# Independence of Path

1. Feb 28, 2009

### gr3g1

The question is:

Show that the given integral is independent of the path.

F(x,y) = (2xy)dx + (x^2)dy

So i take the integral of 2xy w.r.t x and it gives:

x^2*y + g(y)

now I take the partial derivative of that function w.r.t to y and i get:

x^2 + g'(y)

I set it equal to (x^2)

so

x^2 + g'(y) = x^2

so g'(y) = 0

so i would integrate

x^2*y + 0

is that right?

Thank you

2. Feb 28, 2009

### Dick

Yes. Though you could have written x^2*y+C where C is any constant.

3. Feb 28, 2009

### HallsofIvy

Simply to show that the integral is independent of the path, it is sufficient to show that the function you found exists- and to do that just observe that $(2xy)_y= 2x= (x^2)_x$.

4. Feb 28, 2009

### gr3g1

Thanks a lot guys, but the problem is that when I evaluate the integral on the points (2,4) and (1,1)

I get (4)(4) - (1)(1) = 15

Now, the assignment says I must also integrate along any convenient path between the points.

So, I use the function y = x^2 as my function and replace y in:
F(x,y) = (2xy)dx + (x^2)dy

So I get: F(x) = 2x^3 + x^2

I integrate that to:

(1/2) * x^4 + (1/3) * x^3

Now I evaluate using the points 2 and 1 (the x values from the two given points)

I get: 8 + (8/3) - (1/2) - (1/3) which is equal to 9.8333 and not what I am expected to get! I am suppose to get 15

Does anyone see my error?

Thanks again
Greg

5. Mar 1, 2009

### atyy

How about getting dy in terms of x and dx?

6. Mar 1, 2009

### gr3g1

I am not familiar with that..

My book tells me I have to just find a function y that matches the two points, and replace that y into the original equation

7. Mar 1, 2009

### atyy

y=x^2
dy/dx=2x
dy=2xdx

So F(x)=2x^3dx+x^2(2xdx)

8. Mar 1, 2009

### gr3g1

Ok, so now I assume I just have to integrate?

2x^3+2x^3

becomes :

x ^ 4

and yes!! I get the right answer!!

Im not sure as to what 2xdx is.. or what you just did.. There is no mention of this in my material.. I am very grateful for your help... but Im pretty lost.. why is my book only telling me to reply the function with the new y?
I dont understand what you just did or why

9. Mar 1, 2009

### atyy

OK, go ahead and integrate and see if the answer comes out right, so we can figure out if I'm making any sense. If it is I'll explain the 2xdx.

10. Mar 1, 2009

### gr3g1

Your right!! I i just edited my post on top..
Thanks a lot!

Whats with the 2xdx?

11. Mar 1, 2009

### atyy

You wanted to integrate

F(x,y) = (2xy)dx + (x^2)dy

on the path y = x^2.

So every where in F(x,y) you must replace all "y" terms with the appropriate function of x. This includes the "y" in "dy"

I've put that in bold for emphasis.

Ok, so how do you get dy in terms of an appropriate function of x?

On the path, y=x^2

So if you differentiate you get dy/dx=2x

Now multiply both sides with dx and you get dy(dx)/dx=2x(dx).

The dx/dx on the left side becomes 1, so you get dy=2x(dx).

12. Mar 1, 2009

### gr3g1

Thanks alot!!

Just one last question, is this the procedure I must apply to all questions of similar nature? Why is my book just simply replacing y?

13. Mar 1, 2009

### gr3g1

Wait, the examples include functions such as y = x + 1.
this would dy/dx as 1.....

I think I got it now!