Setting up an integral...

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).

Reply 2

OP

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?

(edited 13 years ago)

Reply 3

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).

(edited 13 years ago)

Reply 4

OP

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.

(edited 13 years ago)

Reply 5

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.

Reply 6

OP

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)

Reply 7

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)).

Reply 8