Integral in two dimensions

**jedoob** · February 23rd 2006, 10:56 AM

Hi, I am fairly new to line and curve integrals - I was wondering how to the following question:

Evaluate: The curve integral C of (x/((x^2)-(y^2)))dx + (y/((y^2)-(x^2)))dy
where C is any curve from (1,0) to (5,3) in the xy plane that does not meet the lines y=x and y=-x.

Also, could someone please let me know how to write proper mathematical symbols in this forum like I've seen in some of the threads.

Thanks.

**topsquark** · February 23rd 2006, 11:33 AM

Originally Posted by jedoob

Hi, I am fairly new to line and curve integrals - I was wondering how to the following question:

Evaluate: The curve integral C of (x/((x^2)-(y^2)))dx + (y/((y^2)-(x^2)))dy
where C is any curve from (1,0) to (5,3) in the xy plane that does not meet the lines y=x and y=-x.

Also, could someone please let me know how to write proper mathematical symbols in this forum like I've seen in some of the threads.

Thanks.

$\oint_{C} \frac{x}{x^2-y^2}\, dx + \frac{y}{y^2-x^2}\, dy$

First, how to enter this stuff. Check the following link for how to use LaTeX on the forum: http://www.mathhelpforum.com/math-he...read.php?t=266 . Scroll down a ways and there's a quick tutorial. Once you gain a little expertise, I found a site on the internet with more script names, but I'm told that not all of them may work here: http://meta.wikimedia.org/wiki/Help:Formula . To see the code I input in the integral above, simply left-click on it and a box will pop up.

Okay, since we may choose any (valid) path between the points (1,0) and (5,3), we may as well pick a simple one and choose the line: $y=(3/4)x-(3/4)$ which doesn't cross either y=x or y=-x on the interval between the two chosen points.

What this means is that we are going to convert the line integral to a simple single integral by plugging $y=(3/4)x-(3/4)$ into the integrand. So $\frac{x}{x^2-y^2}=\frac{x}{x^2-[(3/4)x-(3/4)]^2}$
$=\frac{16x}{7x^2+18x-9}$ after some simplifying. You can do the y and dy for the second part of the integral.

Thus $\oint_{C} \frac{x}{x^2-y^2}\, dx + \frac{y}{y^2-x^2}\, dy = \int_1^5 \left ( \frac{16x}{7x^2+18x-9} +... \right ) \, dx$ , where $...$ is the y part I didn't do above.

-Dan

**topsquark** · February 23rd 2006, 11:49 AM

Originally Posted by jedoob

Hi, I am fairly new to line and curve integrals - I was wondering how to the following question:

Evaluate: The curve integral C of (x/((x^2)-(y^2)))dx + (y/((y^2)-(x^2)))dy
where C is any curve from (1,0) to (5,3) in the xy plane that does not meet the lines y=x and y=-x.

Also, could someone please let me know how to write proper mathematical symbols in this forum like I've seen in some of the threads.

Thanks.

By the way, did I mention I seem to like doing things the hard way? There is a MUCH simpler path to take than the one I did in my last post. We don't care how we get from (1,0) to (5,3) so choose the following path: (1,0) to (5,0) then from (5,0) to (5,3). (Recall we can't cross y=x so the path (1,0) to (1,3) to (5,3) is not allowed.)

So, in the first part of the path (call it C1) y is not changing, so dy=0. Thus:
$\oint_{C_1} \frac{x}{x^2-y^2}\, dx + \frac{y}{y^2-x^2}\, dy = \int_1^5 \frac{x}{x^2-(0)^2} \, dx$ since y=0 on this path.

A similar type of result comes from the second part of the path (C2):
$\oint_{C_2} \frac{x}{x^2-y^2}\, dx + \frac{y}{y^2-x^2}\, dy = \int_0^3 \frac{y}{y^2-(5)^2} \, dy$

And then, of course: $\oint_{C} = \oint_{C_1} + \oint_{C_2}$ using an obvious shorthand.

This path should prove to be much easier than the one in my first post, though they should of course turn out to be equal.

-Dan

**jedoob** · February 23rd 2006, 04:44 PM

Great. Thanks.

**jedoob** · February 23rd 2006, 05:09 PM

Thanks topsquark for your answer - quick question - how do you complete the second curve integral C2 and evaluate it between 0 and 3?

**topsquark** · February 24th 2006, 04:35 AM

Originally Posted by jedoob

Thanks topsquark for your answer - quick question - how do you complete the second curve integral C2 and evaluate it between 0 and 3?

$\int_0^3 \frac{y}{y^2-25}\, dx$

I would do this by decomposing the fraction in the integrand into two linear terms. (I think it's called "partial fraction decomposition" or something like that.)
$\frac{y}{y^2-25}=\frac{A}{y+5}+\frac{B}{y-5}=\frac{A(y-5)+B(y+5)}{y^2-25}$
So we know that A+B=1 and -5A+5B=0. Solving for A and B will give you a simpler integrand to integrate.

-Dan

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