Homework Help: Work Integral

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1. Oct 12, 2015

Charge2

1. The problem statement, all variables and given/known data
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.

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

3. 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?

Last edited: Oct 12, 2015
2. Oct 12, 2015

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.)

3. Oct 12, 2015

Charge2

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?.

Last edited: Oct 12, 2015
4. Oct 12, 2015

LCKurtz

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.

5. Oct 13, 2015

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. Im 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,

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

Last edited: Oct 13, 2015
6. Oct 13, 2015

Ray Vickson

This is all correct.

7. Oct 15, 2015

Charge2

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