Can't find $h_1(y,z)$, $h_2(x,z)$, and $h_3(x,y)$.

Question

Given vector field $F(x,y,z)=(3x^2y, x^3+y^3,0)$. Find $f(x,y,z)$ such that $\nabla f=F$.

I have tried as below.

\begin{alignat*}{2} &&\nabla f &= F\\ \iff&&\left(\dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y}, \dfrac{\partial f}{\partial z}\right)&=(3x^2y, x^3+y^3,0) \end{alignat*} We have PDEs $\dfrac{\partial f}{\partial x}=3x^2y$, $\dfrac{\partial f}{\partial y}=x^3+y^3$, and$\dfrac{\partial f}{\partial z}=0$. \begin{align*} \dfrac{\partial f}{\partial x}=3x^2y\iff \partial f=3x^2y\partial x \iff \int\partial f=\int 3x^2y\partial x \iff f=x^3y+h_1(y,z). \end{align*} \begin{align*} \dfrac{\partial f}{\partial y}=x^3+y^3&\iff \partial f=\left(x^3+y^3 \right)\partial y \iff \int\partial f=\left(x^3+y^3 \right)\partial y\\ &\iff f=x^3y+\dfrac{1}{4}y^4+h_2(x,z). \end{align*} \begin{align*} \dfrac{\partial f}{\partial z}=0\iff \partial f=0\partial z \iff \int\partial f=\int 0\partial z \iff f=h_3(x,y). \end{align*} Thus, \begin{align*} x^3y+h_1(y,z)=x^3y+\dfrac{1}{4}y^4+h_2(x,z)\iff h_1(y,z)=\dfrac{1}{4}y^4+h_2(x,z), \end{align*} \begin{align*} x^3y+h_1(y,z)=h_3(x,y), \end{align*} \begin{align*} x^3y+\dfrac{1}{4}y^4+h_2(x,z)=h_3(x,y). \end{align*}

Now I can't find $h_1(y,z)$, $h_2(x,z)$, and $h_3(x,y)$. How to find it?

Answer 1

From $$\frac{\partial f}{\partial x} = 3x^2y$$ we get that $f(x,y,z) = x^3y+g(y,z)$, and then $$x^3+y^3 = \frac{\partial f}{\partial y} = x^3 + \frac{\partial g}{\partial y} \quad \leadsto \quad \frac{\partial g}{\partial y} = y^3 \quad \leadsto \quad g(y,z) = \frac14 y^4+h(z).$$ Thus $$f(x,y,z) = x^3y+\frac14 y^4+h(z)$$ and finally $$0 = \frac{\partial f}{\partial z} = h'(z) \quad \leadsto \quad h(z) = c$$ where $c$ is a constant.