Prove Independence of Path for F(x,y) Integral

In summary: Yes, the book tells you to just find a function y that matches the two points, and replace that y into the original equationy=x^2dy/dx=2xdy=2xdx.
  • #1
gr3g1
71
0
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
 
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  • #2
Yes. Though you could have written x^2*y+C where C is any constant.
 
  • #3
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 [itex](2xy)_y= 2x= (x^2)_x[/itex].
 
  • #4
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
gr3g1 said:
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

How about getting dy in terms of x and dx?
 
  • #6
Hi, thanks for the reply..
I am not familiar with that..

Can you please explain?

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

So F(x)=2x^3dx+x^2(2xdx)
 
  • #8
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 I am pretty lost.. why is my book only telling me to reply the function with the new y?
I don't understand what you just did or why
 
  • #9
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
Your right! I i just edited my post on top..
Thanks a lot!

Whats with the 2xdx?
 
  • #11
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
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
Wait, the examples include functions such as y = x + 1.
this would dy/dx as 1...

I think I got it now!
 

1. What does it mean to prove independence of path for an integral?

Proving independence of path for an integral means showing that the value of the integral does not depend on the path taken to reach the endpoints of integration. In other words, the integral will have the same value regardless of the specific path chosen.

2. Why is it important to prove independence of path for an integral?

Proving independence of path is important because it ensures that the value of the integral is well-defined and does not change depending on the path chosen. This is a fundamental concept in multivariable calculus and has many applications in physics and engineering.

3. How do you determine if an integral is independent of path?

To determine if an integral is independent of path, you can use one of two methods: the fundamental theorem of line integrals or the method of checking for conservative vector fields. The fundamental theorem states that if the integral is evaluated along any two paths between the same endpoints, the values will be equal. The method of conservative vector fields involves checking if the vector field associated with the integral is conservative, which means it can be written as the gradient of a scalar function.

4. Can you give an example of an integral that is independent of path?

One example of an integral that is independent of path is the line integral of a conservative vector field. For instance, the integral of the vector field F(x,y) = (xy, x^2) along any path from (0,0) to (1,1) will have the same value of 1/4, regardless of the specific path chosen.

5. What are some real-world applications of proving independence of path for an integral?

Proving independence of path has many real-world applications, especially in physics and engineering. For example, in fluid dynamics, the circulation of a fluid around a closed loop is independent of the path taken by the fluid, which can be mathematically represented by an integral. This concept is also crucial in understanding electromagnetic fields and their behavior, as well as in maximizing efficiency in engineering design processes.

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