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This morning, I had eggs for breakfast, and I was looking at the pieces of broken shells and thought "What is the surface area of this egg?" The problem is that I have no real idea about how to find the surface area.

I have learned formulas for circles, and I know the equation for an ellipse; however, I don't know how to apply that.

The only idea I can think of is to put an egg on a sheet of paper and trace it, and then measure the outline drawn, and then try to find an equation for that ellipse and rotate that about the $x$-axis. Now, my problem is how I can find the equation of the ellipse from the graph, and will my tracing method really be the edge of the egg? Also, can I use the standard surface area integral? Will I have to use some techniques to solve the integral that are not covered in the AP BC Calculus?

There has to be a better method for finding the surface area. Please, help me understand how to find the surface area of an egg; i.e., how to use my mathematical knowledge for something other than passing exams.

J. M. ain't a mathematician
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yiyi
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    Considering the fact that the answer is almost probably irrational number without any nice expression, I'd suggest to wrap a few eggs with a piece of fabric, then measure the area of the fabric that actually covered each egg, and then average. Of course, this is by no mean "the mathematical method", but it had to be said. :-) – Asaf Karagila May 31 '13 at 01:54
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    The egg's outline is not an ellipse... the ellipse has two axes of symmetry, while the egg's oval only has one. – J. M. ain't a mathematician May 31 '13 at 01:55
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    http://www.poultryscience.org/ps/paperpdfs/05/p0530482.pdf – Amzoti May 31 '13 at 01:57
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    Of course, you can break the egg's shell into nearly flat pieces, taking care not to lose any pieces, and then take the area of those pieces... – J. M. ain't a mathematician May 31 '13 at 01:57
  • @Amzoti where does it show how the surface area was found, ie, how did they make the equation. – yiyi May 31 '13 at 02:05
  • Equations 15 and 16. I thought this paper would put an exclamation point on why this is hard based on some of the comments. – Amzoti May 31 '13 at 02:08
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    @J.M.: Who are you, Archimedes? :-) – Asaf Karagila May 31 '13 at 02:14
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    @Asaf, No, but I can at least hope not to get the results of Banach and Tarski... ;) – J. M. ain't a mathematician May 31 '13 at 02:15
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    if you have a precise enough scale (chem lab, not kitchen), you could weigh the egg shell, then cut a nice regular piece (disc, square...) then measure and weigh it... – DJohnM May 31 '13 at 02:16
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    @User, "cut a nice regular piece" - you make this sound easier than it is... ;) – J. M. ain't a mathematician May 31 '13 at 02:18
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    A tile saw. A tiny, tiny tile saw... – DJohnM May 31 '13 at 02:19
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    Make mold of egg, line mold with thin papier-mache, trim dried paper-mache, slice papier mache so it lies flat and measure. If egg breaks make omelette. – daniel May 31 '13 at 02:20
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    @J.M.: Why not? That would solve the world hunger problem!! – Asaf Karagila May 31 '13 at 02:20
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    @Asaf, no it won't; all you get is many pieces of shells, and I know for a fact that eating the shells is not a healthy thing to do... ;) – J. M. ain't a mathematician May 31 '13 at 02:24
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    @J.M.: You could use it to feed other animals which will later turn into slabs of meat, to be bathed and marinated in garlic-mustard-cherries sauce... :-) – Asaf Karagila May 31 '13 at 04:19
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    Ah, you are looking for an egguation? ;-p – Macke May 31 '13 at 11:52
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    The engineer in me says to measure the length, height and width of the rectangular prism that contains the egg, and use the equation: `A = 2lh + 2lw + 2hw` to calculate the surface area. Slap on error bars of +- 100% and call it a day. – zzzzBov May 31 '13 at 22:26
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    **What do you mean by surface area?** Do you mean the area of a smooth, thin rubber sheet stretched over the egg? Or the area which includes all of the surface imperfections and pores of the egg on a microscopic level? A object that superficially looks like an egg, but with, say, a porous microsurface, could easily have a surface area the size of a football field. – Kaz May 31 '13 at 23:05
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    @Kaz I was thinking of your first idea, but the second would be nice to see. – yiyi Jun 01 '13 at 13:47
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    Now as a followup, you can try to find the surface area of a chicken. – rschwieb Jun 03 '13 at 15:52

