Question : By calculating $\nabla \times V$, show that
$$V~(x, y , z )=\begin{pmatrix} ~yz^2+3\\ xz^2+2z+1\\ 2xyz+2y\end{pmatrix}$$
can be expressed as V $=\nabla \phi$ for some function $\phi$, and find $\phi$.
What I have calculated/ my Answer: $\nabla \times V = Curl(V)$
$$Curl(V) = \begin{bmatrix} \hat{i}& \hat{j} & \hat{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ yz^2+3& xz^2+2z+1& 2xyz+2y\end{bmatrix}$$
$\space$
$Curl(V) = ~\hat{i} { \begin{bmatrix} \frac{\partial}{\partial y }& \frac{\partial}{\partial z} \\ xz^2+2z+1& 2xyz+2xy\\ \end{bmatrix} } $$-\hat{j} { \begin{bmatrix} \frac{\partial}{\partial x }& \frac{\partial}{\partial z} \\ yz^2+3& 2xyz+2xy\\ \end{bmatrix} } $
$+\hat{k} { \begin{bmatrix} \frac{\partial}{\partial x }& \frac{\partial}{\partial y} \\ yz^2+3&x^2+2z+1\\ \end{bmatrix} }$
$\space $
$ Curl(V) = ~\hat{i} { \begin{bmatrix} \frac{\partial}{\partial y }(2xyz+2y) - \frac{\partial}{\partial z}(xz^2+2z+1) \end{bmatrix} } $ $-\hat{j} { \begin{bmatrix} \frac{\partial}{\partial x }(2xyz+2y) - \frac{\partial}{\partial z}(yz^3+3) \end{bmatrix} } $ $\space$
$-\hat{k} { \begin{bmatrix} \frac{\partial}{\partial x }(xz^2+2z+1) - \frac{\partial}{\partial y}(yz^3+3) \end{bmatrix} } $
$\space$
$$Curl(V) = \langle ~0 ~\hat{i}- 0~\hat{j} ~ +0~\hat{k}\rangle = 0$$
Note : we know that curl (gradient vector) $= 0$
$$~Curl(\nabla \phi) = 0 $$
Hence, proved :
$$ V =\nabla \phi$$
Finding :
$$V =\nabla \phi$$
$$(yz^2+3)~\hat{i} ~+ (xz^2+2z+1)~\hat{j} ~+ (2xyz+2y)~\hat{k} = ~ \frac{\partial \phi } {\partial x}~\hat{i} + \frac{\partial \phi } {\partial y}~\hat{j} + \frac{\partial \phi } {\partial z}~\hat{k} $$
Written as :
$$ \frac{\partial \phi } {\partial x} = yz^2+3 $$
$$ \frac{\partial \phi } {\partial y} = xz^2+2z+1$$
$$ \frac{\partial \phi } {\partial z} = 2xyz+2y$$
Un-doing the partial differentiation :
with respect to $~ \partial{x}, ~ \phi = xyz^2 + 3x + C_1(y,z)$
with respect to $~ \partial{y}, ~\phi = xyz^2+ 2yz + y + C_2(x,z)$
with respect to $~ \partial{z}, ~\phi = xyz^2 + 2yz + C_3(x,y) $
Final : $$\phi ~(x,y,z) = xyz^2 + 2yz + 3x + y + C_1(y,z)+ C_2(x,z) + C_3(x,y) $$
Could you guys please verify my steps and my final answer