3
$\begingroup$

Given a sphere $S$ of fixed diameter $D$ (or radius $R=D/2$, it will be convenient to have both, I suppose), and a point $P$ on its surface, let's create a ball $B$ of variable radius $r$ with its centre in $P$. What I have been wondering was - what is the surface area $A(r)$ of the intersection of the sphere and the ball as a function of $r$?

Recently I found a neat solution using the formula for the area of spherical cap - which it will be, after all, I just had to relate the missing variables for the formula to the ones I know, and I managed to get that $A=\pi r^2$ for $0 \leq r \leq D$ and I believe that to be correct (it might not necessarily be, I just was unable to find any limiting cases that would prove the formula wrong).

However, my initial approach - the crux of the question - was different. I replaced B with a sphere $B'$ with the same parameters (i.e. centre in $P$ and variable radius $r$), attempted to find the perimeter of the intersection (circle) $p(r)$ and then let $A(r)=\int_{0}^{r}p(x) dx$

Finding that $$p(r)=2\pi \sqrt{r^2-\frac{r^4}{D^2}}$$ was simple, but the integration was not, and not knowing enough to reliably follow through with it, I asked Wolfram Alpha to do the heavy lifting, which gave me this: $$\int p(x) dx = \frac{2\pi (x^2-1)\sqrt{x^2-x^4}}{3x}+c$$ (I assumed D=1 for simplicity to easier see the curve, perform sanity checks, and believing that I can always multiply the resulting area by $D^2$)

The curve seemed plausible, I just had to compute the exact function $A(r)$ as defined above as a definite integral. Because the function wasn't defined at 0, I had to compute $$\lim_{x \to0^+}2\pi \sqrt{x^2-\frac{x^4}{D^2}}$$ which gave me a puzzling result - $2\pi/3$ for unit sphere.

My question is - why this value? What was incorrect in this approach?

$\endgroup$
  • $\begingroup$ How can you get a non-zero value for the limit? $\endgroup$ – Tunococ Jul 23 '14 at 5:21
  • $\begingroup$ Why would it be zero, if the curve goes into negatives near zero on the right? I'm computing not the limit of the entire definite integral, but just the function after integration. Unless you're wondering why is it acceptable for it to go into negatives in the first place - I'm not computing an integral from minus infinity to r, but from 0 to r, and in the end I'm going to subtract the value of the integral at 0 from the value of the integral at r, so if the former is guaranteed to be no larger than the latter, the end result - area - is positive, and it's all fine. I hope I was clear enough $\endgroup$ – Haxton Fale Jul 23 '14 at 10:31
  • $\begingroup$ I see. I believe you made several typos that you weren't aware of. For example, the equation for $p(r)$ has $x$ instead of $r$ on the right-hand side. You said you had to compute the limit of the integrand, not the antiderivative, but I believe you meant to say you had to compute the limit of the antiderivative. Anyway, I agree with the answer below that the integral for the area was not quite right. As you grow the smaller ball (increase $r$), the intersection (on the surface of the big sphere) does not grow perpendicular to the surface of the small ball. $\endgroup$ – Tunococ Jul 23 '14 at 10:40
  • $\begingroup$ Ah, yes, I apologize for all the typos. I'm not a native English speaker and I haven't studied maths in it either :) Thank you for clarification. $\endgroup$ – Haxton Fale Jul 23 '14 at 10:42
1
$\begingroup$

find the perimeter of the intersection (circle) $p(r)$ and then let $A(r)=\int_0^r p(x)\,dx$

It's not that simple. Area (in 3 dimensions) is generally tricker to compute than volume (also in 3 dimensions), similarly to how length (in 2 dimensions) is harder to deal with than area (in 2 dimensions). When we have an object that fills up a solid chunk of space, we can just cut it into tiny pieces and measure those. But when we measure something more delicate, like a curved surface in three dimensions, the angle at which the surface meets the slicing knife is important.

To set up the integral correctly, you have to figure out the area of a thin belt cut out of the sphere. And it is not $p(x)\,dx$, but rather $\dfrac{p(x)}{\sin \theta}\,dx$ where $\theta$ is the angle between the knife (the thing with which we cut) and the surface that we cut.

Let's do a simpler example: find the area of the disk of radius $1$ in the $xy$-plane. I will slice it by spheres centered at $(0,0,1)$, so the radius of those spheres will vary from $1$ to $\sqrt{2}$. The perimeter cut out by the sphere of radius $x$ is $p(x)=2\pi \sqrt{x^2-1}$. The integral $\int_1^{\sqrt{2}}p(x)\,dx$ is... something crazy with logarithm in it. Definitely not $\pi$.

