Integral curves and Vector field

Question

For a given vector field $X(x,y,z) = (\frac {-y}{x^2+y^2+z^2}, \frac{x}{x^2+y^2+z^2},0)$ on $\mathbb{R} ^3$\ {(0,0,0)}, what is the integral curve $\int_C (X|dx)$ over a closed circle with radius of 1 and being parallel to xy-plane "at height $z_0$." For which $z_0$ is this integral maximum?

Answer 1

HINT:You can solve it using Gauss's theorem (that's what you should usually do) but here I think it's a more simple case. $z_0$ is constant and the third component of the vector field is $0$ so we also got rid of $dz$. so The equation is now in two dimensions $$\int XdX=\int_{c_1}\big(\frac{-y}{x^2+y^2+z_0^2}\big)dx+\big(\frac{x}{x^2+y^2+z_0^2}\big)dy$$ where $C_1:=\{(x,y):x^+y^2=1\}$. Pay attention we couldn't have done it unless $dz=0,z=\text{const.}$. Now One can apply green's theorem, change to polar coordinates and get the desired result. The result will be a function of $z_0$ so define $$F(z_0)=\int_{c_{1,z_0}} X dX$$ where $C_{1,z_0}$ is the unit circle in "height" $z_0$. Derive by $z_0$ and get a maximum.