Calculating area of sphere with constraint on zenith

Question

Let $\mathbb S_R$ be the sphere of radius $R$ centered about the origin. Consider $$A_R= \left\{ (x,y,z)\in \mathbb S_R \mid x^2+y^2+(z-R)^2\leq 1 \right\}.$$

I want to calculate the area of this region of the sphere with radius $R$ about the origin. I already calculated the area of the region of $\mathbb S_R$ of vectors with zenith $\leq \alpha$ is given by $2\pi R^2(1-\cos \alpha)$.

For $A_R$, here's my idea. I want to find the zenith $\alpha$ which satisfies $R\sin \alpha=\sin \alpha+R$, since this is the zenith of points of both $\mathbb S_R$ and the unit sphere translated up $R$ units at points which are of the same height. Then, I just want to integrate the zenith $0\leq \phi\leq \alpha$.

Solving gives $\phi=\arcsin \frac R{R-1}$ and things get a little messy.

On the other hand, another student posted a solution which makes sense. He just writes $\alpha$ should satisfy $2R^2-2R^2\cos\alpha=1$ and then obtains the area of $A_R$ is $\pi$, independently of the radius.

Why is my method wrong?

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Answer 1

The problem seems to be what you mean by "the zenith of points of both $\mathbb S_R$ and the unit sphere translated up $R$ units at points which are of the same height."

The quantity $R \sin \alpha$ is the distance from the $z$ axis of points whose spherical coordinates are $(R, \theta, \alpha)$ for any $\theta,$ that is, the points on the sphere $\mathbb S_R$ that are at a zenith angle of $\alpha$ relative to the center of that sphere. So far, this makes sense.

The quantity $\sin \alpha$ is the distance from the $z$ axis of points on the unit sphere about the origin at a zenith angle of $\alpha.$ This does not change when you translate that sphere upward $R$ units, so it's unclear why you would add $R.$

In fact, since $0 < \sin\alpha \leq 1$ whenver $0 < \alpha < \pi,$ for any such $\alpha$ we would find that $R \sin\alpha \leq R < R + \sin\alpha,$ so it is impossible to have $R \sin\alpha = R + \sin\alpha.$ And that equation is just as impossible for $\alpha = 0$ or $\alpha = \pi$ as long as $R > 0.$ In short, the equation you proposed just does not make sense.

Now there may be some sense in equating the heights ($z$ coordinates) of points at the intersection of $\mathbb S_R$ and the unit sphere around $(0,0,R).$ For that, we use the cosine of the azimuth angle relative to each sphere, and we note that the azimuth angle of these points relative to the sphere around $(0,0,R)$ is greater than their azimuth angle relative to $\mathbb S_R.$ There is a relationship between the two angles, but it's not what your equation says.

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As for the claim that $2R^2-2R^2\cos\alpha=1,$ which the other student presumably did not explain, let $P$ be a point where the two spheres intersect. Then the points $(0,0),$ $(0,R),$ and $P$ form an isoceles triangle with two legs of length $R$ (each a radius of $\mathbb S_R$) and a base of $1$ (the segment from $(0,R)$ to $P,$ which is a radius of the unit sphere around $(0,R).$)

If we drop a perpendicular from either $P$ or $(0,R)$ to the opposite leg, we divide the isoceles triangle into two right triangles, one with hypotenuse $R$ and angle $\alpha$ and one with hypotenuse $1$ and angle $\frac\alpha2.$ If we drop a perpendicular from $(0,0)$ to the base of the isoceles triangle, we get two congruent right triangles with hypotenuse $R$ and angle $\frac\alpha2,$ with leg $\frac12$ opposite the angle $\frac\alpha2.$

In either case we end up having to deal with a right triangle with angle $\frac\alpha2.$ Let's try working with the second pair of right triangles. The construction tells us that $R \sin\frac\alpha2 = \frac12.$ Square both sides and multiply by $2$: $$2R^2 \sin^2\frac\alpha2 = 2\left(\frac14\right) = \frac12.$$ Use the trigonometry identity $2 \sin^2\frac\alpha2 = 1 - \cos\alpha$ (which you can get from either the half-angle sine formula or the double-angle cosine formula) to substitute $1 -\cos\alpha$ for $2 \sin^2\frac\alpha2$: $$R^2(1 - \cos\alpha) = \frac12.$$

You can then distribute the $R^2$ over $1-\cos\alpha$ and multiply both sides by $2$ to get the other student's formula if you want, although I think the formula above is more convenient for just plugging into your known area formula.

Answer 2

I think @David K has addressed your point. Here is how I would proceed:

Parametrize the surface as follows: \begin{cases} x=x\\ y=y\\ z=\sqrt{R^2-x^2-y^2} \end{cases} with $(x,y)\in D$, where $D$ is projection of the intersection of both spheres in the $xy$-plane. When both spheres intersect, we have \begin{cases} x^2+y^2+z^2 = R^2 \\ x^2+y^2+(z-R)^2 = 1 \end{cases} which leads to $x^2+y^2=1-\frac{1}{4R^2}$. In other words, $D$ is a disc with radius $\sqrt{1-\frac{1}{4R^2}}=\frac{\sqrt{4R^2-1}}{2R}$ (assuming $R\ge \frac{1}{{2}}$). This parametrization only works when the radius of the bottom sphere is smaller than (or equal to) the radius of $D$, i.e. when $R\le \frac{\sqrt{4R^2-1}}{2R}$, or $R\ge 1/\sqrt{2}$.

We need the norm of the normal vector to $S$, i.e., the norm of $$\pmatrix{1\\0\\\frac{-x}{\sqrt{R^2-x^2-y^2}}}\times\pmatrix{0\\1\\\frac{-y}{\sqrt{R^2-x^2-y^2}}}=\pmatrix{\frac{x}{\sqrt{R^2-x^2-y^2}} \\ \frac{y}{\sqrt{R^2-x^2-y^2}} \\ 1} $$

which is $$ \sqrt{1+\frac{x^2+y^2}{R^2-x^2-y^2}}=\frac{R}{\sqrt{R^2-x^2-y^2}} $$ It follows that the wanted area equals $$ \iint_D \frac{R}{\sqrt{R^2-x^2-y^2}}\; dA = \int_0^{2\pi}\int_0^{\frac{\sqrt{4R^2-1}}{2R}} \frac{R}{\sqrt{R^2-r^2}}\; r dr d\theta = 2\pi R^2-\pi |2R^2-1|\\ $$

Remember that we assumed that $R\ge 1/\sqrt{2}$, so this expression equals $\pi$. If $0

Note: spherical coordinates are much more interesting here. Equation $(1)$ is simpler to find, to solve, and it is valid for all $R\ge 1/2$.