Is there a generalized method of rotation for curves? I know that we can rotate a curve in $R^2$ about a linear axis, as is common for first year calculus problems involving solids of revolution.  But has anyone come up with a general method to take a real valued function in $R^2$ and rotate about another function in $R^2$ that is not necessarily linear?  I assume the generalized definition would take a point off the curve to the point on the other side of the curve the same distance off the curve along a line perpendicular to the curve at some point.  The alternative definition I thought of was given in an answer below, but that is not what I'm looking for.  I want to be able to rotate a curve about another curve geometrically with or without an established coordinate plane, which is why I assumed the definition above.
Trying to make this more precise:
My definition takes a curve $C$ and finds the slope of the normal line at point $(x_o,y_o)$.  Supposing the slope found is $m$, the normal line is $y=m(x-x_o)+y_o$.  Find the point(s) $(x_o,y_o)$ for which this line intersects the point/curve to be rotated.  Find the distance along this line between $C$ and the point to be rotated.  Then, traversing the line in the opposite direction from $C$, find the point that same distance away from $C$.  This yields the rotated point.
Using the definition I give and use Curve $C$ as an axis:
 1.  Is the relation between a point in $R^2$ and its image after rotation a function?
 2.  Does this depend on whether $C$ represents a real valued function?  For example, rotation about a circle is not a function while rotation about a parabola is?
 3.  Does this yield a well-defined surface of revolution?
 4.  Could such a rotation yield interesting results, e.g., transforming a smiley face into a sad face, or turning a one kind of conic into another?
 5.  Supposing this definition cannot yield a well-defined surface of revolution, as some have suggested, what definition could?
 6.  Are there helpful links or articles that address any of these issues?
A counterexample to the conjecture of(1):  Take $C$ to be the unit circle centered at the origin and rotate $(0,2)$ about $C$.
 A: Depending on the purpose you have in mind, rotation of the function $f$ about the function $g$ may be equivalent to the rotation of $f-g$ about the $x$-axis. 
In particular, this works for finding the volume of the solid defined by the rotation of $f$ about $g$.
A: As the previous answer said, your question is quite general. Assuming you want to stay in the same plane (you said rotate the function in $\mathbb{R}^2$), I will borrow a concept from differential geometry. In diff geom, when we construct a tube (a surface) around a parametrised curve $\gamma$ (which is your $g$, the function around which we rotate), we rotate a constant distance $r$ around $\gamma$. This rotation is carried over from o to $2\pi$. Here is the formula:
$$
T(u,v) = \gamma(u) + r(\mathbf{n}(u)\cos(v)+\mathbf{b}(u)\sin(v))
$$
where $\mathbf{n}$ and $\mathbf{b}$ are unit vectors pointing the in the normal and binormal ($\mathbf{t} \times \mathbf{n}$, note that $\mathbf{t} = \dot{\gamma}$).
Anyway, not to delve too far into differential geometry. In your case, two things need to be modified. The rotation only needs to be for $\pi$. There is no binormal as we are only in $\mathbb{R}^2$, and you can see that $\sin(\pi)$ knocks this part out. Another thing is that instead of rotating a constant distance, this is now a function $f$. So the formula for your question will be something like this:
$$
R(u) = g(u) - (f(u)-g(u))  \mathbf{n}
$$
But this will not work since $f$ and $g$ could be travelling at different speeds ($\|f'\|$ and $\|g'\|$), so $f(u)$ would not necessarily be the point on $f$ encountered when extending the normal of the point $g(u)$. Say we are at a point $g_0=g(u_0)$. We need $f(u)$ in this case to be on the line passing through $g(u_0)$ and parallel to $\mathbf{n}$. Another way to say this is 
$$ \tag{1}
(f(u)-g_0) \cdot \mathbf{n} = \Vert f(u)-g_0\Vert
$$
assuming that $\mathbf{n}$ is a unit vector. This means you will have 
$$\tag{2}
R(u)=g(u)- \{ (f(u)-g(u))\cdot\mathbf{n} \} \mathbf{n} 
$$
with the added requirement that at each particular point $u_0$, we need equation (1) to hold.
This approach would still have problems, I think because you would need to do this for every point in that order--finding $\mathbf{n}(u_0)$ is not hard, and should be done first. Then finding $c$ such that $f(c)-g(u_0)$ is parallel to $\mathbf{n}(u_0)$; there might be more than one such $c$. Then taking 
$$
\left(f(c)-g(u_0)\right) \cdot \mathbf{n}
$$
gives you the length of the vector to reflect. This vector has direction $\mathbf{n}$. Whence we obtain equation (2). However, I can't think of a way to express this order of events explicitly.
I don't think this depends on the function being real-valued. It can be complex valued, but you'd need to be careful with some added structure from $\mathbb{C}$.
