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# Setting up an integral... watch

1. Use Green's Theorem to evaluate .
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...

-- r dr d

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

-- r dr d ?
2. 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).
3. (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...

- (1- ) r dr d (because I used Green's theorem)

Because and cancel out, I did not bother converting x and y into cos(theta) and sin(theta)...

- r dr d...and then I just solved this to get the answer...is my answer correct?
4. If the circle is at (2, 3), when , what are x and y?

Now look at your integral. When , 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).
5. (Original post by DFranklin)
If the circle is at (2, 3), when , what are x and y?

Now look at your integral. When , 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 does not lie on the circle.
6. 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 over the circle, you would NOT be able to do it by finding

, because when the circle is not at the origin.
7. (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 over the circle, you would NOT be able to do it by finding

, because 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)
8. 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)).
9. (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.
10. 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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