I want to prove that a continuously differentiable vector field $V$ is the gradient of some function $f$ on an open rectangular solid $(a,b)\times(c,d)\times(e,f)\subset\mathbb{R}^3$ if it has curl $0$.
I have proved this fact in the $\mathbb{R}^2$ case where an explicit function can be constructed, $i.e.$ if we have two continuously differentiable functions $p(x,y)$ and $q(x,y)$ satisfy that $\partial p/\partial y=\partial q/\partial x$, then we can find a function $f(x,y)=\int_{x_0}^x p(x_0,t)dt+\int_{y_0}^yq(x,s)ds$ which satisfies $\partial f/\partial x=p$ and $\partial f/\partial y=q$.
I tried to use a similar method by putting $f(x,y,z)=\int_{x_0}^xP(t_1,y,z)dt_1+\int_{y_0}^yQ(x_0,t_2,z_0)dt_2+\int_{z_0}^zR(x_0,y,t_3)dt_3$ where $V=(P,Q,R)$, but this does not seem to work. More specifically, $\partial f/\partial x=P$ and $\partial f/\partial y=Q$, but $\partial f/\partial z\neq R$.
Here is the question: Could we construct an explicit expression for the function like we did in the 2D case, or is there another way to prove this? Thank you!
