Applying Green's THM, Polar Coords substitution

[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.

 

[tiny-tim]

welcome to pf!

hi dbkats! welcome to pf!

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

same way as usual …

x² + y² = r²

x = rcosθ, y = rsinθ, dxdy = rsinθdrdθ

show us your integrand in x and y ​

 

[dbkats]

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 ​

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

 

[tiny-tim]

first, let's simplify …

{(2/(1 + x/y)²)-1-(2y/(x²+y²))}

 

[dbkats]

first, let's simplify …

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?

 

[Dick]

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?

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

 

[dbkats]

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

>.< 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

 

[Dick]

>.< 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

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

 

[dbkats]

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

Or that, you hahaha...