Line integral in polar coordinates

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  • #1
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Homework Statement



calculate:
[tex]\oint \frac{2-y}{x^2+(y-2)^2} dx + \frac{x}{x^2+(y-2)^2} dy [/tex]

where [itex] y = \sin{t} + 2, x = \cos{t}, 0 \leq t \leq \pi[/itex]

Homework Equations



Green's Theorem.

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.
 

Answers and Replies

  • #2
arildno
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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
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Ok so if I do the line integral,

[tex]2-y = -\sin{t}[/tex]

[tex]x = \cos{t}[/tex]

[tex]x^2+(y-2)^2 = \cos^2{t} + \sin^2{t} = 1[/tex]

[tex]\frac{dx}{dt} = -\sin{t} , \frac{dy}{dt} = \cos{t}[/tex]

The integral becomes:

[tex]
\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[/tex]

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

[tex]P = \frac{2-y}{x^2+(y-2)^2}, Q = \frac{x}{x^2+(y-2)^2}[/tex]

[tex]\frac{\partial P}{\partial y} = \frac{(y-2)^2-x^2}{(x^2+(y-2)^2)^2}[/tex]
[tex]\frac{\partial Q}{\partial x} = \frac{(y-2)^2-x^2}{(x^2+(y-2)^2)^2}[/tex]

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 [itex]2\pi[/itex].

So what went wrong here?
 
  • #4
arildno
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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
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oh, then it was a mistake in the first post. it should go to [itex]2\pi[/itex]. sorry.
and I also understand why green's theorem fails. thanks. :)
 
  • #6
arildno
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To nit-pick:

Green's theorem did not fail, it was your application of it that was inappropriate
:smile:
 
  • #7
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So Green should've come with a better theorem ;)
 

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