# Volume of a region on the sphere

a) Consider the region on $\mathbb{R}^{3}$ lying between two parallel planes that intersect the sphere. Show that the area of this region depends only on the distance between the two planes.

Is that right? Is there a simple demonstration of it?

b) Show that the result in part a) does not hold in $\mathbb{R}^{n}$ if $n\neq 3$ and "planes" are replaced by "hyperplanes".

Thanks

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Presumably that should be $\Bbb R^n$, not $\Bbb R^3$, in (b). Yes, (a) is correct; it’s a fairly routine volume of revolution problem. –  Brian M. Scott Sep 28 '12 at 12:39
Wouldn't the volume between two planes separated by d that pass through the sphere near the poles be smaller than that near the equator for same d? –  daniel Sep 28 '12 at 12:42

Of course daniel is right that the three-dimensional volume of the part of a sphere in $\mathbb R^3$ between two parallel planes at a distance $d$ is less near the poles. I suspect that what you meant to claim is that the surface area of the sphere between two parallel planes depends only on their distance $d$.

This is because the surface area of an infinitesimal ring between two parallel planes at an infinitesimal distance $\mathrm dx$ at a distance $x$ from the centre of a sphere of radius $r$ is given by the perimeter $2\pi\sqrt{r^2-x^2}$ of the ring times the infinitesimal width $\mathrm dxr/\sqrt{r^2-x^2}$ of the ring, and the product $2\pi r\mathrm dx$ is independent of $x$.

In $n$ dimensions, the perimeter is replaced by an $(n-2)$-dimensional volume proportional to $(r^2-x^2)^{(n-2)/2}$, whereas the width remains the same, so the product is proportional to $(r^2-x^2)^{(n-3)/2}$. This is independent of $x$ only for $n=3$.

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Yes you are rightm im gonna change there. –  Tomás Sep 28 '12 at 14:02
Dear @joriki, I was rereading your answer and I get confused. I think that in the place of $x$ we must have $h$. Is this right? Thank you –  Tomás Jul 25 '13 at 19:43
@Tomás: You're right, sorry about that. I made them all $x$. –  joriki Jul 27 '13 at 9:00

a) After a rotation we can assume, that the planes are defined by $x=a$ and $x=b$ with $-1 \leq a \leq b \leq 1$. The area between these panes is then given (cf. Wikipedia: Surface of revolution) by $$A_{a,b} = 2\pi \int_a^b y(x) \sqrt{1 + y'(x)^2} dx$$ where $y(x) = \sqrt{1 - x^2}$. Evaluating the integral we arrive at $A_{a,b} = 2\pi(b-a)$.

b) Take $a = -1$ and $b = -1 + \epsilon$. It is easy to see, that $A_{a,b} \sim V (2\epsilon)^{\frac{n-1}{2}}$ as $\epsilon \to 0$ where $V$ is the volume of the $(n-1)$-dimensional Ball of radius $1$, whereas for $c = -\epsilon/2$ and $d = \epsilon / 2$ we have $A_{c,d} \sim S \epsilon$ as $\epsilon \to \infty$, where $S$ is the surface area of the $(n-2)$-dimensional unit sphere.

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This is not true. Consider the unit sphere, and planes parallel to the $y,z$-plane, at $x$-coordinates $a$ and $b$ ($-1\le a<b\le 1$). Then, the plane at $x\in (a,b)$ intersects the sphere in a circle (disk) with radius $\sqrt{1-x^2}$, its area is $(1-x^2)\pi$. So the volume of that region of the sphere can be calculated by $$\int_a^b (1-x^2)\pi\, dx = \pi\cdot \left[x- \frac{x^3}3 \right]_{x=a}^b = (b-a)-\frac{(b-a)(b^2+ab+a^2)}3$$ and it doesn't only depend on $b-a$. Neither in higher dimension.

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sorry it was not volume, its was area. –  Tomás Sep 28 '12 at 14:04