10 Answers10

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There is a nice equation describing the equation of an egg curve credit to Nobuo Yamamoto : $$ (x^2+y^2)^2 = ax^3 + \frac{3a}{10}xy^2, \tag{1} $$ where $0\leq x\leq a$, $a$ is the length of the major axis of symmetry for an egg.

In other words, we could get it by cutting a boiled egg in half and measure the distance from tip to the bottom. I just drew it in MATLAB, and the curve looks like the following for $a=1$: egg

I must say this curve fits pretty well with an egg. Now we have a curve, then the method of computing surface area by revolution, which is taught in Calculus II I believe, can be used to computing the surface. We just revolve the curve above around the egg's major axis of symmetry and get a surface, here is what looks like when we revolve it around the $x$-axis by degree $\pi$, we can get the lower half by revolve another $\pi$: eggsurf

First we solve for $y$ in (1) when $y>0$: $$ y = \sqrt{\frac{3ax}{20} - x^2 + x\sqrt{\frac{7ax}{10} + \frac{9a^2}{400}} }, $$ The formula of computing surface area by revolution is: $$ S = 2\pi\int_0^a y\sqrt{1+\left(\frac{dy}{dx}\right)^2} \,dx,\tag{2} $$ The derivative is: \begin{align} \frac{dy}{dx} = \frac{1}{2\sqrt{\frac{3ax}{20} - x^2 + x\sqrt{\frac{7ax}{10} + \frac{9a^2}{400}} }} \left( \frac{3a}{20}- 2x+ \sqrt{\frac{7ax}{10} + \frac{9a^2}{400}} + \frac{7ax}{20\sqrt{\frac{7ax}{10} + \frac{9a^2}{400}}}\right), \end{align} Plugging $dy/dx$ and $y$ into (2), then you could use your favorite tool of numerical integration to perform the computing for you(Octave, MATLAB, Mathematica, etc).


A more tweakable/numerical/experimental approach:

As J. M. suggests in the comments, the shape looks like an egg, but is a real egg being approximated nicely by this curve? I guess the answer is that "it really depends on that specific egg"!

Let's say we still want to use surface of revolution to compute the surface area.

But this time, we handle it more numerically from the very start, instead of looking for a curve to fit one thing for all.

Two assumptions:

  • All eggs are axial symmetric with respect to its major axis, i.e., if $x$-axis is its major axis, its surface can be obtained by revolving a curve $y= f(x)$, for $0\leq x\leq a$.
  • That curve $y = f(x)$ has certain smoothness: $f$ and $f'$ are continuous for $x\in (0,a)$ .

Now we want to compute the integral (2) using Simpson's rule or Trapezoidal rule, which is also taught in Calculus II in most colleges I believe. A remark is that $|f'|\to \infty$ when $x\to 0$ and $x\to a$, it would be much better if we use adaptive qudrature by putting more sample points near $0$ and $a$.

sample

The steps are:

  1. Boil an egg, cut it by half, hold it against a paper, use a pencil to outline its boundary (upper half is enough).

  2. Draw the major axis, set it to be the $x$-axis, and measure its length $a$.

  3. Choose $(n+1)$-sample points (including the end points) so that the points are equidistant to their neighbor on the curve. $n$ is chosen to be even, measure the distance to the major axis ($y$-coordinates) like the above figure.

