Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    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 ?
    Online

    18
    ReputationRep:
    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).
    • Thread Starter
    Offline

    0
    ReputationRep:
    (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?
    Online

    18
    ReputationRep:
    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).
    • Thread Starter
    Offline

    0
    ReputationRep:
    (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.
    Online

    18
    ReputationRep:
    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.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (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)
    Online

    18
    ReputationRep:
    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)).
    • Thread Starter
    Offline

    0
    ReputationRep:
    (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.
    Online

    18
    ReputationRep:
    Yes, you're just fortunate that all the x's and y's cancelled so you didn't actually have to worry about it.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: April 8, 2011

University open days

  1. University of Bradford
    University-wide Postgraduate
    Wed, 25 Jul '18
  2. University of Buckingham
    Psychology Taster Tutorial Undergraduate
    Wed, 25 Jul '18
  3. Bournemouth University
    Clearing Campus Visit Undergraduate
    Wed, 1 Aug '18
Poll
How are you feeling in the run-up to Results Day 2018?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.