Determine whether these extensions of $\mathbb{C}[x]$ are integral In these notes by Gathmann the question is posed

Which of the following extensions of $\mathbb{C}[x]$ are integral?
  
  
*
  
*$\mathbb{C}[x,y,z]/(z^2 - xy)$
  
*$\mathbb{C}[x,y,z]/(z^2 - xy, y^3 - x^2)$
  
*$\mathbb{C}[x,y,z]/(z^2 - xy, x^3 - yz)$
  

So #2 is a composition of integral extensions (pass through $\mathbb{C}[x,y]/(y^3-x^2)$). Extension #1 I believe is not integral because it does not satisfy incomparability: lying over $(x) \triangleleft \mathbb{C}[x]$ is both $(x, z^2-xy)$ and $(x, z^2 - xy, y)$. (Edit: Whoops! Forgot to use prime ideals.)
How am I supposed to see the third one? Heuristics and "how to think about it" answers are much appreciated.
 A: This solution is really pedestrian, using only the definition, there are probably a lot more elegant methods.
If $ S =\mathbb{C}[x,y,z]/(z^2 - xy, x^3 - yz)$ was integral over $R =\mathbb C[x]$, then for $I=(z) \triangleleft S $, $R/(R\cap I) \subset S/I$ would be integral.
Claim: $R \cap I = (x^3)$
Proof: Clearly, $(x^3) \subset R \cap I$. As $R$ is a PID, $R \cap I$ is generated by some divisor of $x^3$, so it suffices to show that $x^2 \notin R\cap I$. If this was false, we would have in $\mathbb{C}[x,y,z]$ $$x^2-zf(x,y,z)=(z^2-xy)g(x,y,z)+(x^3-yz)h(x,y,z)$$ for some $f,g,h \in \mathbb{C}[x,y,z]$.  Pluggin in $z=0$ and 
$y=0$ gives $x^2=x^3h(x,0,0) \Rightarrow 1=xh(x,0,0)$ in  $\mathbb C [x]$ which is absurd.
Now suppose that $\mathbb{C}[x]/(x^3) \subset \mathbb{C}[x,y,z]/(z^2 - xy, x^3 - yz,z) \cong \mathbb{C}[x,y]/(xy,x^3)$ was integral. Then there is some monic non-constant polynomial $f(T,x) \in \mathbb{C}[x]/(x^3)[T]$ such that $f(y,x) = 0$ in $\mathbb C[x,y]/(xy,x^3)$. Since $xy=0$, a lot of terms in this equation cancel out and we may assume that $f$ is of the form $f(T) = g(T)+h(x)$, where $g(T) \in \mathbb C[T]$ is monic and non-constant. Now lifting this to $\mathbb{C}[x,y]$ we get that that $f(y,x)=g(y)+h(x)$ is in $(xy,x^3)$. But every element of $(xy,x^3)$ is divisible by $x$. Now if $x \mid f(y,x)$, then $f(y,0) \equiv 0$, but clearly $f(y,0)=g(y)+h(0)$ is non-constant, as $g$ is non-constant. This finishes the proof.
A: If $S=\mathbb{C}[x,y,z]/(z^2-xy,x^3-yz)$ were integral over $\mathbb{C}[x]$, then $S/(x)$ would be integral over $\mathbb{C}[x]/(x)=\mathbb{C}$.  But $S/(x)=\mathbb{C}[y,z]/(z^2,yz)$ contains $\mathbb{C}[y]$ as a subring, and so is not integral over $\mathbb{C}$.
Here's a more geometric way to think about essentially the same argument.  If $S$ were integral over $\mathbb{C}[x]$, then the induced morphism $\operatorname{Spec} S\to\mathbb{A}^1$ would be finite, and in particular would have finite fibers.  In very concrete terms, this means that if you plug in any particular number for $x$, then the system of equations $z^2-xy=0$ and $x^3-yz=0$ should have only finitely many solutions.  But if you plug in $x=0$, these equations just say $z^2=0$ and $yz=0$, and so $y$ can be anything as long as $z=0$.
