Independence of path of line integral

1. Jan 16, 2012

sharks

The problem statement, all variables and given/known data

The attempt at a solution
$$\frac{\partial P}{\partial y}=\frac{2y}{x^3}$$
$$\frac{\partial Q}{\partial x}=\frac{2y}{x^3}$$
$$\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}$$

According to my notes: Both functions are continuous, except at x=0.
My tentative explanation: This is due to the denominator in $\frac{2y}{x^3}$ which will make the fraction equal to infinity at x=0. Therefore the functions won't be continuous at x=0. Is this explanation correct?

Apparently, at this stage, i am free to choose whatever path i please. I'm going to start from path y=1 (1,1) to (2,1) and then move up the path x=2 (2,1) to (2,2).

Then, i do direct substitution:
$$I=\int^2_1\frac{11}{x^3}\,dx+\int^2_1 -\frac{5}{4}\,ydy$$
Is this correct?
After i did the integration, i get the final answer: $-\frac{87}{8}$ but it's wrong.

2. Jan 16, 2012

The1337gamer

I think if your line integral is path independent, you can use the Fundamental Theorem of Calculus as your vector field is just a gradient field of a scalar function.

3. Jan 16, 2012

sharks

I don't understand what you're talking about.

4. Jan 16, 2012

sharks

For the path i chose in the first post, the graph is like this:

I think you're wrong about the path going "under" the origin, as the line x=0 is the y-axis. So the graph is not allowed to cross the y-axis but can go under the x-axis. Correct?

5. Jan 17, 2012

sharks

If the answer is 9/4, then what is wrong with my calculation from post #1?
Is this expression correct?
$$I=\int^2_1\frac{11}{x^3}\,dx+\int^2_1 -\frac{5}{4}\,ydy$$

6. Jan 17, 2012

sharks

Can anyone help?

7. Jan 17, 2012

ehild

It is correct, and the result is 9/4. How did you get different result? Show details.

ehild

8. Jan 17, 2012

sharks

Hi ehild!
$$\int^2_1\frac{11}{x^3}\,dx=\int^2_111x^{-3}\,dx=-\frac{11}{2x^2}=\frac{33}{8}$$
$$\int^2_1 -\frac{5}{4}\,ydy=-\frac{5y^2}{8}=-\frac{15}{8}$$
$$\frac{33}{8}-\frac{15}{8}=\frac{9}{4}$$
Indeed, the answer is $\frac{9}{4}$. However, in my notes, the answer is $-\frac{9}{8}$ for this problem. It must be a mistake in my notes.

Last edited: Jan 17, 2012
9. Jan 18, 2012

ehild

Your notes are wrong then. It happens...

And the functions are not continuous at x=0, because of the x in the denominator.

ehild

10. Jan 18, 2012

sharks

To confirm my suspicions on this:

For example, if $$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}=\frac{2x}{y^2}$$
Then, the functions are not continuous at y=0.

And if $$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}=\frac{2x}{x+y^2}$$
Then, the functions are not continuous at (0,0).

Are my conclusions correct?

11. Jan 18, 2012

ehild

I do not get you. What functions are you talking about? Of course, the functions P and Q are not continuous where the denominator is zero (x=0). So you need to use such path for integration which does not include x=0.

ehild

12. Jan 18, 2012

sharks

In my post #12 above, i just gave 2 random examples, as i wanted to understand that it's really the denominator from $\frac{\partial P}{\partial y}$ and $\frac{\partial Q}{\partial x}$ which determine whether the functions P and Q are continuous or not. Since there is also x^3 and x^2 in the denominators from the original problem. I'm sorry if this seems a bit confusing.

13. Jan 20, 2012

sharks

Ah OK! Thanks for this clarification, sheriff89.