I have a geometry of a cylinder that is curved along both the lengths where there is generally a height of a cylinder. I am aware of the equation of a cylinder. I wanted to know what could the equation of a curved cylinder be as a function of the x,y, and z coordinates.? Also, could this curved cylinder be called an ellipsoid?
I have attached a geometry herewith along with the dimensions.
On
Consider the blue curve first
In my notation the small radius is $r=0.01$ and the large radius is $R=0.09$
It is an arc, of radius $R$ and center at $x=R - r$
I use the parameter angle $\theta$ for the sweep angle of the arc from the center.
The coordinates of the arc are
$$ \pmatrix{x \\ y \\ z} = \pmatrix{ (R-r) - R \cos \theta \\ R \sin \theta \\ 0} $$
Now apply a rotation about the y-axis to this arc to get the coordinates of the shape. The parameter angle of the rotation is $\varphi$
$$ \pmatrix{x \\ y \\ z} = \begin{bmatrix} \cos \varphi & 0 & \sin \varphi \\ 0 & 1 & 0 \\ -\sin \varphi & 0 & \cos \varphi \end{bmatrix} \pmatrix{ (R-r) - R \cos \theta \\ R \sin \theta \\ 0}$$
or
$${\rm surf}( \varphi, \theta) = \pmatrix{x=-\cos \varphi ( R \cos \theta - R + r) \\ y=R \sin \theta \\ z=\sin \varphi (R \cos \theta -R + r)} \tag{1} $$
The surface is fully described by the parameters $\theta$ and $\varphi$. If you limit the length of the shape to $\ell$ then you find the limits of $\theta$ from
$$ \tfrac{\ell}{2} = R \sin \theta \tag{2} $$
So $\varphi = 0 \ldots 2\pi$ and $\theta = -\arcsin \left( \tfrac{\ell}{2 R} \right) \ldots \arcsin \left( \tfrac{\ell}{2 R} \right)$.
Here are the results of the parametric surface as viewed from the z-axis, for $R=10$ and $r=1$
From the y coordinate use $\theta = \arcsin \left( \tfrac{y}{R} \right)$
From the z coordinaten use $\varphi = - \arctan \left( \tfrac{z}{x} \right)$
the end relationship is
$$ \boxed{ \sqrt{R^2-y^2}-\sqrt{x^2+z^2}-R+r = 0 } \tag{3} $$
That would probably be called a "barrel shape". The equation depends on the type of curve you choose for the barrel. If it's a circular curve, if the $y$ axis is along the axis of symmetry of the barrel and we count the centre of the barrel as the origin. then the equation for the barrel would be something like $$x^2+z^2=r^2$$ for the $x$ and $z$ coordinates, and $$(r+h)^2+y^2=R^2$$ Here $R$ is the radius of curvature of the barrel side and $h<R$ is a constant which expresses an offset from the origin of the centre of the radius of curvature.
So we end up like $$(\sqrt{x^2+z^2}+h)^2+y^2=R^2.$$
In the case of your diagram, it looks like you have $h=80 \text{mm}$ and $R=90 \text{mm}$.
I couldn't persuade Wolfram Alpha to do a 3D plot of this, but here is a 2D version.
In practice you would do this parametrically.