Last line integral problem (hopefully)

1. Apr 30, 2005

PhysicsMajor

Greetings again,

Show that for F(x,y)=<2xy-3, x^(2)+4y^(3)+5> the line integral F(x,y).dr is independant of path. Then evaluate the line integral for any curve C with initial point (-1,2) and the terminal point (2,3).

Thanks again, you all have been very helpful.

2. Apr 30, 2005

Data

has your class covered exact differentials?

3. Apr 30, 2005

PhysicsMajor

i don't believe so unless its known by another name?

4. Apr 30, 2005

Data

I'll post a detailed explanation tomorrow (if no one else has by then). I doubt you would know the subject by a different name.

Last edited: Apr 30, 2005
5. Apr 30, 2005

Oggy

A line integral is independent of path if there exists a function U, that Fdr is it's exact (total) differential. In this case U=x^2y-3x+5y+y^4.
dU=(2xy-3)dx+(x^2+4y^3+5)dy=Fdr.

6. Apr 30, 2005

HallsofIvy

Staff Emeritus
An "exact differential" fdx+ gdy (a physics major may prefer to think of it as a "conservative force field") is, as Oggy said, one such that there exist a function U such that dU= fdx+ gdy. By the "chain rule", $$dU= \frac{\partial U}{\partial x}dx+ \frac{\partial U}{/partial y}$$ so we must have $$f= \frac{\partial U}{/partial x}$$ and [tex]g= \frac{\partial U}{/partial g}. A quick way to test that is to use the "cross derivative test: If the second partials are continuous, that requires that
[tex]\frac{\partial f}{/partial y}= U_{xy}= U_{yx}= \frac{\partial g}{\partial x}[/itex].