# Ellipsoid but not quite

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)?

• What is the new body? Mar 8, 2016 at 12:44
• @user41736: If the thickness is measured along normals to the surface, then the outer surface of the cover may indeed not be an ellipsoid. I'm not sure. It isn't obvious, at any rate; I would have to be convinced by an argument if it is an ellipsoid. The problem is the requirement that the cover be uniformly thick. Just dilating a covered sphere won't do it, because the dilation is larger along the ellipsoid's longer axis than it is along one of its shorter axes.
– MPW
Mar 8, 2016 at 12:59
• This is called an offset surface. You can obtain its parametric equations from the parametric equations of the ellipsoid (sphere in spherical coordinates, stretched), by computing the normal vector and adding it. Unfortunately the analytical expressions aren't simple (obtaining an implicit equation is scary), and chances are low that the volume integral is tractable. Anyway, the extra volume must be close (if not just equal) to the ellipsoid area times the thickness.
– user65203
Mar 8, 2016 at 13:15
• @ZubinMukerjee: Clearly not. Consider the degenerate ellipse formed by $x^2+\frac{y^2}{b^2} = 1$ as $b \to 0$. This tends to a line segment between $(-1, 0)$ and $(1, 0)$, and a uniform blanket around that evidently is not an ellipse. Mar 8, 2016 at 17:21
• @peter.petrov: I suppose what's intuitive varies from person to person. :-) Mar 8, 2016 at 18:59

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.

• Here $\bar{B}(d)$ means the closed ball of radius d centered at the origin? Can you check the integrals? I may try solving them later since I have several concrete values for a,b,c. Actually $(a,b,c)=(3,3,1)$. Maybe that makes the problem somewhat easier? Mar 8, 2016 at 14:06
• @peter.petrov yes. it is the closed ball centered at origin. Mar 8, 2016 at 14:08
• OK... Very interesting (that the new body is equal to that Minkowski sum). Mar 8, 2016 at 14:09
• @achillehui Should I read this to try to understand what's going on? Or is it just the standard sum? I'm confused. Also I think you could copy-paste your last equation as an answer to this question. Mar 8, 2016 at 14:10
• @ZubinMukerjee ellipse and ball are regular enough geometric objects, ordinary Minkowski sum should be enough. For that question, one probably should derive the equation. It comes down to given any unit vector $\vec{u} = (u_1,\ldots,u_n)$ and ellipsoid $\mathcal{E} = \{ (x_1, \ldots, x_n ) : \sum_{i=1}^n (x_i/a_i)^2 \le 1 \}$, the maximum value of $\vec{x}\cdot\vec{u}$ for $\vec{x}\in \mathcal{E}$ equals to $\sqrt{\sum_{i=1}^n (a_i u_i)^2 }$. Mar 8, 2016 at 14:28

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.

• Well, $d$ is not << any of $(a,b,c)$, it can be e.g. $a/2$ or $b/2$ or $2c/3$. Mar 8, 2016 at 14:02

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

• Seems this is more like an answer to the doubts if the new body is an ellipsoid or not, rather than an answer to the original question. Right? Anyway, thanks for taking the time to think and write this note. Mar 8, 2016 at 15:40

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

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

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.

• I thought it was thickened by a constant, not by a factor of $\epsilon$. If that's right then I don't think the second equation you have is correct. Mar 8, 2016 at 14:12
• It was thickened by a constant d indeed. I also don't think the new body is just a new ellipsoid with a'=a+d, b'=b+d, c'=c+d. I tried that already, and got wrong answer. Mar 8, 2016 at 14:12
• @ZubinMukerjee projecteuler.net/problem=449 It's not that, if it was that then the number 60.35475635 would be the difference between the two ellipsoid volumes. Very easy to compute, but does not match that number 60.35475635 (I get 58.643062867 when taking that difference). So I guess something is wrong here. I didn't want to spoil the Project Euler problem though by linking to it from here. Based on the answers so far, seems I will need to read up a bit (which is good) before being able to solve it. Mar 8, 2016 at 14:22
• @ZubinMukerjee: I will post my computations as an edit to this answer. Unless I've done something stupid, I don't think it's wrong. Give me a few minutes to write it up. Please look at it and see if you see a hole in my argument.
– MPW
Mar 8, 2016 at 14:41
• The gradient $\nabla F(x_0,y_0,z_0)$ is $(2x_0/a^2,2y_0/b^2,2z_0/c^2)$, not $(2x_0/a,2y_0/b,2z_0/c)$. (So the segment is not perpendicular to the original surface either, @AndrewD.Hwang.) Mar 8, 2016 at 15:57