Evaluating the Work Integral Between Two Points

[Charge2]

Homework Statement

Evaluate the Work Integral,
$I = \int_\Gamma [ (\frac{y} {x^2 + y^2} + 1) dx - \frac{x} {x^2+y^2}dy]$
between the points (5, 30/pi) and (2,8/pi), using the potential function. Pesent your answer in exact form.

Homework Equations

$W = \int F . dr$
$\int_\Gamma (Pdx+Qdy)$

The Attempt at a Solution

Usually, we are given the curve along the line. This question we have not.

(5 - 2) and (30/pi - 8/pi) is the region between the points is it not?

Or do I rearange the given formula I, for y?

Do we even use the line integral formula, or use Green's Theorem instead?Thanks in advance.

 

[HallsofIvy]

You can't use "Green's theorem" because that requires an integral around a closed path. However, this problem said "using the potential function" which certainly implies that there is a potential function! That is, that there exist a function, F(x,y), such that \frac{\partial F}{\partial x}= \frac{y}{x^2+ y^2}+ 1 and \frac{\partial F}{\partial y}= -\frac{x}{x^2+ y^2}. Find that function and evaluate at the endpoints. (That is, the "work" is the change in potential between the two points.)

 

[Charge2]

You can't use "Green's theorem" because that requires an integral around a closed path. However, this problem said "using the potential function" which certainly implies that there is a potential function! That is, that there exist a function, F(x,y), such that \frac{\partial F}{\partial x}= \frac{y}{x^2+ y^2}+ 1 and \frac{\partial F}{\partial y}= -\frac{x}{x^2+ y^2}. Find that function and evaluate at the endpoints. (That is, the "work" is the change in potential between the two points.)

Ok so I've obtained the partial derivatives,
$\frac{\partial F}{\partial x}= \frac{-2yx}{(x^2+ y^2)^2}$
$\frac{\partial F}{\partial y}= \frac{2yx}{(x^2+ y^2)^2}$

Now, do I integrate from the points,

$I =\int^5_2 [\frac{-2yx}{(x^2+ y^2)^2}+ \frac{2yx}{(x^2+ y^2)^2}] = [\frac{-2y5}{(5^2+ y^2)^2}+ \frac{2y5}{(5^2+ y^2)^2}] - [\frac{-2y2}{(2^2+ y^2)^2}+ \frac{2y2}{(2^2+ y^2)^2}] = 0$

Would that not just equal zero, even with the y values substituted in place? Making this a conservative system? I've most definitely done this the wrong way, haven't I?.

 

[LCKurtz]

Ok so I've obtained the partial derivatives,
$\frac{\partial F}{\partial x}= \frac{-2yx}{(x^2+ y^2)^2}$
$\frac{\partial F}{\partial y}= \frac{2yx}{(x^2+ y^2)^2}$

I've most definitely done this the wrong way, haven't I?.

Yes, that is wrong. You are given $F_x = \frac y {x^2+y^2}+1$ and $F_x =\frac{-x}{x^2+y^2}$. You have differentiated them and called them the same things. To find a potential function you must find an $F(x,y)$ whose partial derivatives agree with what is given. This means you need to find anti-partial derivatives. Integrate, not differentiate. And your answer will be of the form $F(x,y) = ...$ Surely your text must have examples for finding potential functions.

 

[Charge2]

I've practiced very little with line integrals. I've done some further research and I think I have come up with a solution. I am very new to latex so I written a brief summary.

After doing the partial derivatives, they equated to each other. I think this shows that we can find a potential function. A check. Even though it is implied in the question

I then found the potential function by integrating and differentiating with respect to y. Obtaining,

$f(x,y)=\tan^-1 \frac{x}{y} + x$

Inputting the given points, obtaining the answer,

$W = f(2,8/pi) - f(5,30/pi)$
$= tan^-1\frac{2}{8/pi} + 2 - tan^-1\frac{5}{30/pi} - 5$
$= tan^-1\frac{pi}{4} - tan^-1\frac{pi}{6} - 3$

 

[Ray Vickson]

I've practiced very little with line integrals. I've done some further research and I think I have come up with a solution. I am very new to latex so I written a brief summary.

After doing the partial derivatives, they equated to each other. I think this shows that we can find a potential function. A check. Even though it is implied in the question

I then found the potential function by integrating and differentiating with respect to y. Obtaining,

$f(x,y)=\tan^-1 \frac{x}{y} + x$

Inputting the given points, obtaining the answer,

$W = f(2,8/pi) - f(5,30/pi)$
$= tan^-1\frac{2}{8/pi} + 2 - tan^-1\frac{5}{30/pi} - 5$
$= tan^-1\frac{pi}{4} - tan^-1\frac{pi}{6} - 3$

This is all correct.

 

[Charge2]

Yes! Thankyou everyone for the kind and helpful guidance. :-)