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Flux integral

  • Thread starter kidsmoker
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  • #1
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Homework Statement



Calculate the outward flux of the two dimensional vector field

[tex]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}})[/tex]

through the boundary of the ball

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


Homework Equations



Divergence theorem:

[tex]\int F.n ds = \int\int divF dA [/tex]

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

[tex]\int (R/2) dt = (R/2)(2\pi) = \pi R .[/tex]

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

Thanks for any help!
 
Last edited:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
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.
 

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