How do you find the area of an egg shape formed from the union of circles whose diameters are horizontal chords of the unit circle?

Question

This egg is the union of the red circles, whose diameters are horizontal chords of (the upper half of) the unit circle.

How do you find its area?

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Some background information on this egg is shown below:

Answer 1

In short it can be shown that the top part of the egg is bounded by half-ellipse with radii $a,b=1,\sqrt2$, and it is given that the bottom part is a half-circle of radius $r=1$. Use the fact that ellipse area is $\pi ab$ and circle area $\pi r^2$, to get that the total area is: $\frac12(\sqrt2+1)\pi$.

Below is how I solved the problem to obtain the above result.

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Let $R$ be radius of the black circle inside the egg. For the unit circle, set $R=1$.

To recap the problem, I will be solving the problem of the area of the following egg shape: Set defined as union of circles whose diameters are horizontal chords of the upper half of a radius $R$ circle.

We can parameterize the egg by defining the chords as $\overline{AB}$ where:

(I assume you are familiar with sine and cosine on unit circle. We are just scaling it by $R$.)

$$\begin{align} A&=(R\cos(\alpha),R\sin(\alpha))\\ B&=(R\cos(\pi-\alpha),R\sin(\pi-\alpha)) \end{align}$$

Where the family of chords is now given by all $\overline{AB}$ such that $\alpha\in[0,\pi/2]$.

Each chord defines a circle centered at midpoint of it, with radius $R\cos(\alpha).$

The midpoint (circle center) coordinates are: $(0,R\sin(\alpha))$.

This gives the family of circles which defines the egg:

$$ \bigcup_{\alpha\in[0,\pi/2]} \{x^2+(y-R \sin (\alpha))^2 \le R^2 \cos ^2(\alpha) \} $$

We are now solving for the envelope of the egg. That is, we need to find the curve bounding the egg, so we can integrate for the area. Notice that due to symmetry and periodic functions, we can equivalently search for the envelope given by all chords and let $\alpha\in\mathbb R$, and then take the part of it above $x$-axis.

A similar problem has already been solved: "Find $\bigcup_{r\in R}\ \{(x,y): (x-r)^2 + (y+2r)^2 < r^2+1\}$".

The goal is to solve for $\alpha$ from $\frac{\partial f}{\partial \alpha}(x,y,\alpha)=0$ and substitute it back into $f(x,y,\alpha)=0$.

That is, we find the derivative of $f(x,y,\alpha)=0$ in respect to $\alpha$ :

$$\begin{align} f(x,y,\alpha)&=x^2+(y-R \sin (\alpha))^2-R^2 \cos ^2(\alpha)\\ \frac{\partial f}{\partial \alpha}(x,y,\alpha) &= 2 R \cos (\alpha) (2 R \sin (\alpha)-y) \end{align}$$

Solving $\frac{\partial f}{\partial \alpha}(x,y,\alpha)=0$, notice that $2R\cos(\alpha)$ is the length of the chord (twice the radius of initial circle) and hence we can assume it is positive. This means it must be $(2 R \sin (\alpha)-y)=0$, from which we obtain $\alpha=\arcsin(\frac{y}{2R})$.

Substituting $\alpha$ into $f(x,y,\alpha)=0$ and after simplifying, we get:

$$\begin{align} x^2+\frac{y^2}{2}&=R^2\\ \frac{x^2}{R^2}+\frac{y^2}{(\sqrt2 R)^2}&=1 \end{align}$$

Which is an ellipse. I have sketched this and made a Geogebra construction to triple-check my calculation so far. (I've been using Mathematica to double-check the derivation and substitution).

We need to integrate the area above $x$-axis and below the upper part of the obtained envelope:

$$ y=\sqrt{2(R^2-x^2)} $$

To get the upper part of the egg. The area below the $x$-axis is simply half a circle and given by $ \frac12 R^2\pi$.

In total we have:

$$ \text{Area(Egg)}=\left(\int_{-R}^R dx\int_{0}^{\sqrt{2(R^2-x^2)}}dy\right) + \frac12 R^2\pi $$

This integral can be done by substituting in the polar coordinates.

But you can also avoid integration by accepting that the area of the ellipse is $\pi ab$. We have $a=R, b=\sqrt2R$, and we are after half of an ellipse, so we have: $\frac{\sqrt{2}}{2}R^2\pi$. That is:

$$ \text{Area(Egg)}=\frac{\sqrt{2}}{2}R^2\pi + \frac12 R^2\pi $$

By summing half the area of the ellipse for upper part, and half the area of circle for bottom part.

Finally, we have:

$$ \text{Area(Egg)}=\frac{\sqrt{2}+1}{2} R^2 \pi $$

Plug in $R=1$ to obtain the case when the black circle is the unit circle.

Answer 2

The chord at ordinate $a$ generates a circle of radius $\sqrt{1-a^2}$,

$$x^2+(y-a)^2=1-a^2.$$

To obtain the envelope, we eliminate $a$, using the derivative on $a$ of this equation,

$$2(a-y)=-2a$$

and we obtain an ellipse of semi-axis $\sqrt 2,1$, of equation:

$$x^2+\frac{y^2}{2}=1.$$

Hence the total area is the average of the ellipse and circle areas

$$\frac{\pi\sqrt2+\pi}{2}.$$

Answer 3

We take chord mid-point height $h$ as parameter. $$(x-0)^2+(y-h)^2=(\sqrt{R^2-h^2})^2; x^2+y^2-2yh+2h^2 = R^2\tag1 $$ To find envelope we take partial derivative w.r.t.$ h$ $$-2y+4h=0\rightarrow y=2h \tag2$$ and eliminating $h$ between them $$ x^2 + y^2/2=R^2 \tag3$$ which is an ellipse. The two parts of egg (half-ellipse,circle) have area $$\frac12 \pi R \sqrt2R+\,\frac12 \pi R^2 \tag4$$

Parametric plot of (1) in Mathematica showing egg shape:

R = 1; ParametricPlot[{Sqrt[R^2 - h^2] Cos[t], 
  Sqrt[R^2 - h^2] Sin[t] + h}, {t, 0, 2 Pi}, {h, 0, R}, 
 Mesh -> {20, 9}, GridLines -> Automatic, 
 PlotStyle -> {Magenta, Thick}]