  4. Denote the sample point as $(x_i,y_i)$, $x_0=0$, $x_n = a$.

  5. Approximate $$\frac{dy}{dx}\Big|_{x_i} \approx s_i = \frac{1}{2}\left( \frac{y_{i+1} - y_i}{x_{i+1} - x_i} + \frac{y_{i} - y_{i-1}}{x_{i} - x_{i-1}} \right).$$ For two end points: $$ s_0 = \frac{y_1 - y_0}{x_1 - x_0},\quad \text{ and }\quad s_n = \frac{y_n- y_{n-1}}{x_n - x_{n-1}}. $$ This step may be problematic, we can use other methods to approximate $dy/dx$: for example, approximating the curve by cubic spline using sample points $(x_i,y_i)$, but it would be beyond the content of college calculus.

  6. Let $h = x_{i+1} - x_i = a/n$, approximate (2) by computing: $$ \frac{2\pi h}{3}\bigg[g(x_0)+2\sum_{j=1}^{n/2-1}g(x_{2j})+ 4\sum_{j=1}^{n/2}g(x_{2j-1})+g(x_n) \bigg], \quad \text{ where } g(x_i) = y_i \sqrt{1+s_i^2}. $$


Some results comparison:

Amzoti gave a link in his comments above that has two semi-empirical formulas: $$ S_1 = (3.155 − 0.0136a + 0.0115b)ab, \;\text{ and }\;S_2 = \left(0.9658\frac{b}{a}+2.1378\right)ab $$ where $a$ and $b$ are the length for major and minor axis of the real eggs. If there exists an egg's shape like (1), $a = 1$, and $b\approx 0.7242629$, the surface area computed by above formula is: $$ S_1 \approx 2.278215 \;\text{ and }\;S_2 \approx 2.054946. $$ Using the surface by revolution formula (2), we have: $$ S \approx 2.042087. $$

Shuhao Cao
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    Nice, mathematical answer and thanks for including the reference link. – yiyi May 31 '13 at 07:26
  • @Shuhao Can you give a few numbers? (e.g. for $a=1$). Now I'm curious. – lcv May 31 '13 at 07:28
  • I already knew about this equation (on that note, maybe it is more convenient to use the parametric equations); the reason I had decided not to post this is that all we have is a curve that *looks like an egg*, but we don't really know if this is exactly the shape taken by an egg. – J. M. ain't a mathematician May 31 '13 at 08:11
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    It doesn't look to me like an egg, IME eggs are usually longer and thinner. – jwg May 31 '13 at 08:14
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    @jwg, there is in fact a tweakable parameter in the equation that allows you to produce more elongated ovals; nevertheless, we of course don't know if this is the true shape taken by an egg, since it only gives results that are qualitatively similar to an egg. – J. M. ain't a mathematician May 31 '13 at 08:40
  • Could an answer be to find a value of the parameter for which this 'looks' close to the egg you have, then try to measure whether or not they are actually close? – jwg May 31 '13 at 09:08
