Determine z=z(x,y) given one of it's partial derivatives

Question

Looking forward for some help on this one.

Determine $ z=z(x,y) $ if:

$\frac{\partial{z}}{\partial{y}} = \frac{x}{x^2+y^2}$

Answer 1

The RHS of the given equation is undefined at $(0,0)$; but even if you restrict to $\Omega:={\mathbb R}^2\setminus\{(0,0\}$ you cannot find a single function $z:\>\Omega\to{\mathbb R}$ such that $${\partial z\over\partial y}={x\over x^2+y^2}\qquad\bigl((x,y)\in\Omega\bigr)\ .\tag{1}$$ Locally the problem can be solved: Given a point $(x_0,y_0)\in\Omega$ you can find a neighborhood $U$ of $(x_0,y_0)$, e.g., a halfplane, and choose a continuous representant $$(x,y)\mapsto \phi(x,y)\qquad\bigl((x,y)\in U\bigr)$$ of the polar angle there. Furthermore choose an arbitrary $C^1$-function $x\mapsto c(x)$. Then $$z(x,y):=\phi(x,y)+c(x)$$ satisfies $(1)$ in $U$. This is the most general solution of $(1)$ in $U$. But it is impossible to concatenate such local solutions to a single global solution valid in all of $\Omega$. If you remove the negative $x$-axis from $\Omega$ then you can choose the principal value $\Phi(x,y)$ of the polar angle in the resulting domain $\Omega'$ and in this way obtain a global solution in $\Omega'$.