After some trigonometry, I found $\sin \theta = \dfrac{\sqrt{x^2-1}}{x}$, so the correct computation is $$2\pi \int_1^{\sqrt{2}}\sqrt{x^2-1} \frac{x}{\sqrt{x^2-1}}\,dx = \pi$$

Of course this is a ridiculous way to find the area of a disk, but I hope the example shows what is going on.


The same thing is responsible for the length formula $\int \sqrt{1+(y')^2}\,dx$ for the length of the graph. Here $\sqrt{1+(y')^2}$ is the reciprocal of the sine* of the angle at which our (vertical) knife meets the graph.


(*) Yes, I know $1/\sin $ is often called cosecant. I hate it.

$\endgroup$
  • $\begingroup$ I think I see what went wrong, but the exact maths involved is still a bit beyond me. In my case, would theta be the angle between the surface tangential to B' and one tangential to S at any point of contact? (Since I'm assuming that my "knife" would be B' and my "thing which I cut" would be S) $\endgroup$ – Haxton Fale Jul 23 '14 at 10:38
1
$\begingroup$

Start with a sphere of radius 'r', centered at the origin. Consider the circular perimeter that cuts the sphere exactly in two in the plane of the x- and y- axese. Now focus on a symmetrical arc above and below the x-axis that will define the spherical cap we wish to work with. The angle from the centre/origin to the top of this arc we will call 'phi', and the general angle within this sector (above the x-axis) we will call 'theta'. [In the notes that follow, the letter 'd' will stand for delta, as in 'a small change in', and a full stop without a space after it will stand for multiplication. ]

Consider a little bit of this arc, 'da', with its bottom end at some general angle, theta, to the x-axis. If you rotate this arc section around the x-axis then it will trace out an area like a rubber band (the slightly thicker ones) stretched around a ball. The area will be the band's circumference around the x-axis multiplied by da. The radius for this circumference will be $r \sin\theta$ and the arc length da can be expressed as $r d\theta$. This makes the area of the band equal to $2\pi r \sin \theta r\,d\theta$. This can be integrated with respect to theta within the limits $\theta=0$ and $\theta=\phi$. The result is $$\int_{0}^{\phi} -2\pi r^2 \cos\theta \,d\theta$$ which is $-2\pi r^2 \cos \phi - 1$, which we will call formula 'A'.

Now, $\cos \phi$ is given by the usual 'adjacent'/'hypotenuse' relationship where the adjacent side is the perpendicular distance from the base of the spherical cap to the origin of the sphere, and the hypotenuse is the radius of the sphere. We will denote the perpendicular height of the spherical cap from its base by the letter 'h', and so $\cos\phi =r/(r-h)$, or $\cos\phi =1-h/r$, which can be re-arranged as $\cos \phi -1=-h/r$.

If we substitute $-h/r$ into formula 'A', then we get the result for the area of the spherical cap as $Area=2\pi r h$. Done.

$\endgroup$
  • $\begingroup$ Welcome to Math.SE! See math notation guide for the tips on formatting math here. I edited your answer. $\endgroup$ – user147263 Jan 9 '15 at 6:41
0
$\begingroup$

I just want to confirm that the formula you found was correct. Here's how I justify (derive) it.

Suppose the big hollow sphere $S$ is centered at the origin. It turns out that the area of $S \cap ([a, b] \times \mathbb R \times \mathbb R)$ is equal to $2 \pi R(b - a)$ for $-R \le a \le b \le R$. (This was quite surprising to me the first time I discovered it.)

So, if you place the small solid ball $B$ at $(R, 0, 0)$ and the ball has radius $r$, the intersection $S \cap B$ would have area equal to $2\pi R(R - d)$ where $d$ is the $x$-coordinate of $S \cap \partial B$, which is a circle. The problem reduces to finding $d$ from the given $r$.

To simplify the notation, just look at the projection of this problem on the plane $z = 0$ (or $y = 0$, or any other plan that cuts through centers of both spheres/balls). I'll use $z = 0$ so coordinates become $(x, y)$. The sphere $S$ becomes a circle of radius $R$ centered at the origin. $\partial B$ becomes a circle of radius $r$ centered at $(R, 0)$. The intersection of $S$ and $\partial B$ becomes two points $(d, y)$ and $(d, -y)$. The triangle $(-R, 0)$, $(R, 0)$, $(d, y)$ is a right triangle similar to triangle $(d, y)$, $(R, 0)$, $(d, 0)$. Similarity gives $\frac{2R}{r} = \frac{r}{R - d}$, hence $R - d = \frac{r^2}{2R}$. The final answer is $$ A(r) = 2\pi R(R - d) = 2\pi R\frac{r^2}{2R} = \pi r^2. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.