Area of sphere intercepted by given curve the given curve.

Question

I was just messing around in geogebra 3D, when I noticed a strangely shaped portion of the unit sphere at origin, made by intercepting it with $z = e^{-x^2}$.

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So I was wondering wether we could calculate the area of the spherical part intercepted, i.e. the pink butterfly-like part shown in the pictures?

Here is the link to the graph in geogebra

I cant seem to figure out where to start...... any help would be appreciated.

Answer 1

The intersection curve is

$$x^2+y^2+z^2=1,\\z=e^{-x^2}$$

and on the $xy$ plane,

$$x^2+e^{-2x^2}+y^2=1$$ or

$$y=\pm\sqrt{1-x^2-e^{-2x^2}},$$

which is defined for $|x|\le x^*=0.8926433386409\cdots$

Using symmetry, the area of the portion of the sphere so delimited is given by the surface integral

$$4\int_0^{x^*}\int_0^{\sqrt{1-x^2-e^{-2x^2}}}\frac1{\sqrt{1-x^2-y^2}}dy\,dx.$$

The integration on $y$ is possible, giving

$$\left.\arctan\dfrac y{\sqrt{1-x^2-y^2}}\right|_0^{\sqrt{1-x^2-e^{-2x^2}}}=\arctan(e^{x^2}\sqrt{1-x^2-e^{-2x^2}})$$ but the second on $x$ is hopeless. You will need numerical integration.

Answer 2

You can determine the boundary of the region, projected down to the $x$-$y$ plane, by solving

$$ e^{-x^2} = \sqrt{1-x^2-y^2} $$

Squaring both sides, then isolating $y^2$ on one side, gives us

$$ y = \pm \sqrt{1-x^2-e^{-2x^2}} $$

Those values of $y$ can give you the integrand. However, rather than use it directly in a three-d surface integral, as suggested by Yves Daoust in the comments below, I will make use of a pretty fact that ought to be better known than it is:

Suppose you encase a unit sphere in a right cylinder of radius $1$ and height $2$ (so that it exactly circumscribes the sphere). Then cut any slab of the cylinder and the encased sphere (perpendicular to the cylinder's axis). The surface area of the sphere slice and the surface area of the cylinder slice are exactly equal.

This is perhaps a bit difficult to visualize if you haven't seen it before, so here's a picture.

The light blue portions of the surface areas are equal (not the dark blue sections).

What this means, analytically, is that we can first find what angle of the sphere is subtended by this value of $y$ that we computed, and twice that angle (because $y$ goes from the negative square root to the positive square root) is the surface area of the spherical slice in question. For this purpose, we want the arcsin function (inverse sine). So our integrand will be $2 \arcsin \frac{\sqrt{1-x^2-e^{-2x^2}}}{\sqrt{1-x^2}}$.

Now, what are the limits of this integration? We need to find the values of $x$ where the integrand shrinks back down to $0$. That is, we must find $x$ where

$$ 1-x^2-e^{-2x^2} = 0 $$

As WolframAlpha tells us, this is not answerable in elementary functions (at least, I'm pretty confident it can't be); we must make use of the Lambert W function, and obtain a numerical value:

$$ x = \pm \frac{\sqrt{W\!\left(-\frac{2}{e^2}\right)+2}}{\sqrt2} \approx 0.892643 $$

To obtain our area, then, we integrate (numerically) to obtain

$$ A = \int_{x=-0.892643}^{0.892643} 2 \arcsin \frac{\sqrt{1-x^2-e^{-2x^2}}}{\sqrt{1-x^2}} \, dx \approx 1.2334 $$