# Line Integrals

by gnome
Tags: integrals, line
 P: 1,046 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?
 P: 1,046 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 $$\int_C f(x,y)ds = \lim_{n\rightarrow\infty} \sum_{i=1}^n f(x_i^*,y_i^*)\Delta s_i$$ 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 $$\int_C f(x,y)dy = \lim_{n\rightarrow\infty} \sum_{i=1}^n f(x_i^*,y_i^*)\Delta y_i$$ 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.