# Math Help - flux integral help

1. ## flux integral help

Hey,

Let $f(x,y) = ln(x^2 +y^2)$

Let C be the circle $x^2 + y^2 =a^2$ with $a^2 \ne 0$

and evaluate the flux integral:

$\oint_C\nabla f \cdot n ds$

I can't seem to use Greens Theorem on this particular question and don't know why.

Second part:
Now let K be an arbitrary simple closed curve in the plane that does not pass through
(0,0). Use Green's theorem to show that:

$\oint_K\nabla f \cdot n ds$
has, only, two possible values depending on whether (0,0) lies inside or outside of K.
I really need help on this one.

Can someone help me with this?

Thanks

2. Originally Posted by Nguyen
Hey,

Let $f(x,y) = ln(x^2 +y^2)$

Let C be the circle $x^2 + y^2 =a^2$ with $a^2 \ne 0$

and evaluate the flux integral:

$\oint_C\nabla f \cdot n ds$

I can't seem to use Greens Theorem on this particular question and don't know why.

Second part:
Now let K be an arbitrary simple closed curve in the plane that does not pass through
(0,0). Use Green's theorem to show that:

$\oint_K\nabla f \cdot n ds$
has, only, two possible values depending on whether (0,0) lies inside or outside of K.
I really need help on this one.

Can someone help me with this?

Thanks
Note that

$\nabla f(x,y)=\frac{2x}{x^2+y^2}\mathbf{i}+\frac{2y}{x^2+ y^2}\mathbf{j}$

$\mathbf{n}=\frac{1}{\sqrt{x^2+y^2}}\left( x\mathbf{i}+ y\mathbf{j}\right)$

$\nabla f \cdot \mathbf{n}=\frac{2}{\sqrt{x^2+y^2}}$

Now we just need to compute

$\int \frac{2}{\sqrt{x^2+y^2}}ds$

$x=a\cos(t) \quad y=a\sin(t)$

$\int_{0}^{2\pi}\frac{2}{a}(adt)=2\int_{0}^{2\pi}dt =4\pi$

for part 2 if the curve does not contain the origin Greens theorem (Stokes theorem)applies and

$\nabla \times \nabla f=0$

so the integral is zero if it does consider the new domain with a circle of radius a cut out around the origin the greens theorem applies on the domain with the hole and

$\int_{K}\nabla f \cdot \mathbf{n}ds = \int_{D}\nabla f \cdot \mathbf{n}ds + \lim_{a \to 0} \int_{C}\nabla f \cdot \mathbf{n}ds$

By Greens theorem again we get

$0 = \int_{D}\nabla f \cdot \mathbf{n}ds - \lim_{a \to 0} \int_{C}\nabla f \cdot \mathbf{n}ds$

$\int_{D}\nabla f \cdot \mathbf{n}ds = \lim_{a \to 0} \int_{C}\nabla f \cdot \mathbf{n}ds =4\pi$

The last step comes from part 1 as the integral was independent of a.