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    @jwg it depends on the egg. My parents have kept ~ two dozen chickens for the last 20 years; the range of "normal" eggs (excluding ex ones with 2 yolks) I've seen ranges from about 80% of the width of the curve shown to about 60% of the height of it. – Dan Is Fiddling By Firelight May 31 '13 at 12:56
  • @J.M. Thanks, that makes sense. I do think the revolution method does work for egg for it is axial symmetric. – Shuhao Cao May 31 '13 at 16:32
  • Oh, sure, to a good approximation, (chicken) eggs are axially symmetric. Our problem's the outline; so, this solution gives the surface area of an idealized egg, which may or may not correspond to actual eggs. – J. M. ain't a mathematician May 31 '13 at 16:40
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    In any event: for the lazy, here's a set of parametric equations for a generalization of the Yamamoto oval: $$\begin{align*} x&=\left(\frac{a}{2}-\frac{b}{2}\sin^2\frac{u}{2}\right) (1+\cos\,u)\\ y&=\left(\frac{a}{2}-\frac{b}{2}\sin ^2\frac{u}{2}\right)\sin\,u\end{align*}$$ The case discussed in this answer corresponds to $b=7a/10$. – J. M. ain't a mathematician May 31 '13 at 16:46
  • Excellent in-depth answer! Thanks! – Darth Egregious May 31 '13 at 21:14
  • If you revolve the whole curve by $ \pi $ it's going to produce the whole surface and there is no need to rotate it further. – racic Jun 18 '13 at 10:45
  • The length of the two axes could be measured (leaving the egg intact) by sliding two vertical surfaces (like the walls of two shoe boxes) sliding them in parallel with the help of a straight edge, so that the egg is caught between them. By placing the egg laying on the table and making sure that its longer axis is roughly perpendicular to the ruler, one can measure the short axis of the egg as the distance between the surfaces (rotating it around its long axis one can take several measurements and average them too). Repeating the process with the longer axis one obtains the remaining length. – Guillermo BCN May 20 '20 at 09:11
  • Broken link. Perhaps it's now at http://nyjp07.com/index_egg_E.html – lhf Apr 04 '21 at 15:18
  • Try investigating the polar curve $r = a \cos^3 \theta + b \cos \theta$. The edge around $r = 0$ is a bit pointy but is a reasonable approximation. – Toby Mak Jul 12 '21 at 02:00
  • Is [this site](https://goodcalculators.com/egg-surface-area-and-volume-calculator/) also useful? – Snack Exchange Mar 19 '22 at 16:18
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Let's make a problem a little more interesting by generalizing it. I have an arbitrary convex object, and I want to find its surface area. Of the answers posted so far, only the triangulation strategy of Zach L. and Cong Xu works in this case without breaking the object into little pieces. Here's another approach.

