Rotate a function around an axis (not a volume of revolution question) I am trying to understand this problem a little better. 
Rotate the curve $y=x^4-2x^2$ around the $y$-axis, find an equation for the resulting surface.
If I write the curve as a vector-valued function (parameterically), as $\mathbf{x} = [x(t),y(t),z(t)]^T$, where
\begin{align*}
x(t) &= t, \\
y(t) &= t^4-2t^2, \\
z(t) &= 0,
\end{align*}
and write the rotation matrix out,
\begin{equation}
R_{\theta} = \begin{bmatrix} \cos\theta & 0 & -\sin\theta \\ 0 & 1& 0 \\ \sin\theta &0 & \cos\theta \end{bmatrix}
\end{equation}
Then the equations of the surface paramterised by $t$ and $\theta$ are $$ R_{\theta} \mathbf{x} = \begin{bmatrix} \hat{x} \\\hat{y}\\ \hat{z}\end{bmatrix}=\begin{bmatrix} t\cos\theta \\t^4-2t^2\\ t\sin\theta  \end{bmatrix}. $$
Trying to solve for the parameters, I get $t ^{\color{#C00}{2}}=1\pm\sqrt{1-y} \, $ $\color{#C00}{\mbox{(this was the mistake)}}$, and $\theta = \arctan\left(\displaystyle\frac{z}{x}\right)$. Solving would give something like
$$
z= x\tan\arccos\left(\frac{x}{1\pm\sqrt{1+y}}\right),
$$
however neither the parameterised version nor this expression look right when plotted. Where did I go wrong? 
 A: The parametrization in the original post is fine. But a better way of eliminating the variables is to observe that
$$
x^2+z^2=t^2\cos^2\theta+t^2\sin^2\theta=t^2.
$$
Consequently we get that
$$
y=t^4-2t^2=(x^2+z^2)^2-2(x^2+z^2).\qquad(*)
$$
IMHO the best way of describing the surface of revolution is to use this equation to write $y$ as a function of the other coordinates.
But we can actually get to equation $(*)$ without taking the route via parametrization. This is because these tricks always allow us to  eliminate
the parameters in the case of a surface of revolution. If we rotate the graph of $y=f(x),x\ge0,$ about the $y$-axis, the surface we get is always
$$
y=f(\sqrt{x^2+z^2}).
$$
This is because in the $xy$-plane the distance $r$ of a point from the $y$-axis is given by $r=|x|$. In the $xyz$-space the distance of a point from the $y$-axis is given by $r=\sqrt{x^2+z^2}$. The surface of revolution consists of those points were $y=f(r)$. For more reading about the use of $r$ in this way look up cylindrical coordinates. More typically $r$ is the distance from the $z$-axis but that is immaterial.
The same process, of course, works no matter which axis we rotate about. We only need to make sure that the coordinates have appropriate roles. If we rotate the curve $y=f(x)$ about the $x$-axis, we might first rewrite the curve as $x=f^{-1}(y)$ (assuming that this is feasible), and then the equation of the surface of revolution is simply
$$
x=f^{-1}(\sqrt{y^2+z^2})\qquad(**)
$$
as here. Of course, we often don't want to use the inverse function, so we simply apply $f$ to both sides of $(**)$ and get $f(x)=\sqrt{y^2+z^2}$, or probably the more familiar
$$
y^2+z^2=f(x)^2.\qquad(***)
$$
Returning to the example and rotations about $y$-axis.
This time the curve was given in the form $y=g(x)$. If it had been given in the form $x=f(y)$, we would have used the analogue of $(***)$ and written the equation for the surface in the form $x^2+z^2=f(y)^2$. Finding and using the inverse of
$y=f(x)=x^4-2x^2$ is unpleasant, so I used the analogue of $(**)$ instead.
One more point. Here the function $y=f(x)=x^4-2x^2$ is even, so we don't need to restrict to positive values of $x$ - the graph returns to its original form after being rotated 180 degrees. In general the surfaces of revolution arising from the parts $x\ge0$ and $x\le0$ don't match.
Below there are two images. The first is simply a plot of the function $f(x),x\in[-1.5,1.5]$.

The second is the corresponding 3D-surface. Please observe the orientation of the axes. Here the $y$-axis points up. This was produced using the parametrization from the OP with $t\in[-1.5,1.5],\theta\in[0,2\pi]$.

