore
\int_C 2x\, ds = \int_0^1 2x\sqrt{\left (\frac{dx}{dx}\right )^2 + \left (\frac{dy}{dx}\right )^2}dx
\int_C 2x\, ds = \int_0^1 2x\sqrt{1 + (2x)^2}dx
\,\,\,\,= \frac{1}{6}[(1+4x^2)^\frac{3}{2}\,|_0^1
\,\,\,\, = \frac{1}{6}(5\sqrt{5}-1)


The second type is defined as
\int_C f(x,y)dy = \lim_{n\rightarrow\infty} \sum_{i=1}^n f(x_i^*,y_i^*)\Delta y_i<br />
So for the same function and the same curve and limits we can write
dy = 2x dx and then
\int_C 2x\, dy = \int_0^1 2x \times 2x\,dx
\,\,\,\, = \int_0^1 4x^2\,dx
\,\,\,\, = \frac{4}{3}\,x^3\,|_0^1
\,\,\,\, = \frac{4}{3}

I find this second type of integral defined in James Stewart's Multivariable Calculus 5e on page 1101, and in various exercises at the end of that section, in case you have access to a copy. But he gives no discussion as to what it's purpose might be.