assistance required with complex integrals please I'm a bit stuck on a question on complex integrals I have no idea how to go about solving it and would be really grateful if someone could help.
The question is: 2. Let
$$\  F(x,y,z)=(2xy+4xz)i+(x^2+6yz)j+(2x^2+3y^2)k,x,y,z∈R. $$
(a) Find a function $\ ϕ:R^3→R $ 
such that  $\ \vec F⃗ = \nabla ϕ. $
(d) Evaluate the integral  $$\ \int_Γ(\vec F .\vec {dr} ) $$
where Γ is the path
$\ y=x^2,z=0 $
from (0,0,0) to (2,4,0) followed by the line segment from 
(2,4,0) to (1,1,2)
 A: To find the potential function $\phi$, we need only integrate the components of $\vec F$.  To proceed, we begin with the $\hat x$-component to find 
$$\begin{align}
\phi(x,y,z)&=\int F_x(x,y,z)\,dx\\\\
&=\int (2xy+4xz)\,dx\\\\
&=x^2y+2x^2z+C_1(y,z) \tag 1
\end{align}$$
where $C_1(y,z)$ is an integration constant with respect to $x$, but can depend on both $y$ and $z$.
Next, we use $\phi(x,y)$ as given in $(1)$ to find $C_1(y,z)$.  Noting that $\frac{\partial \phi(x,y,z)}{\partial y}=F_y(x,y,z)$ yields
$$\begin{align}
x^2+\frac{\partial C_1(,y,z)}{\partial y}=x^2+6yz
\end{align}$$
from which we find $C_1(y,z)=3y^2z+C_2(z)$, where $C_2(z)$ is a second integration constant.  At this point, we can write the potential function as 
$$\phi(x,y,z)=x^2y+2x^2z+3y^2z+C_2(z) \tag 2$$
Finally, taking the partial of $\phi(x,y,z)$, as given by $(2)$, with respect to $z$, and setting it equal to $F_z(x,y,z)$ yields
$$2x^2+3y^2+C_2'(z)=2x^2+3y^2$$
from which we find that $C_2(z)$ is a constant.  
Therefore, we can write the potential function as 
$$\bbox[5px,border:2px solid #C0A000]{\phi(x,y,z)=x^2y+2x^2z+3y^2z+C}$$
for any constant $C$.

Since we have determined that $\vec F(x,y,z)\nabla \phi(x,y,z)$, $\vec F$ is conservative, and its path integral depends only on the end points of the path.  We can write
$$\begin{align}
\int_{\vec r_1}^{\vec r_2}\vec F(x,y,z)\cdot \vec d\vec \ell&=\int_{\vec r_1}^{\vec r_2} \nabla \phi(x,y,z)\cdot d\vec \ell\\\\
&=\int_{\vec r_1}^{\vec r_2} d\phi\\\\
&=\phi(1,1,2)-\phi(0,0,0)\\\\
&=11\end{align}$$       
