## Calculus (3rd Edition)

$$\phi (x,y) = \frac{1}{2}x^2+ \frac{1}{2}y^2 +C$$
Given $$\mathbf{F}=\langle x, y\rangle$$ We need to find $\phi (x,y)$ such that $$\frac{\partial \phi }{\partial x}=x,\ \ \ \frac{\partial \phi }{\partial y}=y$$ By integration, we get \begin{align*} \phi (x,y)&= \frac{1}{2}x^2+C_1\\ \phi (x,y)&= \frac{1}{2}y^2+C_1\\ \end{align*} Then we can choose $$\phi (x,y) = \frac{1}{2}x^2+ \frac{1}{2}y^2 +C$$