Volume and perimeter of bodies of revolution

[gugo82]

I drew inspiration from this thread in Analisi's room to propose the following question.

***

Let $N \in NN$ greater than $2$; in what follows a point of $RR^N$ will be denoted as $z$.
Let $f \in C_c(RR)\capC_(pw)^1(RR)$* be non negative, $r \in RR^N$ be a straight line through $o=(0,\ldots ,0)$ and $Pi:=r^bot$ the hyperplane through $o$ orthogonal to its direction.

Define the body of revolution described by $f$ around $r$ to be:

(BR) $\quad D=D(f,r):=\{ z \in RR^N: \quad "proj"_r(z) \in "supp "f " and " |"proj"_Pi(z)|<=f("proj"_r(z))\}$**

For example, it's easy to see that a ball $B$ of radius $rho>0$ is the body of revolution described by:

$f(t):=\{(\sqrt(rho^2-t^2), ", if " |t|<=rho),(0, ", otherwise"):}$

around, say, the $N$th axis of the standard Cartesian frame of $RR^N$.

Note that a body of revolution like the ones defined above hasn't pieces of boundary orthogonal to the axis, therefore the previous one is not the most general definition of body of revolution we can give.

***

Problem:

Let $D=D(f,r)$ be a body of revolution as in (BR).
Show rigorously that:

(*) $\quad "Vol"(D)=omega_(N-1)*\int_(-oo)^(+oo) f^(N-1)(t)" d"t$

and at least heuristically that:

(**) $\quad "Per"(D)=(N-1)omega_(N-1)*\int_(-oo)^(+oo) \sqrt(1+[f'(t)]^2)*f^(N-2)(t)" d"t$

where $"Vol"(\cdot )$ denotes the Lebesgue measure in $RR^N$, $"Per"(\cdot )$ stands for the De Giorgi perimeter (which coincides with the usual surface area measure for $(N-1)$-dimensional regular manifolds) and $omega_(N-1)="volume of the unit ball in "RR^(N-1)=(pi^(N/2))/(Gamma(N/2+1))$.

These two formulae generalize the ones known for volume and surface area measure of a body of revolution in $RR^3$.

Hints: Since rotations don't change volumes and perimeters, we can take w.l.o.g. $r$ to be the $N$th axis of the standard Cartesian frame of $RR^N$.
If we let $z=(x,t)$ with $x\in RR^(N-1)$ and $t\in RR$, we can use cylindrical coordinates in $RR^N$ to evaluate $"Vol"(D)$ because we can write:

$\quad D=\{(x,t)\in RR^N: t\in "supp "f " and " |x|<=f(t)\}$.

The explicit expression of $"Per"(D)$ can be found thinking how a piece of the graph of $f$ having "infinitesimal lenght" generate a piece of $\partial D$ of "infinitesimal perimeter" rotating around the $N$th axis.

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* $C_(pw)^1(RR)$ denotes the sets of functions which are piecewise of class $C^1$ in $RR$, i. e. the functions $f:RR to RR$ satisfying the following requirement:

"For each bounded interval $[a,b]$ exists a finite number of points $a=x_0
** Here:

- $"proj"_r(z)$ [resp. $"proj"_Pi(z)$] denotes the orthogonal projection of $z$ on $r$ [resp. on $Pi$];
- $"supp "f$ denotes the support of $f$, i. e. the set $\bar(\{ t\in RR: f(t)!=0\})$.

[gugo82]

Let $t \in "supp" f$ be fixed s.t. $f'(t)$ exists.
A small arc of $"graph" f$ through $(t,f(t))$ has "infinitesimal length" $\sqrt{1+[f'(t)]^2} " d"t$; such an arc can generate an anulus of "infinitesimal perimeter" by rotating around the $N$th axis of the Cartesian frame in $RR^N$: we can think this anulus as the lateral surface of a $N$-circular cylinder having radius $f(t)$ and "infinitesimal height" $\sqrt{1+[f'(t)]^2} " d"t$*. Thus the surface area measure of the anulus generated by rotation is:

$(N-1)omega_(N-1) f^(N-2)(t)*\sqrt{1+[f'(t)]^2} " d"t \quad$,

hence by integration (i.e. adding the contributes in perimeter of all anuli) we're able to find:

$"Per"(D)=\int_(-oo)^(+oo)(N-1)omega_(N-1) f^(N-2)(t)*\sqrt{1+[f'(t)]^2} " d"t$

which is (**).


__________
* A $N$-circular cylinder is a set of the type $C=B^(N-1)(x;r) \times ]a,b[ \subseteq RR^N$, where $B^(N-1)(x;r)$ is a $(N-1)$-open ball and $a0$ are called radius and height of the cylinder.
The volume of a cylinder is $"Vol"(C)=\omega_(N-1)r^(n-1) *h$ (this formula, which generalize a well-known result of Elementary Geometry, can be proved by straightforward computation of an integral).
The lateral surface of a cylinder $C$ is just the set $\bar{C}\setminus B^(N-1)(x;r)\times [a,b]$: this set has surface area measure equal to $(N-1)omega_(N-1)r^(N-2)*h$ (another generalization of an elementary result).

