# Homework Help: 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...?