$\displaystyle \vec{X}(s,t)=(s^2cost,s^2sint,s), -3\leq s\leq 3, 0\leq t\leq 2\pi$.

So I've already found the normal vector at (s,t)=(-1,0) to be (-1,0,-2) and found the tangent plane at (1,0,-1) to be $\displaystyle x+2z=-1$ (feel free to correct me if either of those end up being wrong and important to my question).

The question reads "Find an equation for the image of $\displaystyle \vec{X}$ in the form $\displaystyle F(x,y,z)=0$.

What I initially did was to say that $\displaystyle x=s^2cost$, $\displaystyle y=s^2sint$, and $\displaystyle z(x,y)=(x^2+y^2)^{\frac{1}{4}}$, which then just gives the equation $\displaystyle F(x,y,z)=(x^2+y^2)^{\frac{1}{4}}-z=0$, but this seems suspiciously simplistic. Is there some other way to define F that doesn't require that z be defined explicitly in terms of x and y, or is what I've written really all that there is to the problem?