Suppose you project the object onto a randomly oriented plane, i.e. a plane whose normal is chosen uniformly from the unit sphere. Given that the object is convex, the expected value of the projected area is exactly $1/4$ times the surface area of the object, for essentially the same reasons given by Christian Blatter for the 2D case. (Short version: each differential surface element $dA$ contributes on average $dA/4$ to the projected area; its orientation doesn't matter because we're averaging over all possible directions of projection; there is no double-counting because the object is convex). This immediately suggests a Monte Carlo algorithm:

  1. Rotate the object into a random orientation.
  2. Shine collimated light at the object (e.g. from the sun, or from a point-light/parabolic-mirror combination) and observe its shadow on a plane perpendicular to the light direction.
  3. Record the area of the shadow. Maybe you have graph paper pasted on the plane, or maybe you take a picture with a calibrated camera, binarize the image, and count the number of black pixels.
  4. Repeat lots and lots and lots of times.

The average area of the shadow, times $4$, is the surface area of the object.

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    "Shine collimated light at the object (e.g. from the sun...)" - bonus: done long enough, you should be able to cook your egg, too. Since it will heat up, there might be a little increase in the surface area due to expansion... – J. M. ain't a mathematician May 31 '13 at 08:52
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    @J.M. If a little sunlight falling on eggs was enough to cook them, there probably wouldn't be any chickens around today... –  May 31 '13 at 08:57
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    A little won't cook them; it's the "Repeat lots and lots and lots of times." part that just might... :) – J. M. ain't a mathematician May 31 '13 at 08:59
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    My point remains about the equilibrium temperature of an egg in sunlight and the continued existence of chickens. –  May 31 '13 at 09:03
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    If you have the ability to do this accurately, it might be fruitful to project perpendicular to the egg's axis of rotational symmetry, and measure the perimeter of the shadow. – jwg May 31 '13 at 09:13
  • This obviously works for a sphere: $\pi r^2 \larrow 4\pi r^2$. But how do you ensure that your sampling from different angles is uniform? A bias could creep in. Or, is it possible that it's enough to average just three projections of the egg? Two radial views (identical area) and one axial? I suspect that would work for an ellipsoid. – Kaz May 31 '13 at 23:24
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    @Kaz This is certainly not sufficient for an arbitrary convex object. Consider the intersection of the cylinders $x^2+y^2\le1$, $x^2+z^2\le1$, and $y^2+z^2\le1$. This object has circular projections along all three axes, but it has greater volume and surface area than a sphere. (If you assume it is a convex surface of revolution, then of course one axial projection is sufficient to determine the entire object.) – Mario Carneiro May 31 '13 at 23:56
  • How can you effectively determine the exact silhouette of the necessarily randomly oriented egg without using **black magic** to levitate it? A transparent mounting apparatus? D'oh – Steven Lu Jun 01 '13 at 01:58
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    How do you handle diffraction around the egg? – dfeuer Jun 01 '13 at 02:13
  • @dfeuer: Gamma rays. –  Jun 01 '13 at 06:54
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    @RahulNarain: I think you'll run into problems with gamma rays too—the egg will start acting translucent. – dfeuer Jun 01 '13 at 07:05
  • @StevenLu You could use glass, or some other transparent thing. – PyRulez Feb 06 '18 at 19:05
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Use any convenient method to determine the volume of the entire egg.

Remove contents of egg through a small hole without damaging eggshell. (Egg collectors have a number of techniques).

Fill eggshell with water to determine the volume of the inside of the egg.

The difference of these two measurements is the volume of the eggshell.

Break the egg shell and measure the thickness. $$\frac{\text{∆ Volume}}{\text{Thickness}} \approx \text{Area}$$

PyRulez
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DJohnM
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    Shouldn't you write, that $Volume/Thickness$ approximates the area? The egg shell may of uneven thickness. Even if it is not, the inner surface area and outer one are not the same. – bleh May 31 '13 at 05:30
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    You can simplify the process by determining directly the volume of the shell - break the egg and place the shell in a measuring cup (with some water), and note the difference between water levels before and after putting the shell. – Sam May 31 '13 at 07:30
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    You've replaced one measurement of a large quantity with two measurements of very small quantities - the difference in volumes and the thickness of an eggshell. Dividing one of these by the other is going to make your errors blow up out of control. – jwg May 31 '13 at 08:17
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    this is more of a experimental physics answer (and a poor one at that) than a mathematics answer – jk. May 31 '13 at 11:06
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    @jk I'd argue that this question doesn't really fit into math.SE to begin with -- if you're looking to measure a real object, you've entered the realm of (as you say) experimental physics, or engineering. Both use mathematics, but mathematics itself won't determine the answer. – Chris Gregg May 31 '13 at 13:11
  • @ChrisGregg that's probably fair – jk. May 31 '13 at 13:26
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Get one of these things:

enter image description here

Load it up in your favorite 3D modelling software and let it calculate the surface area for you:

enter image description here

congusbongus
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    This is essentially an automated version of Zach L.'s answer. –  May 31 '13 at 06:25
  • Oh, so then I just measure the area of each triangle and then add them up, and I get the area from the measured string. – yiyi May 31 '13 at 06:33
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Here's a shot:

Place dots on the egg, fairly close together, and connect them to form "triangles" on the egg. Measure the distances between the points of the points in the triangle and figure out the area as if it were a Euclidean triangle. Do this for several eggs, perhaps dividing by the square of the length along an axis to normalize it. Then repeat the process, with smaller and smaller triangles, until (hopefully!) the numbers will sort of agree up to some decimal place of accuracy.

