Calculating Outward Flux of Vector Field on Ball Boundary

[kidsmoker]

Homework Statement

Calculate the outward flux of the two dimensional vector field

f:\Re^{2}\rightarrow\Re^{2} , f(x,y)=(x/2 + y\sqrt{x^{2}+y^{2}},y/2 + x\sqrt{x^{2}+y^{2}})

through the boundary of the ball

\Omega = {(x,y)\in\Re^{2} \left| x^{2}+y^{2} \leq R^{2}} \subset\Re^{2}, R>0 .

Homework Equations

Divergence theorem:

\int F.n ds = \int\int divF dA

where the first integral is round the boundary and the second one is over the area (sorry I can't get latex to display the limits of integration).

The Attempt at a Solution

I calculated divF to be 1 which gives the answer to be just pi*R^2 .

I then thought i'd try it the old fashioned way, dotting the unit normal with the field around the edge of the circle. The unit normal is n=(x,y)/R right? So dotting this with the field gives me (x^2+y^2)/2R . Then I changed from x,y to using the angle t around the circle, so x=Rcost, y=Rsint .

This gives f.n= R^2/(2R) = R/2 so the flux is the integral from zero to 2pi of

\int (R/2) dt = (R/2)(2\pi) = \pi R .

I must have gone wrong somewhere cos my answers are different, but I can't spot where :-(

Thanks for any help!

 

[Dick]

Your path integral shouldn't be dt, it should be ds where ds is arc length along the path. The length of the curve is 2piR.