# Line integral in polar coordinates

1. Sep 22, 2009

### manenbu

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

calculate:
$$\oint \frac{2-y}{x^2+(y-2)^2} dx + \frac{x}{x^2+(y-2)^2} dy$$

where $y = \sin{t} + 2, x = \cos{t}, 0 \leq t \leq \pi$

2. Relevant equations

Green's Theorem.

3. The attempt at a solution

In what order should I do everything?
I need to find the derivaties of P and Q (what comes before the dx and dy), and doing it like this is a bit complicated. So I thought I'll change it to polar, but then I need to take dP/dy but now P is a function of t and not x,y so I can't do that.
In that case, how do I convert dy to polar differentials? I know that dA or dxdy equals rdrdt, but what about only dy?

Or another idea I thought of was to make the denominator 1 (because it is - it's cos2t + sin2t, and then take the derivatives but they are different than what I get if I don't do that.

I don't know which method would be correct, so if someone can help me that'd be great.

2. Sep 22, 2009

### arildno

Why use green's theorem at all?

Green's theorem states an IDENTITY between a calculated line integral and a calculated surface integral.

It is no need to switch to a surface/area integral here, since the indicated line integral is very simple to calculate
(As a general rule, don't ever bother with Green's theorem, the divergence theorem etc, unless the switch warranted by these theorems simplifies your calculations)

3. Sep 22, 2009

### manenbu

Ok so if I do the line integral,

$$2-y = -\sin{t}$$

$$x = \cos{t}$$

$$x^2+(y-2)^2 = \cos^2{t} + \sin^2{t} = 1$$

$$\frac{dx}{dt} = -\sin{t} , \frac{dy}{dt} = \cos{t}$$

The integral becomes:

$$\int_{0}^{2\pi} \frac{-\sin{t}(-\sin{t})}{1} dt + \int_{0}^{2\pi} \frac{\cos{t}\cos{t}}{1} dt = \int_{0}^{2\pi} \cos^2{t} + \sin^2{t} dt = \int_{0}^{2\pi} dt = 2\pi$$

but let's assume I do want to use green's theorem.
So:

$$P = \frac{2-y}{x^2+(y-2)^2}, Q = \frac{x}{x^2+(y-2)^2}$$

$$\frac{\partial P}{\partial y} = \frac{(y-2)^2-x^2}{(x^2+(y-2)^2)^2}$$
$$\frac{\partial Q}{\partial x} = \frac{(y-2)^2-x^2}{(x^2+(y-2)^2)^2}$$

Snd since Green's theorem says that it's the double integral of the difference of those 2 derivaties, I get a double integral of 0, not $2\pi$.

So what went wrong here?

4. Sep 22, 2009

### arildno

First off, t only goes to pie, not two-pie!

Secondly, you cannot use Green's theorem, since the integrand contains a singularity at (0,2)

5. Sep 22, 2009

### manenbu

oh, then it was a mistake in the first post. it should go to $2\pi$. sorry.
and I also understand why green's theorem fails. thanks. :)

6. Sep 22, 2009

### arildno

To nit-pick:

Green's theorem did not fail, it was your application of it that was inappropriate

7. Sep 22, 2009

### manenbu

So Green should've come with a better theorem ;)