Ellipsoid but not quite

Question

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:

1. How can I calculate the volume of the new body?

2. 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)?

Answer 1

Let $\mathcal{E} = \{ (x,y,z) \mid \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \le 1 \}$ be the ellipsoid at hand.

The new body $\mathcal{E}_d$ is the Minkowski sum of $\mathcal{E}$ and $\bar{B}(d)$, the closed ball of radius $d$. ie.,

$$\mathcal{E}_d = \{ p + q : p \in \mathcal{E}, q \in \bar{B}(d) \}$$

Since $\mathcal{E}$ is a convex body, the volume of $\mathcal{E}_d$ has a very simple dependence on $d$. It has the form:

$$\verb/Vol/(\mathcal{E}_d) = V + A d + 2\pi \ell d^2 + \frac{4\pi}{3}d^3\tag{*1}$$

where $V$, $A$ and $\ell$ is the volume, surface area and something known as mean width for $\mathcal{E}$.

The problem is for an ellipsoid, the expression for $A$ and $\ell$ are very complicated integrals.
If I didn't make any mistake, they are: $$\begin{align} A &= abc\int_0^{2\pi} \int_0^{\pi} \sqrt{(a^{-2}\cos^2\phi + b^{-2}\sin^2\phi)\sin^2\theta + c^{-2}\cos^2\theta} \sin\theta d\theta d\phi\\ \ell &= \frac{1}{2\pi} \int_0^{2\pi}\int_0^{\pi}\sqrt{(a^2\cos^2\phi + b^2\sin^2\phi)\sin^2\theta + c^2\cos^2\theta} \sin\theta d\theta d\phi \end{align}\tag{*2}$$

Good luck for actually computing the integral.

Update

When $a = b$, the integral simplify to something elementary.

For the special case $a = b \ge 1, c = 1$, by a change of variable $t = \cos\theta$, we have:

$$\begin{align} A &= 4\pi a\int_0^1\sqrt{(1 + (a^2 - 1)t^2}dt\\ &= \frac{2\pi a}{a^2-1}\left(\sqrt{a^2-1}\sinh^{-1}(\sqrt{a^2-1}) + a(a^2-1)\right) \\ \ell &= 2\int_0^1 \sqrt{a^2 + (1-a^2)t^2}dt = \frac{a^2}{\sqrt{a^2-1}}\sin^{-1}\left(\frac{\sqrt{a^2-1}}{a}\right) + 1 \end{align} $$ For a test case, when $a = b = 2, c = d = 1$, we find

$$\begin{align} \verb/Vol/(\mathcal{E}_1) - V &= A + 2\pi \ell + \frac{4\pi}{3} = \frac{\pi}{3\sqrt{3}}\left( 12 \sinh^{-1}(\sqrt{3}) +8 \pi +34\sqrt{3}\right)\\ &\approx 60.35475634605034 \end{align} $$ Matching the number on Euler project 449 that motivates this question.

IMHO, I don't think Euler project expect one to know

1. the volume formula $(*1)$.
2. or how to compute the integrals in $(*2)$.

There should be a more elementary way to derive the same result for the special case $a = b$.
That part will probably stamp on the foot of Euler project. I better stop here.

Answer 2

If $d$ << any of ( a,b,c ) you need to just find the surface area (using the elliptic integrals in Wikipedia) and multiply by $d$. The loss of accuracy using a paint layer thickness between the ellipsoid and Huygen's wavelet sort of body is of no avail or worth in practical engineering work.

It is just like: For a thin tube of radii $ b-a =t$ we take cross section area to be $2 \pi a t$ or $2 \pi b t$ or $ \pi (a+b) t$ but do not care for the second order differences or inaccuracies that are lost when not considering a more correct:

$$ \pi ( b^2-a^2). $$

EDIT 1:

If I am tasked to compute spray paint volume etc., I would settle for an approximation like:

$$ A \approx \frac{12 \pi a b c }{(a+b+c)} $$

EDIT 2:

If d is not negligibly small, the outside anyhow being not an ellipsoid and we may take average of semi-axes and the approximation may be:

$$ Vol. \approx \frac{12 \pi \bar a \bar b \bar c d }{(\bar a+\bar b+ \bar c)} $$

For a = b =2, c=d=1 it gives V = 54.3737.

Since it is a harmonic mean it would be a lower bound.

Answer 3

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

Answer 4

Hint:

Take a unit sphere of radius $1$ and cover it with a uniform layer of thickness $t$. The total volume is $\dfrac43\pi(1+t)^3$.

