Green's Theorem and annulus at 0,0

[bugatti79]

Homework Statement

Use Green's Theorem to evaluate this line integral

Homework Equations

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

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

 

[HallsofIvy]

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

 

[bugatti79]

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

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

 

[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?

 

[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?

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?

 

[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?

 

[bugatti79]

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?

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...

 

[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...

Any suggestions on this one?

 

[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.

 

[bugatti79]

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.

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

 

[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

Any comments on this one folks...?