TITLE: Ellipsoid but not quite
QUESTION [13 upvotes]: I have an ellipsoid centered at the origin. Assume $a,b,c$ are expressed in millimeters. Say I want to cover it with a uniform  coat/layer that is $d$ millimeters thick (uniformly).
I just realized that in the general case, the new body/solid is not an ellipsoid. I wonder:

How can I calculate the volume of the new body?

What is the equation of its surface?


I guess it's something that can be calculated via integrals but how exactly, I don't know.
Also, I am thinking that this operation can be applied to any other well-known solid (adding a uniform coat/layer around it). Is there a general approach for finding the volume  of the new body (the one that is formed after adding the layer)?

REPLY [3 votes]: Let $(x,y,z)=(a\sin u \cos v, b\sin u \sin v,c\cos u)$ on the ellipse $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$, then the unit normal vector is
$$\mathbf{n}=
\frac{\displaystyle \left(\frac{x}{a^2},\frac{y}{b^2},\frac{z}{c^2} \right)}
     {\displaystyle \sqrt{\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}}}$$
Then new surface will have coordinates of
$$(x',y',z')=(x,y,z)+d\mathbf{n}$$
which no longer to be a quadric anymore.
In particular, if $d <-\frac{1}{\kappa} <0$ where $\kappa$ is one of the principal curvatures, then the inner surface will have self-intersection.
If we try reducing the dimension from three (ellipsoid) to two (ellipse) and setting $a=1.5,b=1$, the unit normal vectors (inward) won't pointing on the straight line (i.e. the degenerate ellipse $\displaystyle \frac{x^{2}}{0.5^{2}}+\frac{y^{2}}{0^{2}}=1$).

And also the discrepancy of another case