Evaluate the integral along the paths

  • Thread starter Thread starter Schwarzschild90
  • Start date Start date
  • Tags Tags
    Integral
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 2K views
Schwarzschild90
Messages
111
Reaction score
1

Homework Statement


opgave formulering.PNG


Homework Equations

and 3. The Attempt at a Solution [/B]
Blundell.PNG


The assignment that I'm struggling with can be seen under the heading titled 1. and my attempt at a solution can be seen in 2. and 3.

Obviously, what I'm doing is wrong. I've surely misunderstood the problem statement. Will someone please help me?

Thank you in advance.

-Schwarzschild
 
Physics news on Phys.org
Schwarzschild90 said:

Homework Statement


View attachment 103046

Homework Equations

and 3. The Attempt at a Solution [/B]
View attachment 103047

The assignment that I'm struggling with can be seen under the heading titled 1. and my attempt at a solution can be seen in 2. and 3.

Obviously, what I'm doing is wrong. I've surely misunderstood the problem statement. Will someone please help me?

Thank you in advance.

-Schwarzschild
There seems to be only two attachments to your post. I think attachment #3 is missing.
 
Hi SteamKing

There's no third attachment, sorry, English is not my first language.

-Schwarzschild
 
Not sure what you doing there, let's take

(I) for the path ##(x_1,y_1)->(x_2,y_1)->(x_2,y_2)## (going through straight line segments) we ll have

##\int\limits_{(x_1,y_1)}^{(x_2,y_2)}2xydx=\int\limits_{(x_1,y_1)}^{(x_2,y_1)}2xydx+\int\limits_{(x_2,y_1)}^{(x_2,y_2)}2xydx=\int\limits_{(x_1,y_1)}^{(x_2,y_1)}2xydx+ 0=(x_2^2-x_1^2)y_1##.

You can work similar for ##\int (x^2+2xy)dy## seeing that it ll be zero for the straight line segment##(x_1,y_1)->(x_2,y_1)## so you need to evaluate it only for the straight line segment ##(x_2,y_1)->(x_2,y_2)##

Then you should calculate same things for the path in (II)

If the answer you get in (I) is different than that in (ii) then we can safely say that it is not an exact differential (why?).