Write the equation for the surface generated by by revolving the curve around the indicated axis. $x=2z^2$, revolved around the $x$ axis.
How exactly do I go about this? I can plot the curve out and see that this is going to be an elliptic paraboloid oriented along the x-axis; however, I still am not sure how to actually write its equation out. I know it will take the form
$\frac{y^2}{a^2}+\frac{z^2}{b^2} = \frac{x}{c}$, and that $y=z$. But that's as far as I've managed to get. The actual answer is $x=2(y^2+z^2)$, but I'm not sure how that was arrived at. 
 A: Let 
$$x = 2 \cdot {z^2}$$
This is a curve in $xz-$plane.
To rotate this curve around $x-$axis, we use an rotation matrix:
$$D(\varphi ) = \left( {\begin{array}{*{20}{c}}
  1&0&0 \\ 
  0&{\cos (\varphi )}&{ - \sin (\varphi )} \\ 
  0&{\sin (\varphi )}&{\cos (\varphi )} 
\end{array}} \right)$$
and a parametrization for given curve. That is, we set:
$$c(t) = \left( {\begin{array}{*{20}{c}}
  {2 \cdot {t^2}} \\ 
  0 \\ 
  t 
\end{array}} \right)$$
Rotation is now generated by applying rotation matrix on the curve.
That means:
$$\left( {\begin{array}{*{20}{c}}
  x \\ 
  y \\ 
  z 
\end{array}} \right) = D(\varphi ) \cdot c(t) = \left( {\begin{array}{*{20}{c}}
  {2 \cdot {t^2}} \\ 
  { - t \cdot \sin (\varphi )} \\ 
  {t \cdot \cos (\varphi )} 
\end{array}} \right)$$
Here $0 \leqslant \varphi  < 2\pi $ and $t \in \mathbb{R}$.
This is what we get:

Because:
$$\begin{gathered}
  x = 2 \cdot {t^2} \hfill \\
   \hfill \\
  y =  - t \cdot \sin (\varphi ) \hfill \\
   \hfill \\
  z = t \cdot \cos (\varphi ) \hfill \\ 
\end{gathered}$$
we have:
$$\begin{gathered}
  {y^2} + {z^2} = {t^2} = \frac{1}{2} \cdot (2 \cdot {t^2}) = \frac{1}{2} \cdot x \hfill \\
   \hfill \\
  {y^2} + {z^2} = \frac{1}{2} \cdot x \hfill \\ 
\end{gathered} $$
or, if we want to:
$$x = 2 \cdot ({y^2} + {z^2})$$
and that's all.
