I'm not sure. It looks overwhelmingly complicated to me.Orodruin said:
You might want to consider the start and end coordinates of the path (corresponding to ##t=0## and ##t= \pi##). You might want to consider whether or not the field is conservative.Graham87 said:
Compare the first terms of each expression. What do you see?Graham87 said:
Looks like some derivative or primitive function variation. But it's not in a series.Orodruin said:
There is really no "magic" too it. Just that ##3 x^2 y## is quite clearly the ##x##-derivative of ##x^3 y## ... of which ##x^3## is the ##y##-derivative, etc etcGraham87 said:
A line integral is a type of integral where a function is evaluated along a curve. In the context of curve γ, it is used to compute quantities like work done by a force field along a path, or to evaluate integrals of scalar and vector fields over a curve in space.
"Simplifying substitutions" refers to the process of making algebraic or trigonometric substitutions to transform the integral into a more manageable form. This often involves parameterizing the curve γ using a simpler variable, which can make the integral easier to evaluate.
To parameterize a curve γ, you express the coordinates of points on the curve as functions of a single parameter, typically denoted as t. For example, if γ is a curve in the xy-plane, you might write it as γ(t) = (x(t), y(t)), where t ranges over some interval. This transformation allows you to rewrite the line integral in terms of t.
Common substitutions include trigonometric substitutions, such as using sine and cosine functions for circular paths, or hyperbolic functions for certain types of curves. Additionally, linear substitutions that reparameterize the curve in terms of a simpler variable are frequently used.
Sure! Consider the line integral ∫γ (x^2 + y^2) ds, where γ is the unit circle x^2 + y^2 = 1. A common simplifying substitution is to parameterize the circle using x = cos(t) and y = sin(t), where t ranges from 0 to 2π. The integral then becomes ∫02π (cos^2(t) + sin^2(t)) dt, which simplifies to ∫02π 1 dt, giving a result of 2π.