Zach L.
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    Could you explain - Measure the distances between the points of the points in the triangle. – yiyi May 31 '13 at 02:56
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    Hopefully you don't end up with something like this: http://mathoverflow.net/questions/89991/naive-definition-of-surface-area-doesnt-work/89994#89994 – Chris Brooks May 31 '13 at 03:42
  • @yiyi I would get some kind of marked string and see what the distances are, measured along the surface of the egg. I'll admit that doing this would be incredibly tedious and unpleasant. But if you have nothing better to do... – Zach L. May 31 '13 at 04:34
  • @Joseph That's quite cool! I don't have time to really think about it at all, but I wonder if the same thing can happen when you use the intrinsic distance and comparison triangles, as in the suggestion. Perhaps if yiyi carries out the experiment, they can tell us if the areas do anything odd! – Zach L. May 31 '13 at 04:41
  • @ZachL. Oh, I am excited to do this; however, just making sure that I fully understand what I should be doing. Far as I understand, is to wrap the string around the edges of the triangles drawn on the egg. – yiyi May 31 '13 at 06:30
  • Yep. I'm not sure how many dots you'll need for a good estimate, though. If you want some idea of what you're doing, take a look at Cong Xu's answer. You'll be doing it by hand, though! – Zach L. May 31 '13 at 06:33
  • You could also measure the linear distance between the vertices using callipers rather than the geodesic distance using string? – jwg May 31 '13 at 14:03
  • If you have calipers handy, yes. It would also be much faster, I'd say. – Zach L. May 31 '13 at 15:50
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Supplies:

  • Egg
  • Letter scale
  • Micrometer
  • Cup
  • Chocolate
  • Eyehook???

Instructions:

  1. Melt the chocolate into the cup.
  2. Record the starting weight (Ws) of the cup of chocolate, .
  3. Record the width of the egg (wE) as measured with the micrometer, .
  4. Dunk the egg and ONLY the egg into and out of the chocolate. An eyehook may come in handy here...
  5. Rotate the egg around as it dries to make the coat as even as possible, making sure all drippings land back in the cup.
  6. Record the final weight (Wf) of the cup of chocolate.
  7. Calculate the weight of the chocolate coat (W): $$W = Ws - Wf$$
  8. Record the width of the chocolate-coated egg (wCE) as measured with the micrometer.
  9. Find the thickness of the chocolate shell (T): $$ T = {\frac{wCE - wE}{2}}.$$
  10. Find the density (D) of the melted chocolate by google or by experiment.
  11. Calculate the approximate surface area (A): $$A = {\frac{W}{T * D}}$$

Note: The thicker the chocolate coat, the larger (and more incorrect) your approximation of the surface area will be, so improvements on the final formula can be made which account for this. For example, one reasonable improvement would be to calculate A as the internal surface area of a spherical shell made with the same shell-thickness and shell-volume as the one covering our experimental chocolate egg, which can be derived with the information we have and a little spherical math.

Briguy37
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    This method has the beneficial side effect that it leaves you with a chocolate egg at the end. :) – joeytwiddle Jun 01 '13 at 13:04
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    You could also estimate the thickness of the coating by measuring the weight loss to an object with known surface area, a ping-pong ball for example. – joeytwiddle Jun 01 '13 at 13:09
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    I thought of this with thick paint (@joeytwiddle's variant), but ok! If you twist my arm, I'll go with chocolate... – Euro Micelli Jun 01 '13 at 15:25
  • Really, you should be using sausage meat and breadcrumbs for this. http://www.janeausten.co.uk/wp-content/uploads/2012/01/scotch_egg.jpg – Lucas Jun 02 '13 at 02:32
  • Ingenious! Plus it involves melted chocolate so it's got my approval. – mikhailcazi Jul 26 '13 at 14:00
6

Joining the frenzy of ways to do it, here is a physics-lab-like method:

  • Take some millimeter paper
  • Break the shell in almost-flat pieces and paint them with ink (sink it in ink).
  • With some tweezers take the pieces, make sure there is no spilling ink, and put them on the millimeter paper, ink face down, with any comfortable distance between pieces
  • Cover the paper with another piece of paper and press smoothly to even out the rounded corners of the shell
  • let the ink dry, take the shells and count the squares covered with ink.

