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    Use Green's Theorem to evaluate \displaystyle\int^._c  F .\ dr.
    F(x,y) = < y-ln(x^2 + y^2) , 2arctan(y/x), C is the circle (x-2)^2 + (y-3)^2 = 1 oriented counterclockwise.

    I know how to solve this...I'm just a bit confused about setting up the bounds for the radius...so...disregarding the integral itself (I'll write "--" instead of it), I'm only concerned about the bounds...

    So...

    \displaystyle\int^{2\pi}_{0} \int^3_2 -- r dr d\theta


    Are my bounds for the radius correct?...or is it supposed to be...

    \displaystyle\int^{2\pi}_{0} \int^1_0 -- r dr d\theta ?
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    I don't think you can simply ignore "the integral itself".

    If you write X = x-2, Y = y-3, then you simply have a circle of radius 1 as in your 2nd integral. But of course you then need to rewrite your integrand F(x,y) in terms of F(X, Y).

    (I don't see any way your first integral could make sense - although you could probably find a way if you really wanted to).
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    (Original post by DFranklin)
    I don't think you can simply ignore "the integral itself".

    If you write X = x-2, Y = y-3, then you simply have a circle of radius 1 as in your 2nd integral. But of course you then need to rewrite your integrand F(x,y) in terms of F(X, Y).

    (I don't see any way your first integral could make sense - although you could probably find a way if you really wanted to).
    This is what I did:

    I graphed the circle (x-2)^2 + (y-3)^2 = 1...so the center is at (2,3) and the radius is 1. I think the first integral that I wrote is wrong, because the radius is always 1, even when the origin of the circle is not at (0,0)...the area doesn't change with location...

    \displaystyle\int^{2\pi}_{0} \int^1_0 \frac{-2y}{y^2 + x^2} - (1- \frac{2y}{y^2 + x^2}) r dr d\theta (because I used Green's theorem)

    Because \frac{-2y}{y^2 + x^2} and \frac{2y}{y^2 + x^2} cancel out, I did not bother converting x and y into cos(theta) and sin(theta)...

    \displaystyle\int^{2\pi}_{0} \int^1_0 - r dr d\theta...and then I just solved this to get the answer...is my answer correct?
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    If the circle is at (2, 3), when \theta = 0, r = 1, what are x and y?

    Now look at your integral. When \theta = 0, r = 1, what are x and y?

    Do you see the problem?

    Edit: If everything really cancels except 1, then of course it doesn't actually matter what x and y are. Is that actually the case? (At first glance I wouldn't expect it all to cancel, but I could well be wrong - I can't diff things like arctan(y/x) in my head).
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    (Original post by DFranklin)
    If the circle is at (2, 3), when \theta = 0, r = 1, what are x and y?

    Now look at your integral. When \theta = 0, r = 1, what are x and y?

    Do you see the problem?
    But why do I need to consider looking at what x and y are...I'm using polar coordinates...yes I think I do see the problem, because the point at \theta = 0, r = 1 does not lie on the circle.
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    Having done some calcs, I think everything does cancel, so you don't actually need to get the values of x and y right.

    However, I do hope you realise that if you were, say, wanting to integrate x over the circle, you would NOT be able to do it by finding

    \int_0^{2\pi} \int_0^1 (r \sin \theta) r\, dr\,d\theta, because x \neq r \sin \theta when the circle is not at the origin.
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    (Original post by DFranklin)
    Having done some calcs, I think everything does cancel, so you don't actually need to get the values of x and y right.

    However, I do hope you realise that if you were, say, wanting to integrate x over the circle, you would NOT be able to do it by finding

    \int_0^{2\pi} \int_0^1 (r \sin \theta) r\, dr\,d\theta, because x \neq r \sin \theta when the circle is not at the origin.
    Thank you for answering...can you tell me how I could do it if it did involve x? (If you don't mind)
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    Well, for your circle you have x = 2 + r sin theta, y = 3 + r cos theta (or x = 2 + r cos theta etc. depending on how you're going from (r, theta) to (x, y)).
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    (Original post by DFranklin)
    Well, for your circle you have x = 2 + r sin theta, y = 3 + r cos theta (or x = 2 + r cos theta etc. depending on how you're going from (r, theta) to (x, y)).
    So if we look at the original question that I solved...then I would do the exact same thing...I would just write y= 3 + sin theta and x = 2 + cos theta...right? Thank you for answering, your answer was really helpful.
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    Yes, you're just fortunate that all the x's and y's cancelled so you didn't actually have to worry about it.
 
 
 
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