If you dilate the sphere non-uniformly with factors $a<b<c$, you get the ellipsoid with a layer of a thickness comprised between $at$ and $ct$. With $t=\frac da$, the layer is everywhere thicker than $d$; with $t=\frac dc$, the layer is everywhere thinner than $d$.

From these considerations you can deduce the bracketing

$$\dfrac43\pi abc(1+\frac dc)^3<V<\dfrac43\pi abc(1+\frac da)^3.$$

Answer 5

Edit: This proof is incorrect, and the proposed answer is wrong. As pointed out by @TonyK in the comments, the computation of the gradient is wrong.

I will leave the answer here, nonetheless, since seeing an incorrect proof is often as instructive as seeing a correct one.

Hint: As it turns out, the "thickened" ellipsoid is indeed another ellipsoid. You can show with some effort that if the original ellipsoid is $$\left(\frac xa\right)^2 + \left(\frac yb\right)^2+ \left(\frac zc\right)^2=1$$ then the externally ellipsoid externally thicked by $\epsilon$ has an outer surface $$\left(\frac x{a+\epsilon}\right)^2 + \left(\frac y{b+\epsilon}\right)^2+ \left(\frac z{c+\epsilon}\right)^2=1$$ So, the volume computation for the new surface is essentially the same as the computation for the original surface, assuming you can do that. The two ellipsoids are not similar unless $a=b=c$.

Addendum: This is the computation I used. Originally, I just did it for an ellipse. But the computation is just as easy for an ellipsoid. So suppose we are given an ellipsoid $E$ $$\left(\frac xa\right)^2 + \left(\frac yb\right)^2+ \left(\frac zc\right)^2=1\tag{1}$$ and let us consider the ellipsoid $E'$ $$\left(\frac x{a+\epsilon}\right)^2 + \left(\frac y{b+\epsilon}\right)^2+ \left(\frac z{c+\epsilon}\right)^2=1\tag{2}$$ Consider the point $P(x_0,y_0,z_0)$ on $E$ (WLOG, we work in the first octant where all coordinates are positive) and form the point $$Q((1+\tfrac{\epsilon}a)x_0,(1+\tfrac{\epsilon}b)y_0,(1+\tfrac{\epsilon}c)z_0)$$

Claim 1: $P$ and $Q$ are $\epsilon$ units apart.

Proof: $$|Q-P|^2 = |((1+\tfrac{\epsilon}a)x_0,(1+\tfrac{\epsilon}b)y_0,(1+\tfrac{\epsilon}c)z_0)-(x_0,y_0,z_0)|^2$$ $$=|((\tfrac{\epsilon}a)x_0,(\tfrac{\epsilon}b)y_0,(\tfrac{\epsilon}c)z_0)|^2$$ $$=\epsilon^2|(\tfrac{x_0}a,\tfrac{y_0}b,\tfrac{z_0}c)|^2$$ $$=\epsilon^2((\tfrac{x_0}a)^2 + (\tfrac{y_0}b)^2 + (\tfrac{z_0}c))^2 = \epsilon^2$$ since $P$ lies on $E$. $\blacksquare$

Claim 2: Q lies on $E'$.

Proof: Write $Q$ as $$Q=((1+\tfrac{\epsilon}a)x_0,(1+\tfrac{\epsilon}b)y_0,(1+\tfrac{\epsilon}c)z_0)$$ $$= \left((a+\epsilon)\tfrac{x_0}a,(b+\epsilon)\tfrac{y_0}b,(c+\epsilon)\tfrac{z_0}c\right)$$

Then $Q$ satisfies equation $(2)$ because $P$ satisfies equation $(1)$ since by hypothesis $P$ lies on $E$. $\blacksquare$

Claim 3: $Q$ lies on the line normal to $E$ at $P$.

Proof: Regarding $(1)$ as $F(x,y,z)=1$, we may form the gradient vector $\nabla F(x_0,y_0,z_0) = (2x_0/a,2y_0/b,2z_0/c)$ which is normal to $E$ at $(x_0,y_0,z_0)$. Thus an equation of the line normal to $E$ at $P$ is

$$(x,y,z) = (x_0,y_0,z_0) + t(\tfrac{x_0}a,\tfrac{y_0}b,\tfrac{z_0}c)$$ which may be written $$(x,y,z) = \left( (1+\tfrac ta)x_0,(1+\tfrac tb)y_0,(1+\tfrac tc)z_0\right)$$

Taking $t=\epsilon$, we recover $Q$, so $Q$ lies on this line. $\blacksquare$

Claims 1, 2, and 3 together show that $E'$ is obtained by thickening $E$ by $\epsilon$ units. Each point on $E$ is moved outward by $\epsilon$ units along the normal at that point to form $E'$. This concludes our proof.