There will be some approximation, given that it's hard to estimate half-squares, but I don't see why it's worse than approximating the thickness of a shell as being uniform.

EDIT

This may be formulated as hands-on, distructive "triangulation". Though it may not fit the accurate mathematical definition of the notion.

Sam
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3

Since some others have included empirical solutions, I will suggest a method I have used in the past to get the surface area of irregularly shaped rocks.

  1. Determine the weight of a known area of aluminum foil (e.g., 1 cm$^2$)

  2. Carefully cover the egg with strips of foil so that there is no overlap or folding of the foil.

  3. Remove the foil from the egg and determine the weight of the total amount of foil required to cover the egg in a single layer.

  4. Calculate the surface area as:

$A_e = \frac{A_k}{M_k} M_e$

Where:

  • $A_k$ = the known surface area of a piece of foil
  • $M_k$ = the known mass of a piece of foil
  • $A_e$ = the surface area of the egg
  • $M_e$ = the mass of the foil required to cover the egg in a single layer of foil
DQdlM
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1

in this link you can see the development of what has been discussed here: http://www.mapleprimes.com/questions/148150-How-Can-I-Find-The-Surface-Area-Of-A-Normal-Chicken-Egg#comment148174

0

A few steps should allow to determine the surface of a given egg

  • Take a measuring cup as accurate as possible.

  • Pour some water in the cup, say half the cup: note the current volume $v_1$

  • Put the egg inside the cup and it should sink

  • Note the new volume $v_2$, where $v_2 - v_1$ gives the volume of the egg

  • consider that to be the equivalent volume of a sphere, where $$v_2 - v_1 = \frac{4}{3}\pi r^3$$ and get $r$ from that.

  • the surface is $4\pi r^2$

Michael Albanese
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Déjà vu
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    Hmm, this seems to be a poor estimate. Eggs make bad spheres – davidlowryduda May 31 '13 at 04:41
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    @mixedmath Hmm, the volume of the *equivalent* sphere is the same as the volume of the egg... thus that should work, no? – Déjà vu May 31 '13 at 04:44
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    The volume-surface area relationship for a sphere is not guaranteed to be the same as that of an egg... not to mention $\pi r^2$ does NOT yield the surface area of a sphere. – obataku May 31 '13 at 04:51
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    @oldrinb it was $4\pi r^2$, thanks. Intuitively, the shape of the egg should not matter, since its volume is / should be accurate (as water follows closely the shape of the egg). But I'm not a math expert, so ... I believe you. – Déjà vu May 31 '13 at 04:55
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    @mixedmath: Except spherical eggs in vacuum, which is more or less the sort of eggs you'd get from a spherical chicken in vacuum... :-) – Asaf Karagila May 31 '13 at 05:02
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    Hi, ringo, welcome to MSE. Say a sphere has the same volume as an ellipsoid, but their surface areas may differ by a large margin. – Shuhao Cao May 31 '13 at 05:03
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    @ringo, I am afraid your assumption of equivalence is incorrect. A sphere is known to be the shape with the smallest surface area relative to its volume (the "most efficient container", beating cylinders, cubes and pyramids). *Any other shape* is guaranteed to have a larger relative surface area. To conceptualise this, use the technique of "taking an extreme": imagine continuing to elongate the egg more and more away from sphere shape. It will eventually begin to seem more like a cylinder, and will certainly gain relative surface area along the way. – joeytwiddle Jun 01 '13 at 13:15
  • @joeytwiddle Thanks. This is indeed what I did after reading all these comments (limits case, wiki etc...). But I think, still, it is a rough estimation - an egg being not that far from a sphere... Actually I'd be curious to get an idea of the % of difference between the exact solution, and the volume-to-sphere conversion. – Déjà vu Jun 01 '13 at 13:24
  • @davidlowryduda ostriches eggs are round ;-) – Déjà vu Aug 15 '20 at 09:13