# Applying Green's THM, Polar Coords substitution

1. Apr 9, 2012

### dbkats

Use Green's THM to calculate the line integral ∫C(F<dot> dx), where C is the circle (x-2)2 + (y - 3)2=1 oriented counterclockwise, and F(x,y)=(y+ln(x2+y2), 2tan-1(x/y)).

Green's THM
∂SF<dot>dx=∫∫S(∂F2/∂x) - ∂F1/∂y)

I tried doing it by brute force. I took the partials and put them under the integral. I also computed the bounds of integration and split it into the multiple integral. However, the bounds were pretty messy:
x from 1 to 3, y from 3 to √(1-(x-2)2)+3

When I evaluated the first integral with respect to y, I got an intractible function to take an integral over.

I feel like I should be using polar coordinates, but I am not sure how to substitute them in this case.

2. Apr 9, 2012

### tiny-tim

welcome to pf!

hi dbkats! welcome to pf!
same way as usual …

x2 + y2 = r2

x = rcosθ, y = rsinθ, dxdy = rsinθdrdθ

show us your integrand in x and y

3. Apr 9, 2012

### dbkats

Re: welcome to pf!

I feel that the conventional polar coordinates will not do well because the disk is not centered at the origin. So when we transform the disk region, we do not get a nice square. Or did I misunderstand how that works?

My integral with respect to x and y:

2∫13dx∫3√(1-(x-2)2)+3{(2/(1 + x/y)2)-1-(2y/(x2+y2))}dy

Last edited: Apr 9, 2012
4. Apr 9, 2012

### tiny-tim

Re: welcome to pf!

first, let's simplify …

5. Apr 9, 2012

### dbkats

Re: welcome to pf!

So, I actually messed up writing it out here. It should actually be:
Integrand = (2/(1 + (x/y)2))-1-(2y/(x2+y2))

I feel very stupid for not seeing that it drastically simplifies when I work out the common denominator.

Integrand = (y2 - 2y - x2)/(x2 + y2)

Barring any other stupid arithmetic mistakes. Okay, so I have this very pretty integrand, but the bounds of integration are still really weird. Should I try to transition to polar?

EDIT: As an aside, is there a way to enter my fractions here so they are human-readable?

6. Apr 9, 2012

### Dick

Re: welcome to pf!

It will simplify even more if you get it right. You forgot to use the chain rule on arctan(x/y). It will simplify so much your coordinate problems will be over. And you could check out using LaTex https://www.physicsforums.com/showthread.php?t=8997

Last edited: Apr 9, 2012
7. Apr 9, 2012

### dbkats

Re: welcome to pf!

>.< I feel silly. The integrand evaluates to -1, so I can now use whatever parametrization I feel like. I think that (x-2)=cosθ, (y-3)=sinθ would be a good bet.

I guess once I saw the disk, all I could think about was "how am I going to parametrize this ugly thing"...

Anyway, thanks so much for the help

8. Apr 9, 2012

### Dick

Re: welcome to pf!

You're welcome. You don't need coordinates at all. To integrate a constant you just need to know the area of the disk.

9. Apr 9, 2012

### dbkats

Re: welcome to pf!

Or that, ya hahaha....