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Let $f(x,y)=\frac{1}{x^2+y^2}$, calculate the flux of the vector field ${\rm Grad}\ (f)$ across the boundary of the region $S$ bounded between the circles centered at the origin and radius $1$ and $2$ and in the first quadrant in two ways: directly as a line integral, (the flux of $F$ across the curve $C$ is $\int_C F\cdot {\bf n} ds$) and once using green's theorem.

So first ${\rm Grad}\ (f)=\bigg(\frac{-2x}{(x^2+y^2)^2}, \frac{-2y}{(x^2+y^2)^2}\bigg)$, but then how to calculate the vector ${\bf n}$ which is a normal vectore to the surface $S$. The textbook tells me to parametrized the surface by $G(u,v)$ then find the vector ${\bf n}$ by cross product of $\frac{dG}{du}$ and $\frac{dG}{dv}$.

I would like to know how to parametrize the surface in this case

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Since $S$ is flat space and not curved, then the upwards unit normal is $(0,0,1)$. But if you really need the parametrization then use polar coordinates $r:[1,2]$ and $\theta:[0,\frac{\pi}{2}]$. With $x=rcos(\theta), y=rsin(\theta), z=0$

EDIT: You will need a $z$ component in your Grad($f$), therefore you can express $z=f(x,y)$ as the level set $$g(x,y,z)=0=z-f(x,y)$$ Therefore you get $$Grad(f)=\left(\frac{2x}{{x^2+y^2}^2},\frac{2y}{{x^2+y^2}^2},1\right)$$ Now you can dot your vector with the normal.

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  • $\begingroup$ Thanks a lot! Then the vector n is equal to (0,0,1). so $\int_C F\cdot {\bf n} ds$=$\int_0^{pi/2}$$\int_1^2$1 dr d$\theta$ and the answer is pi/2? $\endgroup$ – user108297 Mar 23 '14 at 15:15
  • $\begingroup$ But when I use the green's theorem, the answer is pi... $\endgroup$ – user108297 Mar 23 '14 at 15:16

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