[gugo82]

"fu^2":anyway, the second part is more difficult, now i don't find a solution. I'm thinking that a possible way is using the Minkosky's definition of the perimeter, that is if $Omega$ is a surface, we can define $Omega(epsilon)={x\inRR^d:d(x,Omega)0$ fix.

So $|Omega|_{d-1}=lim_{epsilon->0}|Omega(epsilon)|_d/(2epsilon)$* (where $|*|_d=m_d(*)$), but now i haven't yet done this calculus, what do you think about this way?

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*For example $Omega={x\inRR^2 : |x|=1}$, we have that $Omega(epsilon)={x:1-epsilon<|x|<1+epsilon}$, so the perimeter result $|Omega|_1=lim_{epsilon->0}|Omega(epsilon)|/(2epsilon)=(pi(1+epsilon)^2-pi(1-epsilon)^2)/(2epsilon)=pi$.

Too technical... It isn't what I had in mind.

It's more convenient to think about how a piece of graph of "infinitesimal lenght" can generate an anulus of "infinitesimal perimeter" rotating around the axis.
In fact one can note that an infinitesimal piece of graph around the point $(t,f(t))$ has infinitesimal lenght equal to $\sqrt(1+[f'(t)]^2)" d"t$ and that the latter quantity is under integral sign in (**)... Then all the work reduces to write down the right expression for the "infinitesimal perimeter" of the forementioned anulus.

Maybe it isn't formally correct (yes, it's just urang-utang©), but it's the easier way to obtain (**).

[fu^2]

"Gugo82":Your solution is quite correct, but I can't figure out the meaning of:
[quote="fu^2"]$E=\{(z_1,...,z_N)|z_N\in K,\sum_{i=1}^{N-1}z_i^{N-1}<=f^{N-1}(z_N)\}$

Could you explain it?
I'm almost sure it's a typo cause you wanted to write $\sum_{i=1}^{N-1}z_i^2<=f^2(z_N)$ instead of $\sum_{i=1}^{N-1}z_i^{N-1}<=f^{N-1}(z_N)$...

Anyway, any ideas to solve the second part?[/quote]

yes, of course! now i change, while i was writing the code, there were too $N$

anyway, the second part is more difficult, now i don't find a solution. I'm thinking that a possible way is using the Minkosky's definition of the perimeter, that is if $Omega$ is a surface, we can define $Omega(epsilon)={x\inRR^d:d(x,Omega)0$ fix.

So $|Omega|_{d-1}=lim_{epsilon->0}|Omega(epsilon)|_d/(2epsilon)$* (where $|*|_d=m_d(*)$), but now i haven't yet done this calculus, what do you think about this way?

-----------------
*For example $Omega={x\inRR^2 : |x|=1}$, we have that $Omega(epsilon)={x:1-epsilon<|x|<1+epsilon}$, so the perimeter result $|Omega|_1=lim_{epsilon->0}|Omega(epsilon)|/(2epsilon)=(pi(1+epsilon)^2-pi(1-epsilon)^2)/(2epsilon)=pi$.

[gugo82]

Your solution is quite correct, but I can't figure out the meaning of:

"fu^2":$E=\{(z_1,...,z_N)|z_N\in K,\sum_{i=1}^{N-1}z_i^{N-1}<=f^{N-1}(z_N)\}$

Could you explain it?
I'm almost sure it's a typo cause you wanted to write $\sum_{i=1}^{N-1}z_i^2<=f^2(z_N)$ instead of $\sum_{i=1}^{N-1}z_i^{N-1}<=f^{N-1}(z_N)$...

Anyway, any ideas to solve the second part?

[fu^2]

I think that maybe i have found the solution of the first point.

(*) we'll show that $Vol(D)=\omega_{N-1}\intf^{N-1}(t)dt$*. Without loss of generality we can suppose that $r$ is the $N^{th}$-axis, by the Gugo's hint.

all the proof did over the follows theorem:

(T) Let $E\in RR^N$ to be a misurable set of $RR^N$, then $m_N(E)=intm_{N-1}(E_z)dz$, where $E_z={(z_1,...,z_{N-1})|(z_1,...,z_{N-1},z)\in E}$.

let now $f:RR->RR$ as in the hp of the problem,
Let $E={(z_1,...,z_N)|z_N\in K,\sum_{i=1}^{N-1}z_i^{2}<=f^{2}(z_N)}$, where $K={t\in RR|f(t)!=0}$. The projection over the $N-1$ axis is a ball of radius $f(z_N)$ (after that we have fixed $z_N$), so $m_{N-1}(E_{z_N})=\omega_{N-1}f^{N-1}(z_N)$ and using the theorem (T) the sentence follows.

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*$\int=\int_{-\infty}^{+\infty}$

if i write the orrible things, you don't kill me!