Green's Theorem and annulus at 0,0

1. Jan 21, 2012

bugatti79

1. The problem statement, all variables and given/known data

Use Green's Theorem to evaluate this line integral

2. Relevant equations

$\int xe^{-2x}dx+(x^4+2x^2y^2)dy$ for the annulus $1 \le x^2+y^2 \le 4$

3. The attempt at a solution

$\displaystyle \int_c f(x,y) dx + g(x,y)dy+ \int_s f(x,y) dx + g(x,y)dy = \int \int _D1 (G_x-G_y) dA=0 \implies \int_c=- \int_s= \int_{-s}$

Let c be the out circle of radius 2 counterclockwise and s the inner radius of 1 clockwise

x=r cos $\theta$, y=r sin $\theta$ substituting

evaluating the last term on RHS, ie

$\displaystyle \int_{-s}= \int_0^{2 \pi} r cos \theta (e^{-2r cos \theta})(-r sin \theta d \theta) +(r^4 cos^4 \theta +2r^2 cos^2 \theta r^2 sin^2 \theta) r cos \theta d \theta$

Is this right so far? Thanks

2. Jan 21, 2012

HallsofIvy

If the problem is, as you say, to use Green's theorem, why are you not using Green's theorem?

3. Jan 21, 2012

bugatti79

I guess because I was following my notes blindly which I think was just demonstrating a principle.

For the annulus then do I use green's to calculate the inner circle and subtract it from the outer circle?

Thanks

4. Jan 21, 2012

bugatti79

Correction,

I think I will split the annulus into a top and bottom and write

$\displaystyle \int \int_R (g_x-f_y) dA = \int \int_T (g_x-f_y )dA + \int \int_B (g_x-f_y) d A$...right?

5. Jan 21, 2012

bugatti79

I calculate that the top and bottom integrals are both 0....? I thought they should be non 0 since the annulus is centred about 0?

6. Jan 21, 2012

HallsofIvy

No, just integrate on the annulus itself. Using polar coordinates as you suggested before, take $\theta$ from 0 to $2\pi$ and r from 1 to 2.

Now, what is $g_x- f_y$?

7. Jan 21, 2012

bugatti79

It is $g_x-f_y =(4x^3 + 4xy^2)-(0)$

If we are using green's to evaluate the above integral which does not include the origin then we expect to get a non 0 answer right? But I keep getting 0 for the above integral...

8. Jan 24, 2012

bugatti79

Any suggestions on this one?

9. Jan 24, 2012

HallsofIvy

Then you are doing the integral wrong!
The integral is
$$4\int_0^1 4r^4dr\int_0^{2\pi}(cos^3(\theta)+ sin(\theta)cos^2(\theta))d\theta$$
(the fourth power of r is due to the differential of area in polar coordinates being "$rdrd\theta$", but it is the $\theta$ integral that is important.)

The $\theta$ integral can be written
$$\int_0^{2\pi} cos(\theta)(1- sin^2(\theta))d\theta+ \int_0^{2\pi} sin(\theta)cos^2(\theta)d\theta$$

Let $u= sin(x)$ in the first integral and let $v= cos(\theta)$ in the second equation.

10. Jan 24, 2012

bugatti79

I calculate $\displaystyle \int \int (g_x -f_y) dA = \int \int (4x^3+4xy^2) dA = \int_1^2 \int_0^{2 \pi}(4r^4 cos^3 \theta +4r^4 cos \theta sin^2 \theta) d \theta dr$....

I am just wondering how you arrived at yours especially the limits...? Have you got the sin theta cos^2 theta switched? should it be cos theta sin^2 theta

11. Jan 25, 2012

bugatti79

Any comments on this one folks...?